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  • NHibernate Pitfalls: Fetch and Paging

    - by Ricardo Peres
    This is part of a series of posts about NHibernate Pitfalls. See the entire collection here. NHibernate allows you to force loading additional references (many to one, one to one) or collections (one to many, many to many) in a query. You must know, however, that this is incompatible with paging. It’s easy to see why. Let’s say you want to get 5 products starting on the fifth, you can issue the following LINQ query: 1: session.Query<Product>().Take(5).Skip(5).ToList(); Will product this SQL in SQL Server: 1: SELECT 2: TOP (@p0) product1_4_, 3: name4_, 4: price4_ 5: FROM 6: (select 7: product0_.product_id as product1_4_, 8: product0_.name as name4_, 9: product0_.price as price4_, 10: ROW_NUMBER() OVER( 11: ORDER BY 12: CURRENT_TIMESTAMP) as __hibernate_sort_row 13: from 14: product product0_) as query 15: WHERE 16: query.__hibernate_sort_row > @p1 17: ORDER BY If, however, you wanted to bring as well the associated order details, you might be tempted to try this: 1: session.Query<Product>().Fetch(x => x.OrderDetails).Take(5).Skip(5).ToList(); Which, in turn, will produce this SQL: 1: SELECT 2: TOP (@p0) product1_4_0_, 3: order1_3_1_, 4: name4_0_, 5: price4_0_, 6: order2_3_1_, 7: product3_3_1_, 8: quantity3_1_, 9: product3_0__, 10: order1_0__ 11: FROM 12: (select 13: product0_.product_id as product1_4_0_, 14: orderdetai1_.order_detail_id as order1_3_1_, 15: product0_.name as name4_0_, 16: product0_.price as price4_0_, 17: orderdetai1_.order_id as order2_3_1_, 18: orderdetai1_.product_id as product3_3_1_, 19: orderdetai1_.quantity as quantity3_1_, 20: orderdetai1_.product_id as product3_0__, 21: orderdetai1_.order_detail_id as order1_0__, 22: ROW_NUMBER() OVER( 23: ORDER BY 24: CURRENT_TIMESTAMP) as __hibernate_sort_row 25: from 26: product product0_ 27: left outer join 28: order_detail orderdetai1_ 29: on product0_.product_id=orderdetai1_.product_id 30: ) as query 31: WHERE 32: query.__hibernate_sort_row > @p1 33: ORDER BY 34: query.__hibernate_sort_row; However, because of the JOIN, what happens is that, if your products have more than one order details, you will get several records – one per order detail – per product, which means that pagination will be broken. There is an workaround, which forces you to write your LINQ query in another way: 1: session.Query<OrderDetail>().Where(x => session.Query<Product>().Select(y => y.ProductId).Take(5).Skip(5).Contains(x.Product.ProductId)).Select(x => x.Product).ToList() Or, using HQL: 1: session.CreateQuery("select od.Product from OrderDetail od where od.Product.ProductId in (select p.ProductId from Product p skip 5 take 5)").List<Product>(); The generated SQL will then be: 1: select 2: product1_.product_id as product1_4_, 3: product1_.name as name4_, 4: product1_.price as price4_ 5: from 6: order_detail orderdetai0_ 7: left outer join 8: product product1_ 9: on orderdetai0_.product_id=product1_.product_id 10: where 11: orderdetai0_.product_id in ( 12: SELECT 13: TOP (@p0) product_id 14: FROM 15: (select 16: product2_.product_id, 17: ROW_NUMBER() OVER( 18: ORDER BY 19: CURRENT_TIMESTAMP) as __hibernate_sort_row 20: from 21: product product2_) as query 22: WHERE 23: query.__hibernate_sort_row > @p1 24: ORDER BY 25: query.__hibernate_sort_row); Which will get you what you want: for 5 products, all of their order details.

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  • Backup Meta-Data

    - by BuckWoody
    I'm working on a PowerShell script to show me the trending durations of my backup activities. The first thing I need is the data, so I looked at the Standard Reports in SQL Server Management Studio, and found a report that suited my needs, so I pulled out the script that it runs and modified it to this T-SQL Script. A few words here - you need to be in the MSDB database for this to run, and you can add a WHERE clause to limit to a database, timeframe, type of backup, whatever. For that matter, I won't use all of the data in this query in my PowerShell script, but it gives me lots of avenues to graph: SELECT distinct t1.name AS 'DatabaseName' ,(datediff( ss,  t3.backup_start_date, t3.backup_finish_date)) AS 'DurationInSeconds' ,t3.user_name AS 'UserResponsible' ,t3.name AS backup_name ,t3.description ,t3.backup_start_date ,t3.backup_finish_date ,CASE WHEN t3.type = 'D' THEN 'Database' WHEN t3.type = 'L' THEN 'Log' WHEN t3.type = 'F' THEN 'FileOrFilegroup' WHEN t3.type = 'G' THEN 'DifferentialFile' WHEN t3.type = 'P' THEN 'Partial' WHEN t3.type = 'Q' THEN 'DifferentialPartial' END AS 'BackupType' ,t3.backup_size AS 'BackupSizeKB' ,t6.physical_device_name ,CASE WHEN t6.device_type = 2 THEN 'Disk' WHEN t6.device_type = 102 THEN 'Disk' WHEN t6.device_type = 5 THEN 'Tape' WHEN t6.device_type = 105 THEN 'Tape' END AS 'DeviceType' ,t3.recovery_model  FROM sys.databases t1 INNER JOIN backupset t3 ON (t3.database_name = t1.name )  LEFT OUTER JOIN backupmediaset t5 ON ( t3.media_set_id = t5.media_set_id ) LEFT OUTER JOIN backupmediafamily t6 ON ( t6.media_set_id = t5.media_set_id ) ORDER BY backup_start_date DESC I'll munge this into my Excel PowerShell chart script tomorrow. Script Disclaimer, for people who need to be told this sort of thing: Never trust any script, including those that you find here, until you understand exactly what it does and how it will act on your systems. Always check the script on a test system or Virtual Machine, not a production system. Yes, there are always multiple ways to do things, and this script may not work in every situation, for everything. It’s just a script, people. All scripts on this site are performed by a professional stunt driver on a closed course. Your mileage may vary. Void where prohibited. Offer good for a limited time only. Keep out of reach of small children. Do not operate heavy machinery while using this script. If you experience blurry vision, indigestion or diarrhea during the operation of this script, see a physician immediately. Share this post: email it! | bookmark it! | digg it! | reddit! | kick it! | live it!

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  • Backup Meta-Data

    - by BuckWoody
    I'm working on a PowerShell script to show me the trending durations of my backup activities. The first thing I need is the data, so I looked at the Standard Reports in SQL Server Management Studio, and found a report that suited my needs, so I pulled out the script that it runs and modified it to this T-SQL Script. A few words here - you need to be in the MSDB database for this to run, and you can add a WHERE clause to limit to a database, timeframe, type of backup, whatever. For that matter, I won't use all of the data in this query in my PowerShell script, but it gives me lots of avenues to graph: SELECT distinct t1.name AS 'DatabaseName' ,(datediff( ss,  t3.backup_start_date, t3.backup_finish_date)) AS 'DurationInSeconds' ,t3.user_name AS 'UserResponsible' ,t3.name AS backup_name ,t3.description ,t3.backup_start_date ,t3.backup_finish_date ,CASE WHEN t3.type = 'D' THEN 'Database' WHEN t3.type = 'L' THEN 'Log' WHEN t3.type = 'F' THEN 'FileOrFilegroup' WHEN t3.type = 'G' THEN 'DifferentialFile' WHEN t3.type = 'P' THEN 'Partial' WHEN t3.type = 'Q' THEN 'DifferentialPartial' END AS 'BackupType' ,t3.backup_size AS 'BackupSizeKB' ,t6.physical_device_name ,CASE WHEN t6.device_type = 2 THEN 'Disk' WHEN t6.device_type = 102 THEN 'Disk' WHEN t6.device_type = 5 THEN 'Tape' WHEN t6.device_type = 105 THEN 'Tape' END AS 'DeviceType' ,t3.recovery_model  FROM sys.databases t1 INNER JOIN backupset t3 ON (t3.database_name = t1.name )  LEFT OUTER JOIN backupmediaset t5 ON ( t3.media_set_id = t5.media_set_id ) LEFT OUTER JOIN backupmediafamily t6 ON ( t6.media_set_id = t5.media_set_id ) ORDER BY backup_start_date DESC I'll munge this into my Excel PowerShell chart script tomorrow. Script Disclaimer, for people who need to be told this sort of thing: Never trust any script, including those that you find here, until you understand exactly what it does and how it will act on your systems. Always check the script on a test system or Virtual Machine, not a production system. Yes, there are always multiple ways to do things, and this script may not work in every situation, for everything. It’s just a script, people. All scripts on this site are performed by a professional stunt driver on a closed course. Your mileage may vary. Void where prohibited. Offer good for a limited time only. Keep out of reach of small children. Do not operate heavy machinery while using this script. If you experience blurry vision, indigestion or diarrhea during the operation of this script, see a physician immediately. Share this post: email it! | bookmark it! | digg it! | reddit! | kick it! | live it!

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  • Metro: Promises

    - by Stephen.Walther
    The goal of this blog entry is to describe the Promise class in the WinJS library. You can use promises whenever you need to perform an asynchronous operation such as retrieving data from a remote website or a file from the file system. Promises are used extensively in the WinJS library. Asynchronous Programming Some code executes immediately, some code requires time to complete or might never complete at all. For example, retrieving the value of a local variable is an immediate operation. Retrieving data from a remote website takes longer or might not complete at all. When an operation might take a long time to complete, you should write your code so that it executes asynchronously. Instead of waiting for an operation to complete, you should start the operation and then do something else until you receive a signal that the operation is complete. An analogy. Some telephone customer service lines require you to wait on hold – listening to really bad music – until a customer service representative is available. This is synchronous programming and very wasteful of your time. Some newer customer service lines enable you to enter your telephone number so the customer service representative can call you back when a customer representative becomes available. This approach is much less wasteful of your time because you can do useful things while waiting for the callback. There are several patterns that you can use to write code which executes asynchronously. The most popular pattern in JavaScript is the callback pattern. When you call a function which might take a long time to return a result, you pass a callback function to the function. For example, the following code (which uses jQuery) includes a function named getFlickrPhotos which returns photos from the Flickr website which match a set of tags (such as “dog” and “funny”): function getFlickrPhotos(tags, callback) { $.getJSON( "http://api.flickr.com/services/feeds/photos_public.gne?jsoncallback=?", { tags: tags, tagmode: "all", format: "json" }, function (data) { if (callback) { callback(data.items); } } ); } getFlickrPhotos("funny, dogs", function(data) { $.each(data, function(index, item) { console.log(item); }); }); The getFlickr() function includes a callback parameter. When you call the getFlickr() function, you pass a function to the callback parameter which gets executed when the getFlicker() function finishes retrieving the list of photos from the Flickr web service. In the code above, the callback function simply iterates through the results and writes each result to the console. Using callbacks is a natural way to perform asynchronous programming with JavaScript. Instead of waiting for an operation to complete, sitting there and listening to really bad music, you can get a callback when the operation is complete. Using Promises The CommonJS website defines a promise like this (http://wiki.commonjs.org/wiki/Promises): “Promises provide a well-defined interface for interacting with an object that represents the result of an action that is performed asynchronously, and may or may not be finished at any given point in time. By utilizing a standard interface, different components can return promises for asynchronous actions and consumers can utilize the promises in a predictable manner.” A promise provides a standard pattern for specifying callbacks. In the WinJS library, when you create a promise, you can specify three callbacks: a complete callback, a failure callback, and a progress callback. Promises are used extensively in the WinJS library. The methods in the animation library, the control library, and the binding library all use promises. For example, the xhr() method included in the WinJS base library returns a promise. The xhr() method wraps calls to the standard XmlHttpRequest object in a promise. The following code illustrates how you can use the xhr() method to perform an Ajax request which retrieves a file named Photos.txt: var options = { url: "/data/photos.txt" }; WinJS.xhr(options).then( function (xmlHttpRequest) { console.log("success"); var data = JSON.parse(xmlHttpRequest.responseText); console.log(data); }, function(xmlHttpRequest) { console.log("fail"); }, function(xmlHttpRequest) { console.log("progress"); } ) The WinJS.xhr() method returns a promise. The Promise class includes a then() method which accepts three callback functions: a complete callback, an error callback, and a progress callback: Promise.then(completeCallback, errorCallback, progressCallback) In the code above, three anonymous functions are passed to the then() method. The three callbacks simply write a message to the JavaScript Console. The complete callback also dumps all of the data retrieved from the photos.txt file. Creating Promises You can create your own promises by creating a new instance of the Promise class. The constructor for the Promise class requires a function which accepts three parameters: a complete, error, and progress function parameter. For example, the code below illustrates how you can create a method named wait10Seconds() which returns a promise. The progress function is called every second and the complete function is not called until 10 seconds have passed: (function () { "use strict"; var app = WinJS.Application; function wait10Seconds() { return new WinJS.Promise(function (complete, error, progress) { var seconds = 0; var intervalId = window.setInterval(function () { seconds++; progress(seconds); if (seconds > 9) { window.clearInterval(intervalId); complete(); } }, 1000); }); } app.onactivated = function (eventObject) { if (eventObject.detail.kind === Windows.ApplicationModel.Activation.ActivationKind.launch) { wait10Seconds().then( function () { console.log("complete") }, function () { console.log("error") }, function (seconds) { console.log("progress:" + seconds) } ); } } app.start(); })(); All of the work happens in the constructor function for the promise. The window.setInterval() method is used to execute code every second. Every second, the progress() callback method is called. If more than 10 seconds have passed then the complete() callback method is called and the clearInterval() method is called. When you execute the code above, you can see the output in the Visual Studio JavaScript Console. Creating a Timeout Promise In the previous section, we created a custom Promise which uses the window.setInterval() method to complete the promise after 10 seconds. We really did not need to create a custom promise because the Promise class already includes a static method for returning promises which complete after a certain interval. The code below illustrates how you can use the timeout() method. The timeout() method returns a promise which completes after a certain number of milliseconds. WinJS.Promise.timeout(3000).then( function(){console.log("complete")}, function(){console.log("error")}, function(){console.log("progress")} ); In the code above, the Promise completes after 3 seconds (3000 milliseconds). The Promise returned by the timeout() method does not support progress events. Therefore, the only message written to the console is the message “complete” after 10 seconds. Canceling Promises Some promises, but not all, support cancellation. When you cancel a promise, the promise’s error callback is executed. For example, the following code uses the WinJS.xhr() method to perform an Ajax request. However, immediately after the Ajax request is made, the request is cancelled. // Specify Ajax request options var options = { url: "/data/photos.txt" }; // Make the Ajax request var request = WinJS.xhr(options).then( function (xmlHttpRequest) { console.log("success"); }, function (xmlHttpRequest) { console.log("fail"); }, function (xmlHttpRequest) { console.log("progress"); } ); // Cancel the Ajax request request.cancel(); When you run the code above, the message “fail” is written to the Visual Studio JavaScript Console. Composing Promises You can build promises out of other promises. In other words, you can compose promises. There are two static methods of the Promise class which you can use to compose promises: the join() method and the any() method. When you join promises, a promise is complete when all of the joined promises are complete. When you use the any() method, a promise is complete when any of the promises complete. The following code illustrates how to use the join() method. A new promise is created out of two timeout promises. The new promise does not complete until both of the timeout promises complete: WinJS.Promise.join([WinJS.Promise.timeout(1000), WinJS.Promise.timeout(5000)]) .then(function () { console.log("complete"); }); The message “complete” will not be written to the JavaScript Console until both promises passed to the join() method completes. The message won’t be written for 5 seconds (5,000 milliseconds). The any() method completes when any promise passed to the any() method completes: WinJS.Promise.any([WinJS.Promise.timeout(1000), WinJS.Promise.timeout(5000)]) .then(function () { console.log("complete"); }); The code above writes the message “complete” to the JavaScript Console after 1 second (1,000 milliseconds). The message is written to the JavaScript console immediately after the first promise completes and before the second promise completes. Summary The goal of this blog entry was to describe WinJS promises. First, we discussed how promises enable you to easily write code which performs asynchronous actions. You learned how to use a promise when performing an Ajax request. Next, we discussed how you can create your own promises. You learned how to create a new promise by creating a constructor function with complete, error, and progress parameters. Finally, you learned about several advanced methods of promises. You learned how to use the timeout() method to create promises which complete after an interval of time. You also learned how to cancel promises and compose promises from other promises.

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  • Dual boot windows 8 and Ubuntu with Windows 8 Boot manager

    - by Mevin Babu
    I have two partitions on my hard-didk , I have installed ubuntu on my 1st partition and windows 8 later on another partition.Now i can only boot into windows 8 because it doesn't recognize Ubuntu. How would i dual boot my PC without using grub . I would like using Windows 8 boot manager as its pretty neat. I tried using easyBCD but it doesn't work.It causes the boot manager to switch to windows 7 Boot Manager .Is there anyother work around or solution Any help would be appreciated. Note: The windows 8 boot Manager is sky blue color interactive menu with mouse and other options and windows 7 boot manager is the normal black and white one where you can only use your keyboard

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  • SQL SERVER – Puzzle to Win Print Book – Explain Value of PERCENTILE_CONT() Using Simple Example

    - by pinaldave
    From last several days I am working on various Denali Analytical functions and it is indeed really fun to refresh the concept which I studied in the school. Earlier I wrote article where I explained how we can use PERCENTILE_CONT() to find median over here SQL SERVER – Introduction to PERCENTILE_CONT() – Analytic Functions Introduced in SQL Server 2012. Today I am going to ask question based on the same blog post. Again just like last time the intention of this puzzle is as following: Learn new concept of SQL Server 2012 Learn new concept of SQL Server 2012 even if you are on earlier version of SQL Server. On another note, SQL Server 2012 RC0 has been announced and available to download SQL SERVER – 2012 RC0 Various Resources and Downloads. Now let’s have fun following query: USE AdventureWorks GO SELECT SalesOrderID, OrderQty, ProductID, PERCENTILE_CONT(0.5) WITHIN GROUP (ORDER BY ProductID) OVER (PARTITION BY SalesOrderID) AS MedianCont FROM Sales.SalesOrderDetail WHERE SalesOrderID IN (43670, 43669, 43667, 43663) ORDER BY SalesOrderID DESC GO The above query will give us the following result: The reason we get median is because we are passing value .05 to PERCENTILE_COUNT() function. Now run read the puzzle. Puzzle: Run following T-SQL code: USE AdventureWorks GO SELECT SalesOrderID, OrderQty, ProductID, PERCENTILE_CONT(0.9) WITHIN GROUP (ORDER BY ProductID) OVER (PARTITION BY SalesOrderID) AS MedianCont FROM Sales.SalesOrderDetail WHERE SalesOrderID IN (43670, 43669, 43667, 43663) ORDER BY SalesOrderID DESC GO Observe the result and you will notice that MidianCont has different value than before, the reason is PERCENTILE_CONT function has 0.9 value passed. For first four value the value is 775.1. Now run following T-SQL code: USE AdventureWorks GO SELECT SalesOrderID, OrderQty, ProductID, PERCENTILE_CONT(0.1) WITHIN GROUP (ORDER BY ProductID) OVER (PARTITION BY SalesOrderID) AS MedianCont FROM Sales.SalesOrderDetail WHERE SalesOrderID IN (43670, 43669, 43667, 43663) ORDER BY SalesOrderID DESC GO Observe the result and you will notice that MidianCont has different value than before, the reason is PERCENTILE_CONT function has 0.1 value passed. For first four value the value is 709.3. Now in my example I have explained how the median is found using this function. You have to explain using mathematics and explain (in easy words) why the value in last columns are 709.3 and 775.1 Hint: SQL SERVER – Introduction to PERCENTILE_CONT() – Analytic Functions Introduced in SQL Server 2012 Rules Leave a comment with your detailed answer by Nov 25's blog post. Open world-wide (where Amazon ships books) If you blog about puzzle’s solution and if you win, you win additional surprise gift as well. Prizes Print copy of my new book SQL Server Interview Questions Amazon|Flipkart If you already have this book, you can opt for any of my other books SQL Wait Stats [Amazon|Flipkart|Kindle] and SQL Programming [Amazon|Flipkart|Kindle]. Reference: Pinal Dave (http://blog.SQLAuthority.com) Filed under: Pinal Dave, PostADay, SQL, SQL Authority, SQL Function, SQL Puzzle, SQL Query, SQL Scripts, SQL Server, SQL Tips and Tricks, T SQL, Technology

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  • Hello Operator, My Switch Is Bored

    - by Paul White
    This is a post for T-SQL Tuesday #43 hosted by my good friend Rob Farley. The topic this month is Plan Operators. I haven’t taken part in T-SQL Tuesday before, but I do like to write about execution plans, so this seemed like a good time to start. This post is in two parts. The first part is primarily an excuse to use a pretty bad play on words in the title of this blog post (if you’re too young to know what a telephone operator or a switchboard is, I hate you). The second part of the post looks at an invisible query plan operator (so to speak). 1. My Switch Is Bored Allow me to present the rare and interesting execution plan operator, Switch: Books Online has this to say about Switch: Following that description, I had a go at producing a Fast Forward Cursor plan that used the TOP operator, but had no luck. That may be due to my lack of skill with cursors, I’m not too sure. The only application of Switch in SQL Server 2012 that I am familiar with requires a local partitioned view: CREATE TABLE dbo.T1 (c1 int NOT NULL CHECK (c1 BETWEEN 00 AND 24)); CREATE TABLE dbo.T2 (c1 int NOT NULL CHECK (c1 BETWEEN 25 AND 49)); CREATE TABLE dbo.T3 (c1 int NOT NULL CHECK (c1 BETWEEN 50 AND 74)); CREATE TABLE dbo.T4 (c1 int NOT NULL CHECK (c1 BETWEEN 75 AND 99)); GO CREATE VIEW V1 AS SELECT c1 FROM dbo.T1 UNION ALL SELECT c1 FROM dbo.T2 UNION ALL SELECT c1 FROM dbo.T3 UNION ALL SELECT c1 FROM dbo.T4; Not only that, but it needs an updatable local partitioned view. We’ll need some primary keys to meet that requirement: ALTER TABLE dbo.T1 ADD CONSTRAINT PK_T1 PRIMARY KEY (c1);   ALTER TABLE dbo.T2 ADD CONSTRAINT PK_T2 PRIMARY KEY (c1);   ALTER TABLE dbo.T3 ADD CONSTRAINT PK_T3 PRIMARY KEY (c1);   ALTER TABLE dbo.T4 ADD CONSTRAINT PK_T4 PRIMARY KEY (c1); We also need an INSERT statement that references the view. Even more specifically, to see a Switch operator, we need to perform a single-row insert (multi-row inserts use a different plan shape): INSERT dbo.V1 (c1) VALUES (1); And now…the execution plan: The Constant Scan manufactures a single row with no columns. The Compute Scalar works out which partition of the view the new value should go in. The Assert checks that the computed partition number is not null (if it is, an error is returned). The Nested Loops Join executes exactly once, with the partition id as an outer reference (correlated parameter). The Switch operator checks the value of the parameter and executes the corresponding input only. If the partition id is 0, the uppermost Clustered Index Insert is executed, adding a row to table T1. If the partition id is 1, the next lower Clustered Index Insert is executed, adding a row to table T2…and so on. In case you were wondering, here’s a query and execution plan for a multi-row insert to the view: INSERT dbo.V1 (c1) VALUES (1), (2); Yuck! An Eager Table Spool and four Filters! I prefer the Switch plan. My guess is that almost all the old strategies that used a Switch operator have been replaced over time, using things like a regular Concatenation Union All combined with Start-Up Filters on its inputs. Other new (relative to the Switch operator) features like table partitioning have specific execution plan support that doesn’t need the Switch operator either. This feels like a bit of a shame, but perhaps it is just nostalgia on my part, it’s hard to know. Please do let me know if you encounter a query that can still use the Switch operator in 2012 – it must be very bored if this is the only possible modern usage! 2. Invisible Plan Operators The second part of this post uses an example based on a question Dave Ballantyne asked using the SQL Sentry Plan Explorer plan upload facility. If you haven’t tried that yet, make sure you’re on the latest version of the (free) Plan Explorer software, and then click the Post to SQLPerformance.com button. That will create a site question with the query plan attached (which can be anonymized if the plan contains sensitive information). Aaron Bertrand and I keep a close eye on questions there, so if you have ever wanted to ask a query plan question of either of us, that’s a good way to do it. The problem The issue I want to talk about revolves around a query issued against a calendar table. The script below creates a simplified version and adds 100 years of per-day information to it: USE tempdb; GO CREATE TABLE dbo.Calendar ( dt date NOT NULL, isWeekday bit NOT NULL, theYear smallint NOT NULL,   CONSTRAINT PK__dbo_Calendar_dt PRIMARY KEY CLUSTERED (dt) ); GO -- Monday is the first day of the week for me SET DATEFIRST 1;   -- Add 100 years of data INSERT dbo.Calendar WITH (TABLOCKX) (dt, isWeekday, theYear) SELECT CA.dt, isWeekday = CASE WHEN DATEPART(WEEKDAY, CA.dt) IN (6, 7) THEN 0 ELSE 1 END, theYear = YEAR(CA.dt) FROM Sandpit.dbo.Numbers AS N CROSS APPLY ( VALUES (DATEADD(DAY, N.n - 1, CONVERT(date, '01 Jan 2000', 113))) ) AS CA (dt) WHERE N.n BETWEEN 1 AND 36525; The following query counts the number of weekend days in 2013: SELECT Days = COUNT_BIG(*) FROM dbo.Calendar AS C WHERE theYear = 2013 AND isWeekday = 0; It returns the correct result (104) using the following execution plan: The query optimizer has managed to estimate the number of rows returned from the table exactly, based purely on the default statistics created separately on the two columns referenced in the query’s WHERE clause. (Well, almost exactly, the unrounded estimate is 104.289 rows.) There is already an invisible operator in this query plan – a Filter operator used to apply the WHERE clause predicates. We can see it by re-running the query with the enormously useful (but undocumented) trace flag 9130 enabled: Now we can see the full picture. The whole table is scanned, returning all 36,525 rows, before the Filter narrows that down to just the 104 we want. Without the trace flag, the Filter is incorporated in the Clustered Index Scan as a residual predicate. It is a little bit more efficient than using a separate operator, but residual predicates are still something you will want to avoid where possible. The estimates are still spot on though: Anyway, looking to improve the performance of this query, Dave added the following filtered index to the Calendar table: CREATE NONCLUSTERED INDEX Weekends ON dbo.Calendar(theYear) WHERE isWeekday = 0; The original query now produces a much more efficient plan: Unfortunately, the estimated number of rows produced by the seek is now wrong (365 instead of 104): What’s going on? The estimate was spot on before we added the index! Explanation You might want to grab a coffee for this bit. Using another trace flag or two (8606 and 8612) we can see that the cardinality estimates were exactly right initially: The highlighted information shows the initial cardinality estimates for the base table (36,525 rows), the result of applying the two relational selects in our WHERE clause (104 rows), and after performing the COUNT_BIG(*) group by aggregate (1 row). All of these are correct, but that was before cost-based optimization got involved :) Cost-based optimization When cost-based optimization starts up, the logical tree above is copied into a structure (the ‘memo’) that has one group per logical operation (roughly speaking). The logical read of the base table (LogOp_Get) ends up in group 7; the two predicates (LogOp_Select) end up in group 8 (with the details of the selections in subgroups 0-6). These two groups still have the correct cardinalities as trace flag 8608 output (initial memo contents) shows: During cost-based optimization, a rule called SelToIdxStrategy runs on group 8. It’s job is to match logical selections to indexable expressions (SARGs). It successfully matches the selections (theYear = 2013, is Weekday = 0) to the filtered index, and writes a new alternative into the memo structure. The new alternative is entered into group 8 as option 1 (option 0 was the original LogOp_Select): The new alternative is to do nothing (PhyOp_NOP = no operation), but to instead follow the new logical instructions listed below the NOP. The LogOp_GetIdx (full read of an index) goes into group 21, and the LogOp_SelectIdx (selection on an index) is placed in group 22, operating on the result of group 21. The definition of the comparison ‘the Year = 2013’ (ScaOp_Comp downwards) was already present in the memo starting at group 2, so no new memo groups are created for that. New Cardinality Estimates The new memo groups require two new cardinality estimates to be derived. First, LogOp_Idx (full read of the index) gets a predicted cardinality of 10,436. This number comes from the filtered index statistics: DBCC SHOW_STATISTICS (Calendar, Weekends) WITH STAT_HEADER; The second new cardinality derivation is for the LogOp_SelectIdx applying the predicate (theYear = 2013). To get a number for this, the cardinality estimator uses statistics for the column ‘theYear’, producing an estimate of 365 rows (there are 365 days in 2013!): DBCC SHOW_STATISTICS (Calendar, theYear) WITH HISTOGRAM; This is where the mistake happens. Cardinality estimation should have used the filtered index statistics here, to get an estimate of 104 rows: DBCC SHOW_STATISTICS (Calendar, Weekends) WITH HISTOGRAM; Unfortunately, the logic has lost sight of the link between the read of the filtered index (LogOp_GetIdx) in group 22, and the selection on that index (LogOp_SelectIdx) that it is deriving a cardinality estimate for, in group 21. The correct cardinality estimate (104 rows) is still present in the memo, attached to group 8, but that group now has a PhyOp_NOP implementation. Skipping over the rest of cost-based optimization (in a belated attempt at brevity) we can see the optimizer’s final output using trace flag 8607: This output shows the (incorrect, but understandable) 365 row estimate for the index range operation, and the correct 104 estimate still attached to its PhyOp_NOP. This tree still has to go through a few post-optimizer rewrites and ‘copy out’ from the memo structure into a tree suitable for the execution engine. One step in this process removes PhyOp_NOP, discarding its 104-row cardinality estimate as it does so. To finish this section on a more positive note, consider what happens if we add an OVER clause to the query aggregate. This isn’t intended to be a ‘fix’ of any sort, I just want to show you that the 104 estimate can survive and be used if later cardinality estimation needs it: SELECT Days = COUNT_BIG(*) OVER () FROM dbo.Calendar AS C WHERE theYear = 2013 AND isWeekday = 0; The estimated execution plan is: Note the 365 estimate at the Index Seek, but the 104 lives again at the Segment! We can imagine the lost predicate ‘isWeekday = 0’ as sitting between the seek and the segment in an invisible Filter operator that drops the estimate from 365 to 104. Even though the NOP group is removed after optimization (so we don’t see it in the execution plan) bear in mind that all cost-based choices were made with the 104-row memo group present, so although things look a bit odd, it shouldn’t affect the optimizer’s plan selection. I should also mention that we can work around the estimation issue by including the index’s filtering columns in the index key: CREATE NONCLUSTERED INDEX Weekends ON dbo.Calendar(theYear, isWeekday) WHERE isWeekday = 0 WITH (DROP_EXISTING = ON); There are some downsides to doing this, including that changes to the isWeekday column may now require Halloween Protection, but that is unlikely to be a big problem for a static calendar table ;)  With the updated index in place, the original query produces an execution plan with the correct cardinality estimation showing at the Index Seek: That’s all for today, remember to let me know about any Switch plans you come across on a modern instance of SQL Server! Finally, here are some other posts of mine that cover other plan operators: Segment and Sequence Project Common Subexpression Spools Why Plan Operators Run Backwards Row Goals and the Top Operator Hash Match Flow Distinct Top N Sort Index Spools and Page Splits Singleton and Range Seeks Bitmaps Hash Join Performance Compute Scalar © 2013 Paul White – All Rights Reserved Twitter: @SQL_Kiwi

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  • It's Official, I'm a Geek

    - by andyleonard
    I'm honored to join Glen Gordon ( Blog - @glengordon ) and G. Andrew Duthie ( Blog - @devhammer ) today at 3:00 PM EDT for an MSDN Webcast entitled GeekSpeak: Inside SQL Server Integration Services (SSIS). This is a LiveMeeting and you can join in the fun as an attendee here . It's a live show, so bring your questions! :{> Andy Share this post: email it! | bookmark it! | digg it! | reddit! | kick it! | live it!...(read more)

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  • how to access inaccessible mac os x hard drive via ubuntu

    - by jon
    Background: My intention was to load a Virtual Machine (VM) on my Mac OS X Snow Leopard. My Mac had just enough room for a VM (my thought process was that VM was the same as partition) However, I burned the newest version of Ubuntu onto a CD, thinking that partitioning and running a virtual machine would be the same. I would restart my computer, booting up Ubuntu installer. The installation would not allow me to partition, forcing me to force shutdown my laptop. when I turn on my laptop, I see that my computer is "missing operating system". So, can someone help me fix my a) bootcamp, b) getting files and if a and b are fixed c) to install ubuntu as a VM?

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  • NTFS Signature is missing

    - by LuckyBearing
    So I got a new secondary hard drive caddy and formatted it to NTFS. I forgot to partition the drive after formatting, I rebooted and now I can't access the drive that has around 400gb of data because the NTFS Signature is missing. "Error mounting...NTFS signature is missing. Failed to mount '/dev/sdb2': Invalid argument The device '/dev/sdb2' doesn't seem to have a valid NTFS. Maybe the wrong device is used? Or the whole disk instead of a partition (e.g. /dev/sda, not /dev/sda1)? Or the other way around?" Says the same thing for sdb1. But I think the sdb2 has most, if not all of the copied data. How do I retrieve my data back? I read somewhere I can install ntfs-progs and it will fix the issue. here

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  • Can't get TRIM test to work

    - by Matthew Marcus
    So I'm attempting to install TRIM using the walkthrough here: How to enable TRIM? But everytime I attempt to run the hdparm command, I get the following when I try to run it w/ sda: reading sector 5805056: FAILED: Input/output error and I get this when running it with sda1: /dev/sda1: Device /dev/sda1 has non-zero LBA starting offset of 2048. Please use an absolute LBA with the /dev/ entry for the full device, rather than a partition name. /dev/sda1 is probably a partition of /dev/sda (?) The absolute LBA of sector 5807104 from /dev/sda1 should be 5809152 Aborting. I'm running Natty in a VBox on Windows 7. Someone PLEASE help.. I keep getting this "consistency check" message on boot of my machine and I think it's because Ubuntu is writing to the same sectors on the VHD too much.. need to get trim working on this thing.. Thanks.

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  • Changing the default installation path to a newly installed hard disk

    - by mgj
    Hi, I am currently working on a dual-booted PC. I am using Windows XP and Ubuntu 10.04 Lucid Lynx released in April 2010. The allocated partition to Ubuntu that I am making use of has almost exhausted. Current memory allocations on the PC wrt Ubuntu OS looks like this: bodhgaya@pc146724-desktop:~$ df -h Filesystem Size Used Avail Use% Mounted on /dev/sda2 8.6G 8.0G 113M 99% / none 998M 268K 998M 1% /dev none 1002M 580K 1002M 1% /dev/shm none 1002M 100K 1002M 1% /var/run none 1002M 0 1002M 0% /var/lock none 1002M 0 1002M 0% /lib/init/rw /dev/sda1 25G 16G 9.8G 62% /media/C /dev/sdb1 37G 214M 35G 1% /media/ubuntulinuxstore bodhgaya@pc146724-desktop:~$ cd /tmp I am trying to mount a 40GB(/dev/sdb1 - given below) new hard disk along with my existing Ubuntu system to overcome with hard disk space related issues. I referred to the following tutorial to mount a new hard disk onto the system:- http://www.smorgasbord.net/how-to-in...untu-linux%20/ I was able to successfully mount this hard disk for Ubuntu 0S. I have this new hard disk setup in /media/ubuntulinuxstore directory. The current partition in my system looks like this: bodhgaya@pc146724-desktop:/media/ubuntulinuxstore$ sudo fdisk -l [sudo] password for bodhgaya: Disk /dev/sda: 40.0 GB, 40000000000 bytes 255 heads, 63 sectors/track, 4863 cylinders Units = cylinders of 16065 * 512 = 8225280 bytes Sector size (logical/physical): 512 bytes / 512 bytes I/O size (minimum/optimal): 512 bytes / 512 bytes Disk identifier: 0x446eceb5 Device Boot Start End Blocks Id System /dev/sda1 * 2 3264 26210047+ 7 HPFS/NTFS /dev/sda2 3265 4385 9004432+ 83 Linux /dev/sda3 4386 4863 3839535 82 Linux swap / Solaris Disk /dev/sdb: 40.0 GB, 40000000000 bytes 255 heads, 63 sectors/track, 4863 cylinders Units = cylinders of 16065 * 512 = 8225280 bytes Sector size (logical/physical): 512 bytes / 512 bytes I/O size (minimum/optimal): 512 bytes / 512 bytes Disk identifier: 0xfa8afa8a Device Boot Start End Blocks Id System /dev/sdb1 1 4862 39053983+ 7 HPFS/NTFS bodhgaya@pc146724-desktop:/media/ubuntulinuxstore$ Now, I have a concern wrt the "location" where the new softwares will be installed. Generally softwares are installed via the terminal and by default a fixed path is used to where the post installation set up files can be found (I am talking in context of the drive). This is like the typical case of Windows, where softwares by default are installed in the C: drive. These days people customize their installations to a drive which they find apt to serve their purpose (generally based on availability of hard disk space). I am trying to figure out how to customize the same for Ubuntu. As we all know the most softwares are installed via commands given from the Terminal. My road block is how do I redirect the default path set on the terminal where files get installed to this new hard disk. This if done will help me overcome space constraints I am currently facing wrt the partition on which my Ubuntu is initially installed. I would also by this, save time on not formatting my system and reinstalling Ubuntu and other softwares all over again. Please help me with this, your suggestions are much appreciated.

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  • NTFS Issues in Windows 7 and 2008 R2 - 'Is it a Bug?'

    - by renewieldraaijer
    I have been using the various versions of the Microsoft Windows product line since NT4 and I really thought I knew the ins and outs about the NTFS filesystem by now. There were always a few rules of thumb to understand what happens if you move data around. These rules were: "If you copy data, the copied data will inherit the permissions of the location it is being copied to. The same goes for moving data between disk partitions. Only when you move data within the same partition, the permissions are kept."  Recently I was asked to assist in troubleshooting some NTFS related issues. This forced me to have another good look at this theory. To my surprise I found out that this theory does not completely stand anymore. Apparently some things have changed since the release of Windows Vista / Windows 2008. Since the release of these Operating Systems, a move within the same disk partition results in the data inheriting the permissions of the location it is being copied into. A major change in the NTFS filesystem you would think!  Not quite! The above only counts when the move operation is being performed by using Windows Explorer. A move by using the 'move' command from within a cmd prompt for example, retains the NTFS permissions, just like before in Windows XP and older systems. Conclusion: The Windows Explorer is responsible for changing the ACL's of the moved data. This is a remarkable change, but if you follow this theory, the resulting ACL after a move operation is still predictable.  We could say that since Windows Vista and Windows 2008, a new rule set applies: "If you copy data, the copied data will inherit the permissions of the location it is being copied to. Same goes for moving data between disk partitions and within disk partitions. Only when you move data within the same partition by using something else than the Windows Explorer, the permissions are kept." The above behavior should be unchanged in Windows 7 / Windows 2008 R2, compared to Windows Vista / 2008. But somehow the NTFS permissions are not so predictable in Windows 7 and Windows 2008 R2. Moving data within the same disk partition the one time results in the permissions being kept and the next time results in inherited permissions from the destination location. I will try to demonstrate this in a few examples: Example 1 (Incorrect behavior): Consider two folders, 'Folder A' and 'Folder B' with the following permissions configured.                    Now we create the test file 'test file 1.txt' in 'Folder A' and afterwards move this file to 'Folder B' using Windows Explorer.                       According to the new theory, the file should inherit the permissions of 'Folder B' and therefore 'Group B' should appear in the ACL of 'test file 1.txt'. In the screenshot below the resulting permissions are displayed. The permissions from the originating location are kept, while the permissions of 'Folder B' should be inherited.                   Example 2 (Correct behavior): Again, consider the same two folders. This time we make a small modification to the ACL of 'Folder A'. We add 'Group C' to the ACL and again we create a file in 'Folder A' which we name 'test file 2.txt'.                    Next, we move 'test file 2.txt' to 'Folder B'.                       Again, we check the permissions of 'test file 2.txt' at the target location. We can now see that the permissions are inherited. This is what should be happening, and can be considered 'correct behavior' for Windows Vista / 2008 / 7 / 2008 R2. It remains uncertain why this behavior is so inconsistent. At this time, this is under investigation with Microsoft Support. The investigation has been going for the last two weeks and it is beginning to look like there is no rational reason for this, other than a bug in the Windows Explorer in Windows 7 and 2008 R2. As soon as there is any certainty on this, I will note it here in this blog.                   The examples above are harmless tests, by using my own laptop. If you would create the same set of folders and groups, and configure exactly the same permissions, you will see exactly the same behavior. Be sure to use Windows 7 or Windows 2008 R2.   Initially the problem arose at a customer site where move operations on data on the fileserver by users would result in unpredictable results. This resulted in the wrong set of people having àccess permissions on data that they should not have permissions to. Off course this is something we want to prevent at all costs.   I have also done several tests with move operations by using the move command in a cmd prompt. This way the behavior is always consistent. The inconsistent behavior is only exposed when using the Windows Explorer to initiate the move operation, and only when using Windows 7 or Windows 2008 R2 systems. It is evident that this behavior changes when the ACL of a folder has been changed, for example by adding an extra entry. The reason for this remains uncertain though. To be continued…. A dutch version of this post can be found at: http://blogs.platani.nl/?p=612

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  • Chrome Apps Office Hours: Storage API Deep Dive

    Chrome Apps Office Hours: Storage API Deep Dive Ask and vote for questions at: goo.gl Join us next week as we take a deeper dive into the new storage APIs available to Chrome Packaged Apps. We've invited Eric Bidelman, author of the HTML5 File System API book to join Paul Kinlan, Paul Lewis, Pete LePage and Renato Dias for our weekly Chrome Apps Office Hours in which we will pick apart some of the sample Chrome Apps and explain how we've used the storage APIs and why we made the decisions we did. From: GoogleDevelopers Views: 0 0 ratings Time: 00:00 More in Science & Technology

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  • Can I install two Ubuntu versions on the same machine?

    - by Abh
    Hello, I have Ubuntu 10.10 32 bit already installed on my machine..I am using MongoDB and it does not work properly with 32 bit machine. So I want to install 64 bit Ubuntu 10.10 on my system on another partition (so that I can have both 32 bit and 64 bit versions). Is it okay to install both 32 bit and 64 bit? I mean will it give any problems? On which partition should I install the 64 bit version? My partitions are as follows: Filesystem Size Used Avail Use% Mounted on /dev/sda1 37G 11G 25G 30% / none 1.4G 260K 1.4G 1% /dev none 1.4G 776K 1.4G 1% /dev/shm none 1.4G 244K 1.4G 1% /var/run none 1.4G 0 1.4G 0% /var/lock /dev/sda6 129G 73G 50G 60% /home /dev/sda7 127G 76G 45G 64% /vol Waiting for your replies.

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  • Cannot resolve the collation conflict between "Latin1_General_CI_AS" and "SQL_Latin1_General_CP1_CI_

    - by Patrick Olurotimi Ige
    I was writing a store proc for a report and i needed some data from another server so i added a linked server to connect to this new db server. when i do a select like below its all fine select a,b,c from Server.DatabaseName.dbo.table But when i use the table in a join i get the error "Cannot resolve the collation conflict between "Latin1_General_CI_AS" and "SQL_Latin1_General_CP1_CI_AS" in the equal to operation." I did check the collation set on the 2 databases and it was actually the same and had mo idea why i'm getting the error. I later found out that you could specifically tell it to use a COLLATE Just rewrite your join like this on a.name COLLATE Latin1_General_CI_AS = eaobjname Hope that helps and saves your precious time Patrick

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  • SQL SERVER – Pending IO request in SQL Server – DMV

    - by pinaldave
    I received following question: “How do we know how many pending IO requests are there for database files (.mdf, .ldf) individually?” Very interesting question and indeed answer is very interesting as well. Here is the quick script which I use to find the same. It has to be run in the context of the database for which you want to know pending IO statistics. USE DATABASE GO SELECT vfs.database_id, df.name, df.physical_name ,vfs.FILE_ID, ior.io_pending FROM sys.dm_io_pending_io_requests ior INNER JOIN sys.dm_io_virtual_file_stats (DB_ID(), NULL) vfs ON (vfs.file_handle = ior.io_handle) INNER JOIN sys.database_files df ON (df.FILE_ID = vfs.FILE_ID) I keep this script handy as it works like magic every time. If you use any other script please post here and I will post it with due credit. Reference: Pinal Dave (http://blog.SQLAuthority.com) Filed under: PostADay, SQL, SQL Authority, SQL DMV, SQL Query, SQL Server, SQL Tips and Tricks, T SQL, Technology

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  • JCP EC Nominations and Meet the Candidates Call

    - by heathervc
    The Nominations period for the 2012 JCP EC Elections closes tomorrow, 11 October at midnight pacific time.  Eligible JCP Members (all current JSPA 2 signers) may nominate themselves.  You will need your Elections credentials to complete the nomination, which were sent to the primary contacts of all eligible JCP Members via email last week. This year all ratified (there are 4 proposed ratified candidates) and elected (there are 7 candidates so far) will appear on one ballot; the top 2 candidates will win elected seats. This year, the selected EC Members will serve a single year term.  Following the 2012 Elections, there will be one merged EC (approved through JSR 355), and a new JCP version, JCP 2.9 will be in effect.  In 2013, all EC members will stand for election to complete the merge process described in the JCP 2.9 process document. All of the candidates' nominations materials are now available. The ratified candidates are:  Cinterion, Credit Suisse, Fujitsu and HP.The elected candidates are:  Cisco Systems, CloudBees, Giuseppe Dell'Abate, London Java Community, MoroccoJUG, Software AG, and Zero Turnaround. Next week, 18 October, we will hold an open teleconference for the Java Community to meet the candidates and ask questions regarding their nomination.  We hope you will be able to participate in the call.  Should the time be inconvenient, a recording will be made available for download, and candidate questions may be posted on this blog entry or sent to [email protected]. Topic: Meet the EC Candidates Date: Thursday, October 18, 2012 Time: 9:30 am, Pacific Daylight Time (San Francisco, GMT-07:00) Meeting Number: 807 818 225 Meeting Password: MeetEC ------------------------------------------------------- To join the online meeting (Now from mobile devices) ------------------------------------------------------- 1. Go to https://jcp.webex.com/jcp/j.php?ED=186721592&UID=0&PW=NMmUzNjY5ZTMw&RT=MiM0 2. If requested, enter your name and email address. 3. If a password is required, enter the meeting password: MeetEC 4. Click "Join". To view in other time zones or languages, please click the link: https://jcp.webex.com/jcp/j.php?ED=186721592&UID=0&PW=NMmUzNjY5ZTMw&ORT=MiM0 ------------------------------------------------------- To join the audio conference only -------------------------------------------------------     +1 (866) 682-4770     Outside the US: global access numbers  https://www.intercallonline.com/portlets/scheduling/viewNumbers/listNumbersByCode.do?confCode=6279803 or +1 (408) 774-4073     Conference code: 9454597     Security code: JCPEC (52732)------------------------------------------------------- For assistance ------------------------------------------------------- 1. Go to https://jcp.webex.com/jcp/mc 2. On the left navigation bar, click "Support".

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  • how to reinstall/repair ubuntu 12.04 after dual boot installation fails with windows 7

    - by Rini
    I have installed Ubuntu 12.04 on my preinstalled windows 7 Sony vaio s series laptop following instructions here: http://www.linuxbsdos.com/2012/05/17/how-to-dual-boot-ubuntu-12-04-and-windows-7/ Everything went well and I am able to boot in to windows after complete installation of Ubuntu. Now following instructions on web I tried to add Ubuntu to my BIOS using Easy BCD (but forget to add windows 7 entry). As a result, I loose windows 7 OS and can't boot in to either OS then I successfully repaired windows 7 using recovery CD. Now my problem is that I can't reinstall Ubuntu 12.04 using Live CD it halts every time before disk partition step giving error. "ubi-partman crashed". "ubi-partman failed with exit code 141. further information may be found in /var/log/syslog. Do you want to try running this step again before continuing? If you do not, your installation may fail entirely or may be broken." and, any choice to continue will result in the same error. I looked in to /var/log/syslog but not able to understand what is error. Then, I ran sudo fdisk -l to view my partitions and it shows me only one partition. Probably, all the partitions I created for Ubuntu 12.04 are lost while running windows 7 recovery CD. So, I don't know whether the Ubuntu is still there or probably corrupted. My boot-info URL is: http://paste.ubuntu.com/1202146/ Please tell me how to remove this error so that I can reinstall/repair Ubuntu 12.04 Thanks in advance. R Shukla My boot-info URL is: http://paste.ubuntu.com/1202146/ Please tell me how to remove this error so that I can reinstall/repair Ubuntu 12.04 Thanks for your help! I tried to boot from the CD but I every time it give me error before disk partitioning step. Also, I am unable to start Gparted. "ubi-partman crashed". "ubi-partman failed with exit code 141. further information may be found in /var/log/syslog. Do you want to try running this step again before continuing? If you do not, your installation may fail entirely or may be broken." and, any choice to continue will result in the same error. I looked in to /var/log/syslog but not able to understand what is error. Then, I ran sudo fdisk -l to view my partitions and it shows me only ne partition. Probaply, all the partitions I created for Ubuntu 12.04 are lost while running windows 7 recovery CD. Please tell me how to remove this error. Best Regards, R S

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  • Problem booting Windows after failed installation Ubuntu 12.04 alongside Windows 7

    - by Tassos
    I tried to install in my laptop Ubuntu 12.04 so that I can dual-boot with Windows 7. I made some mistakes during this process and I didn't manage to install Ubuntu. But my real problem now is that I'm afraid that I also destroyed the installation of Windows 7. Your help would be precious for me. Here are the details of what I did: 1) I followed these instructions to dual boot Ubuntu 12.04 and Windows 7: http://www.linuxbsdos.com/2012/05/17/how-to-dual-boot-ubuntu-12-04-and-windows-7/ The only difference from what is described above, is that in my case the device names where: /dev/mapper/isw_fdjdhbadc_Volume0* instead of: /dev/sda* Note that I had created a bootable USB stick to do that. 2) The installation proceeded normally, but in the end I got a fatal error because the grub-install failed. 3) Then, after googling this problem, I runned ubuntu from the USB stick and run this command: sudo grub-install --root-directory=/home/ubuntu/temp /dev/mapper/isw_fdjdhbadc_Volume0p5 (/isw_fdjdhbadc_Volume0p5 was the partition that I had made for /boot) but this command also failed. 4) Then, I did something stupid (I think): I run the above command as: sudo grub-install --root-directory=/home/ubuntu/temp /dev/mapper/isw_fdjdhbadc_Volume0 namely I tried to install grub in the device isw_fdjdhbadc_Volume0 instead of the boot partition isw_fdjdhbadc_Volume0p5 The above command did not fail and was executed ok. 5) After that, I tried to boot my laptop, but it seemed that I had no operation system. Not even windows were detected. 6) I thought that I should uninstall grub from isw_fdjdhbadc_Volume0. So following some online instructions that I found, I booted again Ubuntu from the USB stick and run the following command (this was stupid since the instructions were for a totally different case than mine): sudo dd if=/dev/zero of=/dev/mapper/isw_fdjdhbadc_Volume0 bs=446 count=1 Afte that, I was still unable to boot Windows. I realize that I deleted something that I shouldn't, but I'm hope that this is not crucial and I can recover somehow. When I boot Ubuntu from the USB, I can see that the partition with Windows is still there, with all the directories, Windows files, my data etc. So, my question is: Is there a way to undo the mistakes that I desribed above and recover Windows 7? This is my major question. After solving that, I'd also like to know what I did wrong with the installation of Ubuntu. Thanks in advance for you valuable help!

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  • Ubuntu Installation-Allocate drive space/Boot Loader

    - by user10134
    When I try to install ubuntu 10.10 from the official livedisc I got in the mail, when I get to the "Allocate Disk Space" step I cannot get it to work. I shrank my win7 partition so I have unallocated space, then I tried using the space while it is formatted in NTFS, but the partitions will not show up in the box. /dev/sda is selected under boot loader, and I can't select anything else, but the partition box is blank so when I click "install ubuntu" it just says: "No root file system is defined. Please correct this from the partitioning menu." -I am trying to dual-boot win7 and ubuntu, but I was never asked in the install process whether I would like to install just ubuntu or dual-boot?

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  • Chainload boot of Ubuntu installed on 32GB SD card from legacy Grub boot on USB

    - by Gary Darsey
    I have Ubuntu installed on a 32 GB SD card (in the Storage Expansion slot on an Acer Aspire One) with Grub2 installed in the same partition. I boot into legacy Grub on a USB drive and would like to boot by chainloading Grub2 from Grub (kernel/initrd or symlink booting would also be fine), but I haven't figured out how to do this from legacy Grub CLI. Output from blkid for this partition is /dev/mmcblk0p1: LABEL="Ubuntu" UUID="7ceb9fa7-238c-4c5d-bb8e-2c655652ddec" TYPE='ext4" / fdisk -lu information Boot indicator ID 83. Related entries in grub.cfg: search --no-floppy --fs-uuid --set-root 7ceb9fa7-238c-4c5d-bb8e-2c655652ddec linux /boot/vmlinuz-3.5.0-17-generic root=UUID=7ceb9fa7-238c-4c5d-bb8e-2c655652ddec... initrd /boot/initrd.img-3.5.0-17-generic I can't seem to replicate this in legacy Grub. Is there any way get Grub2 to chainload? How do I set root with UUID in legacy Grub? I prefer to boot from USB. Would Grub2 on USB (copying the grub.cfg generated during installation) be an option?

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  • How to repair an external harddrive?

    - by dodohjk
    I would like to reformat my hard disk, and if possible recover the (somewhat unimportant) contents if possible. I have a Western Digital 1TB hard drive which had a NTFS partition. I unplugged the drive without safely removing it first. At first a pop up was asking me to use a Windows OS to run the chkdsk /f command, however, in the effort to keep using a Linux OS I used the ntfsfix command on the ubuntu terminal Now, when I try to access the hard drive, it doesn't show up anymore in Nautilus. I tried reformatting it using Disk Utility, but it gives me an error message, and Gparted would hang on the "Scanning devices" step infinitely. Please comment any output that you would like to see and I will add it to my question. EDIT disk utility tells me is on /dev/sdb the command sudo fdisk -l gives dodohjk@DodosPC:~$ sudo fdisk -l [sudo] password for dodohjk: Disk /dev/sda: 250.1 GB, 250059350016 bytes 255 heads, 63 sectors/track, 30401 cylinders, total 488397168 sectors Units = sectors of 1 * 512 = 512 bytes Sector size (logical/physical): 512 bytes / 512 bytes I/O size (minimum/optimal): 512 bytes / 512 bytes Disk identifier: 0x0006fa8c Device Boot Start End Blocks Id System /dev/sda1 * 4094 482344959 241170433 5 Extended /dev/sda2 482344960 488396799 3025920 82 Linux swap / Solaris /dev/sda5 4096 31461127 15728516 83 Linux /dev/sda6 31463424 52434943 10485760 83 Linux /dev/sda7 52436992 62923320 5243164+ 83 Linux /dev/sda8 62924800 482344959 209710080 83 Linux Disk /dev/sdb: 1000.2 GB, 1000202043392 bytes 255 heads, 63 sectors/track, 121600 cylinders, total 1953519616 sectors Units = sectors of 1 * 512 = 512 bytes Sector size (logical/physical): 512 bytes / 512 bytes I/O size (minimum/optimal): 512 bytes / 512 bytes Disk identifier: 0x6e697373 This doesn't look like a partition table Probably you selected the wrong device. Device Boot Start End Blocks Id System /dev/sdb1 ? 1936269394 3772285809 918008208 4f QNX4.x 3rd part /dev/sdb2 ? 1917848077 2462285169 272218546+ 73 Unknown /dev/sdb3 ? 1818575915 2362751050 272087568 2b Unknown /dev/sdb4 ? 2844524554 2844579527 27487 61 SpeedStor Partition table entries are not in disk order I wrote something wrong here, however here the output of fsck /dev/sbd is dodohjk@DodosPC:~$ sudo fsck /dev/sdb fsck from util-linux 2.20.1 e2fsck 1.42.5 (29-Jul-2012) ext2fs_open2: Bad magic number in super-block fsck.ext2: Superblock invalid, trying backup blocks... fsck.ext2: Bad magic number in super-block while trying to open /dev/sdb The superblock could not be read or does not describe a correct ext2 filesystem. If the device is valid and it really contains an ext2 filesystem (and not swap or ufs or something else), then the superblock is corrupt, and you might try running e2fsck with an alternate superblock: e2fsck -b 8193 <device&gt;

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  • Webcast - Social BPM: Integrating Enterprise 2.0 with Business Applications

    - by peggy.chen
    In today's fast-paced marketplace, successful companies rely on agile business processes and collaborative work environments to stay ahead of the competition. By making your application-based business processes visible, shareable, and flexible through dynamic, process-aware user interfaces, you can ensure that your team's best ideas are heard-and implemented quickly. Join us for this complimentary live Webcast and learn how Oracle's business process management (BPM) solution with integrated Enterprise 2.0 capabilities will enable your team to: Embed ad hoc collaboration into your structured processes and gain a unified view of enterprise information-across business functions-for effective and efficient decision-making Reach out to an expanded network for expert input in resolving exceptions in business workflows Add social feedback loops to your enterprise applications and continuously improve business processes Join us for this LIVE Webcast tomorrow as we discuss how business process management with integrated Enterprise 2.0 collaboration improves business responsiveness and enhances overall enterprise productivity. Take your business to the next level with a unified solution that fosters process-based collaboration between employees, partners, and customers. Register for the webcast now!

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