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  • Java ORM related question - SQL Vs Google DB (Big Table?) GAE

    - by StackerFlow
    I was wondering about the following two options when one is not using SQL tables but ORM based DBs (Example - when you are using GAE) Would the second option be less efficient? Requirement: There is an object. The object has a collection of similar items. I need to store this object. Example, say the object is a tree and it has a collection of leaves. Option 1: Traditional SQL type structure: Table for the Tree (with TreeId as the identifier for a row in the Table.) Table for the Leaves (where each leaf has a TreeId and to show the leaves of a tree, I query all leaves where the TreeId is the Id of the tree.) Here, the Tree structure DOES NOT have a field with leaves. Option 2: ORM / GAE Tables: Using the same example above, I have an object for Tree where the object has a collection (Set/List in Java/C++) of leaves. I store and retrieve the Tree together with the leaves (as the leaves are implemented as a Set in the Tree object) My question is, will the second one be less efficient that the first option? If so, why? Are there other alternatives? Thank you!

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  • How could I compare these database values in PHP?

    - by Dan
    So a user selects from a drop down list a value. I take this value, put it into a variable, then select from the database the ID value of that table A holding the selected value also. So now I'm trying to use that ID value to get to a many-to-many relationship table that has the selected value from table A to a different table B. The many-to-many relationship table has both IDs. How can I compare this using PHP? So it would be like: $A = $_POST['a']; $sql = "SELECT a, aID from TABLEA WHERE a = $A"; What do I do then to compare the aID with the many-to-many relationships table, then get the other ID in that table and then take that ID to get values from table B?

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  • Large recovery partitions

    - by Unsigned
    Is there any good reason as to why factory restore partitions are generally much larger than they need to be? Examples I have found in my own experience: Dell XPS laptop Partition: 13.67 GB Used: 6.68 GB Dell Inspiron laptop Partition: 14.7 GB Used: 7.2 GB Toshiba laptop Partition: 15.3 GB Used: 9 GB In all cases, shrinking the partition to only slightly more than the Used space had no ill effects on future factory restorations. Why the exorbitant amount of extra space, given that neither of the three computers ever writes any data to the recovery partition? Is there a good reason I'm overlooking?

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  • What are advantages of using a one-to-one table relationship? (MySQL)

    - by byronh
    What are advantages of using a one-to-one table relationship as opposed to simply storing all the data in one table? I understand and make use of one-to-many, many-to-one, and many-to-many all the time, but implementing a one-to-one relationship seems like a tedious and unnecessary task, especially if you use naming conventions for relating (php) objects to database tables. I couldn't find anything on the net or on this site that could supply a good real-world example of a one-to-one relationship. At first I thought it might be logical to separate 'users', for example, into two tables, one containing public information like an 'about me' for profile pages and one containing private information such as login/password, etc. But why go through all the trouble of using unnecessary JOINS when you can just choose which fields to select from that table anyway? If I'm displaying the user's profile page, obviously I would only SELECT id,username,email,aboutme etc. and not the fields containing their private info. Anyone care to enlighten me with some real-world examples of one-to-one relationships?

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  • Are there any reasons to duplicate table in the same database ?

    - by bob
    Let says we have several MySQL server, one master and some slaves. A member table which contains more than 5.000.000 peoples. Are there any reasons (performance, atomicity, etc..) to use duplicate tables like member_1, member_2, member_3 and then switch randomly when doing operation on it ? (especialy SELECT query) ?

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  • How to return corresponding row number in a table if a value falls within the bounds specified? [closed]

    - by Eshwar
    Possible Duplicate: Looking up a value, depending on which set of dates another date falls between Basically I have an excel table with 3 Columns - Month, Start, Finish - where Start and Finish are lower and upper bounds for transaction numbers and Month is a string. In another cell I have a transaction number that I want to find the corresponding month for. e.g. Jan 01 10 Feb 11 15 And if I want to find 12, I should get Feb out. (No VB, macros, etc. Please)

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  • The database engine couldn't lock the table, because it is already in use by another person or proce

    - by tintincute
    Hi I tried to do "Split a Database" and after I clicked on the "Split" button, here is what I got: "The database engine couldn't lock the table, because it is already in use by another person or process" Any idea? Thanks additional question: is it possible to split your database many times? first i'm trying it at home and the following day i would like to try it at the office if it works. i already tried the split and if i do it tomorrow at the office would that be a problem? Thanks

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  • How do you find all dependencies of a database table?

    - by Carlos
    In MS SQL 2005, is it possible to find out which tables/columns are being used either as keys in another table, or as part of a stored procedure? The reason is I'm trying to clean up some old stored procs and tables, some of which can be removed, some of which can have columns pruned. But obviously I don't want to remove stuff which is being used.

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  • SQL SERVER – Curious Case of Disappearing Rows – ON UPDATE CASCADE and ON DELETE CASCADE – Part 1 of 2

    - by pinaldave
    Social media has created an Always Connected World for us. Recently I enrolled myself to learn new technologies as a student. I had decided to focus on learning and decided not to stay connected on the internet while I am in the learning session. On the second day of the event after the learning was over, I noticed lots of notification from my friend on my various social media handle. He had connected with me on Twitter, Facebook, Google+, LinkedIn, YouTube as well SMS, WhatsApp on the phone, Skype messages and not to forget with a few emails. I right away called him up. The problem was very unique – let us hear the problem in his own words. “Pinal – we are in big trouble we are not able to figure out what is going on. Our product details table is continuously loosing rows. Lots of rows have disappeared since morning and we are unable to find why the rows are getting deleted. We have made sure that there is no DELETE command executed on the table as well. The matter of the fact, we have removed every single place the code which is referencing the table. We have done so many crazy things out of desperation but no luck. The rows are continuously deleted in a random pattern. Do you think we have problems with intrusion or virus?” After describing the problems he had pasted few rants about why I was not available during the day. I think it will be not smart to post those exact words here (due to many reasons). Well, my immediate reaction was to get online with him. His problem was unique to him and his team was all out to fix the issue since morning. As he said he has done quite a lot out in desperation. I started asking questions from audit, policy management and profiling the data. Very soon I realize that I think this problem was not as advanced as it looked. There was no intrusion, SQL Injection or virus issue. Well, long story short first - It was a very simple issue of foreign key created with ON UPDATE CASCADE and ON DELETE CASCADE.  CASCADE allows deletions or updates of key values to cascade through the tables defined to have foreign key relationships that can be traced back to the table on which the modification is performed. ON DELETE CASCADE specifies that if an attempt is made to delete a row with a key referenced by foreign keys in existing rows in other tables, all rows containing those foreign keys are also deleted. ON UPDATE CASCADE specifies that if an attempt is made to update a key value in a row, where the key value is referenced by foreign keys in existing rows in other tables, all of the foreign key values are also updated to the new value specified for the key. (Reference: BOL) In simple words – due to ON DELETE CASCASE whenever is specified when the data from Table A is deleted and if it is referenced in another table using foreign key it will be deleted as well. In my friend’s case, they had two tables, Products and ProductDetails. They had created foreign key referential integrity of the product id between the table. Now the as fall was up they were updating their catalogue. When they were updating the catalogue they were deleting products which are no more available. As the changes were cascading the corresponding rows were also deleted from another table. This is CORRECT. The matter of the fact, there is no error or anything and SQL Server is behaving how it should be behaving. The problem was in the understanding and inappropriate implementations of business logic.  What they needed was Product Master Table, Current Product Catalogue, and Product Order Details History tables. However, they were using only two tables and without proper understanding the relation between them was build using foreign keys. If there were only two table, they should have used soft delete which will not actually delete the record but just hide it from the original product table. This workaround could have got them saved from cascading delete issues. I will be writing a detailed post on the design implications etc in my future post as in above three lines I cannot cover every issue related to designing and it is also not the scope of the blog post. More about designing in future blog posts. Once they learn their mistake, they were happy as there was no intrusion but trust me sometime we are our own enemy and this is a great example of it. In tomorrow’s blog post we will go over their code and workarounds. Feel free to share your opinions, experiences and comments. Reference: Pinal Dave (http://blog.sqlauthority.com) Filed under: PostADay, SQL, SQL Authority, SQL Query, SQL Server, SQL Tips and Tricks, T SQL, Technology

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  • How to overlay a div (or any element) over a table row (tr)?

    - by slolife
    I'd like to overlay a div (or any element that'll work) over a table row (tr tag) that happens to have more than one column. I have tried a few methods, which don't seem to work. I've posted my current code below. I do get an overlay, but not directly over just the row. I tried setting the overlay top to $divBottom.css('top'), but that is always 'auto'. So, am I on the right track, or is there a better way of doing it? Utilizing jQuery is fine as you can see. If I am on the right track, how do I get the div placed correctly? Is the offsetTop an offset in the containing element, the table, and I need to do some math? Any other gotchas I'll run into with that? <html> <head> <title>Overlay Tests</title> <style> #rowBottom { outline:red solid 2px } #divBottom { margin:1em; font-size:xx-large; position:relative; } #divOverlay { background-color:Silver; text-align:center; position:absolute; z-index:10000; opacity:0.5; } </style> </head> <body> <p align="center"><a id="lnkDoIt" href="#">Do it!</a></p> <table width="100%" border="0" cellpadding="10" cellspacing="3" style="position:relative"> <tr> <td><p>Lorem ipsum dolor sit amet, consectetur adipiscing elit.</p></td> </tr> <tr id="rowBottom"> <td><div id="divBottom"><p align="center">This is the bottom text</p></div></td> </tr> <tr> <td><p>Lorem ipsum dolor sit amet, consectetur adipiscing elit.</p></td> </tr> </table> <div id="divOverlay" style=""><p>This is the overlay div.</p><p id="info"></p></div> <script src="../includes/javascript/jquery/jquery-1.4.2.js" type="text/javascript"></script> <script type="text/javascript"> $(document).ready(function() { $('#lnkDoIt').click(function() { var $divBottom = $('#rowBottom'); var $divOverlay = $('#divOverlay'); var bottomTop = $divBottom.attr('offsetTop'); var bottomLeft = $divBottom.attr('offsetLeft'); var bottomWidth = $divBottom.css('width'); var bottomHeight = $divBottom.css('height'); $divOverlay.css('top', bottomTop); $divOverlay.css('left', bottomLeft); $divOverlay.css('width', bottomWidth); $divOverlay.css('height', bottomHeight); $('#info').text('Top: ' + bottomTop + ' Left: ' + bottomLeft); }); }); </script> </body> </html>

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  • When is a SQL function not a function?

    - by Rob Farley
    Should SQL Server even have functions? (Oh yeah – this is a T-SQL Tuesday post, hosted this month by Brad Schulz) Functions serve an important part of programming, in almost any language. A function is a piece of code that is designed to return something, as opposed to a piece of code which isn’t designed to return anything (which is known as a procedure). SQL Server is no different. You can call stored procedures, even from within other stored procedures, and you can call functions and use these in other queries. Stored procedures might query something, and therefore ‘return data’, but a function in SQL is considered to have the type of the thing returned, and can be used accordingly in queries. Consider the internal GETDATE() function. SELECT GETDATE(), SomeDatetimeColumn FROM dbo.SomeTable; There’s no logical difference between the field that is being returned by the function and the field that’s being returned by the table column. Both are the datetime field – if you didn’t have inside knowledge, you wouldn’t necessarily be able to tell which was which. And so as developers, we find ourselves wanting to create functions that return all kinds of things – functions which look up values based on codes, functions which do string manipulation, and so on. But it’s rubbish. Ok, it’s not all rubbish, but it mostly is. And this isn’t even considering the SARGability impact. It’s far more significant than that. (When I say the SARGability aspect, I mean “because you’re unlikely to have an index on the result of some function that’s applied to a column, so try to invert the function and query the column in an unchanged manner”) I’m going to consider the three main types of user-defined functions in SQL Server: Scalar Inline Table-Valued Multi-statement Table-Valued I could also look at user-defined CLR functions, including aggregate functions, but not today. I figure that most people don’t tend to get around to doing CLR functions, and I’m going to focus on the T-SQL-based user-defined functions. Most people split these types of function up into two types. So do I. Except that most people pick them based on ‘scalar or table-valued’. I’d rather go with ‘inline or not’. If it’s not inline, it’s rubbish. It really is. Let’s start by considering the two kinds of table-valued function, and compare them. These functions are going to return the sales for a particular salesperson in a particular year, from the AdventureWorks database. CREATE FUNCTION dbo.FetchSales_inline(@salespersonid int, @orderyear int) RETURNS TABLE AS  RETURN (     SELECT e.LoginID as EmployeeLogin, o.OrderDate, o.SalesOrderID     FROM Sales.SalesOrderHeader AS o     LEFT JOIN HumanResources.Employee AS e     ON e.EmployeeID = o.SalesPersonID     WHERE o.SalesPersonID = @salespersonid     AND o.OrderDate >= DATEADD(year,@orderyear-2000,'20000101')     AND o.OrderDate < DATEADD(year,@orderyear-2000+1,'20000101') ) ; GO CREATE FUNCTION dbo.FetchSales_multi(@salespersonid int, @orderyear int) RETURNS @results TABLE (     EmployeeLogin nvarchar(512),     OrderDate datetime,     SalesOrderID int     ) AS BEGIN     INSERT @results (EmployeeLogin, OrderDate, SalesOrderID)     SELECT e.LoginID, o.OrderDate, o.SalesOrderID     FROM Sales.SalesOrderHeader AS o     LEFT JOIN HumanResources.Employee AS e     ON e.EmployeeID = o.SalesPersonID     WHERE o.SalesPersonID = @salespersonid     AND o.OrderDate >= DATEADD(year,@orderyear-2000,'20000101')     AND o.OrderDate < DATEADD(year,@orderyear-2000+1,'20000101')     ;     RETURN END ; GO You’ll notice that I’m being nice and responsible with the use of the DATEADD function, so that I have SARGability on the OrderDate filter. Regular readers will be hoping I’ll show what’s going on in the execution plans here. Here I’ve run two SELECT * queries with the “Show Actual Execution Plan” option turned on. Notice that the ‘Query cost’ of the multi-statement version is just 2% of the ‘Batch cost’. But also notice there’s trickery going on. And it’s nothing to do with that extra index that I have on the OrderDate column. Trickery. Look at it – clearly, the first plan is showing us what’s going on inside the function, but the second one isn’t. The second one is blindly running the function, and then scanning the results. There’s a Sequence operator which is calling the TVF operator, and then calling a Table Scan to get the results of that function for the SELECT operator. But surely it still has to do all the work that the first one is doing... To see what’s actually going on, let’s look at the Estimated plan. Now, we see the same plans (almost) that we saw in the Actuals, but we have an extra one – the one that was used for the TVF. Here’s where we see the inner workings of it. You’ll probably recognise the right-hand side of the TVF’s plan as looking very similar to the first plan – but it’s now being called by a stack of other operators, including an INSERT statement to be able to populate the table variable that the multi-statement TVF requires. And the cost of the TVF is 57% of the batch! But it gets worse. Let’s consider what happens if we don’t need all the columns. We’ll leave out the EmployeeLogin column. Here, we see that the inline function call has been simplified down. It doesn’t need the Employee table. The join is redundant and has been eliminated from the plan, making it even cheaper. But the multi-statement plan runs the whole thing as before, only removing the extra column when the Table Scan is performed. A multi-statement function is a lot more powerful than an inline one. An inline function can only be the result of a single sub-query. It’s essentially the same as a parameterised view, because views demonstrate this same behaviour of extracting the definition of the view and using it in the outer query. A multi-statement function is clearly more powerful because it can contain far more complex logic. But a multi-statement function isn’t really a function at all. It’s a stored procedure. It’s wrapped up like a function, but behaves like a stored procedure. It would be completely unreasonable to expect that a stored procedure could be simplified down to recognise that not all the columns might be needed, but yet this is part of the pain associated with this procedural function situation. The biggest clue that a multi-statement function is more like a stored procedure than a function is the “BEGIN” and “END” statements that surround the code. If you try to create a multi-statement function without these statements, you’ll get an error – they are very much required. When I used to present on this kind of thing, I even used to call it “The Dangers of BEGIN and END”, and yes, I’ve written about this type of thing before in a similarly-named post over at my old blog. Now how about scalar functions... Suppose we wanted a scalar function to return the count of these. CREATE FUNCTION dbo.FetchSales_scalar(@salespersonid int, @orderyear int) RETURNS int AS BEGIN     RETURN (         SELECT COUNT(*)         FROM Sales.SalesOrderHeader AS o         LEFT JOIN HumanResources.Employee AS e         ON e.EmployeeID = o.SalesPersonID         WHERE o.SalesPersonID = @salespersonid         AND o.OrderDate >= DATEADD(year,@orderyear-2000,'20000101')         AND o.OrderDate < DATEADD(year,@orderyear-2000+1,'20000101')     ); END ; GO Notice the evil words? They’re required. Try to remove them, you just get an error. That’s right – any scalar function is procedural, despite the fact that you wrap up a sub-query inside that RETURN statement. It’s as ugly as anything. Hopefully this will change in future versions. Let’s have a look at how this is reflected in an execution plan. Here’s a query, its Actual plan, and its Estimated plan: SELECT e.LoginID, y.year, dbo.FetchSales_scalar(p.SalesPersonID, y.year) AS NumSales FROM (VALUES (2001),(2002),(2003),(2004)) AS y (year) CROSS JOIN Sales.SalesPerson AS p LEFT JOIN HumanResources.Employee AS e ON e.EmployeeID = p.SalesPersonID; We see here that the cost of the scalar function is about twice that of the outer query. Nicely, the query optimizer has worked out that it doesn’t need the Employee table, but that’s a bit of a red herring here. There’s actually something way more significant going on. If I look at the properties of that UDF operator, it tells me that the Estimated Subtree Cost is 0.337999. If I just run the query SELECT dbo.FetchSales_scalar(281,2003); we see that the UDF cost is still unchanged. You see, this 0.0337999 is the cost of running the scalar function ONCE. But when we ran that query with the CROSS JOIN in it, we returned quite a few rows. 68 in fact. Could’ve been a lot more, if we’d had more salespeople or more years. And so we come to the biggest problem. This procedure (I don’t want to call it a function) is getting called 68 times – each one between twice as expensive as the outer query. And because it’s calling it in a separate context, there is even more overhead that I haven’t considered here. The cheek of it, to say that the Compute Scalar operator here costs 0%! I know a number of IT projects that could’ve used that kind of costing method, but that’s another story that I’m not going to go into here. Let’s look at a better way. Suppose our scalar function had been implemented as an inline one. Then it could have been expanded out like a sub-query. It could’ve run something like this: SELECT e.LoginID, y.year, (SELECT COUNT(*)     FROM Sales.SalesOrderHeader AS o     LEFT JOIN HumanResources.Employee AS e     ON e.EmployeeID = o.SalesPersonID     WHERE o.SalesPersonID = p.SalesPersonID     AND o.OrderDate >= DATEADD(year,y.year-2000,'20000101')     AND o.OrderDate < DATEADD(year,y.year-2000+1,'20000101')     ) AS NumSales FROM (VALUES (2001),(2002),(2003),(2004)) AS y (year) CROSS JOIN Sales.SalesPerson AS p LEFT JOIN HumanResources.Employee AS e ON e.EmployeeID = p.SalesPersonID; Don’t worry too much about the Scan of the SalesOrderHeader underneath a Nested Loop. If you remember from plenty of other posts on the matter, execution plans don’t push the data through. That Scan only runs once. The Index Spool sucks the data out of it and populates a structure that is used to feed the Stream Aggregate. The Index Spool operator gets called 68 times, but the Scan only once (the Number of Executions property demonstrates this). Here, the Query Optimizer has a full picture of what’s being asked, and can make the appropriate decision about how it accesses the data. It can simplify it down properly. To get this kind of behaviour from a function, we need it to be inline. But without inline scalar functions, we need to make our function be table-valued. Luckily, that’s ok. CREATE FUNCTION dbo.FetchSales_inline2(@salespersonid int, @orderyear int) RETURNS table AS RETURN (SELECT COUNT(*) as NumSales     FROM Sales.SalesOrderHeader AS o     LEFT JOIN HumanResources.Employee AS e     ON e.EmployeeID = o.SalesPersonID     WHERE o.SalesPersonID = @salespersonid     AND o.OrderDate >= DATEADD(year,@orderyear-2000,'20000101')     AND o.OrderDate < DATEADD(year,@orderyear-2000+1,'20000101') ); GO But we can’t use this as a scalar. Instead, we need to use it with the APPLY operator. SELECT e.LoginID, y.year, n.NumSales FROM (VALUES (2001),(2002),(2003),(2004)) AS y (year) CROSS JOIN Sales.SalesPerson AS p LEFT JOIN HumanResources.Employee AS e ON e.EmployeeID = p.SalesPersonID OUTER APPLY dbo.FetchSales_inline2(p.SalesPersonID, y.year) AS n; And now, we get the plan that we want for this query. All we’ve done is tell the function that it’s returning a table instead of a single value, and removed the BEGIN and END statements. We’ve had to name the column being returned, but what we’ve gained is an actual inline simplifiable function. And if we wanted it to return multiple columns, it could do that too. I really consider this function to be superior to the scalar function in every way. It does need to be handled differently in the outer query, but in many ways it’s a more elegant method there too. The function calls can be put amongst the FROM clause, where they can then be used in the WHERE or GROUP BY clauses without fear of calling the function multiple times (another horrible side effect of functions). So please. If you see BEGIN and END in a function, remember it’s not really a function, it’s a procedure. And then fix it. @rob_farley

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  • 12c - flashforward, flashback or see it as of now...

    - by noreply(at)blogger.com (Thomas Kyte)
    Oracle 9i exposed flashback query to developers for the first time.  The ability to flashback query dates back to version 4 however (it just wasn't exposed).  Every time you run a query in Oracle it is in fact a flashback query - it is what multi-versioning is all about.However, there was never a flashforward query (well, ok, the workspace manager has this capability - but with lots of extra baggage).  We've never been able to ask a table "what will you look like tomorrow" - but now we do.The capability is called Temporal Validity.  If you have a table with data that is effective dated - has a "start date" and "end date" column in it - we can now query it using flashback query like syntax.  The twist is - the date we "flashback" to can be in the future.  It works by rewriting the query to transparently the necessary where clause and filter out the right rows for the right period of time - and since you can have records whose start date is in the future - you can query a table and see what it would look like at some future time.Here is a quick example, we'll start with a table:ops$tkyte%ORA12CR1> create table addresses  2  ( empno       number,  3    addr_data   varchar2(30),  4    start_date  date,  5    end_date    date,  6    period for valid(start_date,end_date)  7  )  8  /Table created.the new bit is on line 6 (it can be altered into an existing table - so any table  you have with a start/end date column will be a candidate).  The keyword is PERIOD, valid is an identifier I chose - it could have been foobar, valid just sounds nice in the query later.  You identify the columns in your table - or we can create them for you if they don't exist.  Then you just create some data:ops$tkyte%ORA12CR1> insert into addresses (empno, addr_data, start_date, end_date )  2  values ( 1234, '123 Main Street', trunc(sysdate-5), trunc(sysdate-2) );1 row created.ops$tkyte%ORA12CR1>ops$tkyte%ORA12CR1> insert into addresses (empno, addr_data, start_date, end_date )  2  values ( 1234, '456 Fleet Street', trunc(sysdate-1), trunc(sysdate+1) );1 row created.ops$tkyte%ORA12CR1>ops$tkyte%ORA12CR1> insert into addresses (empno, addr_data, start_date, end_date )  2  values ( 1234, '789 1st Ave', trunc(sysdate+2), null );1 row created.and you can either see all of the data:ops$tkyte%ORA12CR1> select * from addresses;     EMPNO ADDR_DATA                      START_DAT END_DATE---------- ------------------------------ --------- ---------      1234 123 Main Street                27-JUN-13 30-JUN-13      1234 456 Fleet Street               01-JUL-13 03-JUL-13      1234 789 1st Ave                    04-JUL-13or query "as of" some point in time - as  you can see in the predicate section - it is just doing a query rewrite to automate the "where" filters:ops$tkyte%ORA12CR1> select * from addresses as of period for valid sysdate-3;     EMPNO ADDR_DATA                      START_DAT END_DATE---------- ------------------------------ --------- ---------      1234 123 Main Street                27-JUN-13 30-JUN-13ops$tkyte%ORA12CR1> @planops$tkyte%ORA12CR1> select * from table(dbms_xplan.display_cursor);PLAN_TABLE_OUTPUT-------------------------------------------------------------------------------SQL_ID  cthtvvm0dxvva, child number 0-------------------------------------select * from addresses as of period for valid sysdate-3Plan hash value: 3184888728-------------------------------------------------------------------------------| Id  | Operation         | Name      | Rows  | Bytes | Cost (%CPU)| Time     |-------------------------------------------------------------------------------|   0 | SELECT STATEMENT  |           |       |       |     3 (100)|          ||*  1 |  TABLE ACCESS FULL| ADDRESSES |     1 |    48 |     3   (0)| 00:00:01 |-------------------------------------------------------------------------------Predicate Information (identified by operation id):---------------------------------------------------   1 - filter((("T"."START_DATE" IS NULL OR              "T"."START_DATE"<=SYSDATE@!-3) AND ("T"."END_DATE" IS NULL OR              "T"."END_DATE">SYSDATE@!-3)))Note-----   - dynamic statistics used: dynamic sampling (level=2)24 rows selected.ops$tkyte%ORA12CR1> select * from addresses as of period for valid sysdate;     EMPNO ADDR_DATA                      START_DAT END_DATE---------- ------------------------------ --------- ---------      1234 456 Fleet Street               01-JUL-13 03-JUL-13ops$tkyte%ORA12CR1> @planops$tkyte%ORA12CR1> select * from table(dbms_xplan.display_cursor);PLAN_TABLE_OUTPUT-------------------------------------------------------------------------------SQL_ID  26ubyhw9hgk7z, child number 0-------------------------------------select * from addresses as of period for valid sysdatePlan hash value: 3184888728-------------------------------------------------------------------------------| Id  | Operation         | Name      | Rows  | Bytes | Cost (%CPU)| Time     |-------------------------------------------------------------------------------|   0 | SELECT STATEMENT  |           |       |       |     3 (100)|          ||*  1 |  TABLE ACCESS FULL| ADDRESSES |     1 |    48 |     3   (0)| 00:00:01 |-------------------------------------------------------------------------------Predicate Information (identified by operation id):---------------------------------------------------   1 - filter((("T"."START_DATE" IS NULL OR              "T"."START_DATE"<=SYSDATE@!) AND ("T"."END_DATE" IS NULL OR              "T"."END_DATE">SYSDATE@!)))Note-----   - dynamic statistics used: dynamic sampling (level=2)24 rows selected.ops$tkyte%ORA12CR1> select * from addresses as of period for valid sysdate+3;     EMPNO ADDR_DATA                      START_DAT END_DATE---------- ------------------------------ --------- ---------      1234 789 1st Ave                    04-JUL-13ops$tkyte%ORA12CR1> @planops$tkyte%ORA12CR1> select * from table(dbms_xplan.display_cursor);PLAN_TABLE_OUTPUT-------------------------------------------------------------------------------SQL_ID  36bq7shnhc888, child number 0-------------------------------------select * from addresses as of period for valid sysdate+3Plan hash value: 3184888728-------------------------------------------------------------------------------| Id  | Operation         | Name      | Rows  | Bytes | Cost (%CPU)| Time     |-------------------------------------------------------------------------------|   0 | SELECT STATEMENT  |           |       |       |     3 (100)|          ||*  1 |  TABLE ACCESS FULL| ADDRESSES |     1 |    48 |     3   (0)| 00:00:01 |-------------------------------------------------------------------------------Predicate Information (identified by operation id):---------------------------------------------------   1 - filter((("T"."START_DATE" IS NULL OR              "T"."START_DATE"<=SYSDATE@!+3) AND ("T"."END_DATE" IS NULL OR              "T"."END_DATE">SYSDATE@!+3)))Note-----   - dynamic statistics used: dynamic sampling (level=2)24 rows selected.All in all a nice, easy way to query effective dated information as of a point in time without a complex where clause.  You need to maintain the data - it isn't that a delete will turn into an update the end dates a record or anything - but if you have tables with start/end dates, this will make it much easier to query them.

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  • What causes "A disk read error occurred, Press Ctrl + Alt + Del to restart"?

    - by Mehrdad
    I have a virtual machine containing Windows XP SP3. When I resized the VHD file (and the embedded partition), and tried booting, I got: A disk read error occurred Press Ctrl + Alt + Del to restart Some notes: FixBoot and FixMBR don't help. ChkDsk doesn't help. The partition is indeed active. The partition starts at sector 63 (it also did so before the problem) of cylinder 1, head 1, and is marked as type 0x07 (NTFS) My host OS reads the VHD and the partition completely fine I'm interested in knowing the cause rather than the fix. So "re-format the disk", "reinstall Windows", etc. aren't valid solutions. It's a virtual machine after all... I have nothing to lose, so I don't care about fixing it. I just want to know what's causing this problem, in case I run into it again on a physical machine (which I have done before). More info: The layout of the original, dynamic VHD (which works correctly): +-----------------------------------------------------------------------------+ ¦ Disk: 3 MBR/GPT: MBR ¦ ¦ Size: 127.00GB CHS: 16578 255 63 ¦ ¦ Sectors: 266338304 Disk Signature: 0xEE3EEE3E ¦ ¦ Partitions: 1 Partition Order: 1 ¦ ¦ Media Type: Fixed Interface: SCSI ¦ ¦ Description: Msft Virtual Disk ¦ +-----------------------------------------------------------------------------¦ ¦Pos Idx Type/Name Size Boot Hide Start Sector Total Sectors DL Vol Label ¦ +--- --- --------- ---- ---- ---- -------------- -------------- -- -----------¦ ¦ 1 1 07-NTFS 1.5G Yes No 63 3,148,677 F: <None> ¦ +-----------------------------------------------------------------------------+ The layout of the resized, fixed-size VHD (which doesn't work): +-----------------------------------------------------------------------------+ ¦ Disk: 3 MBR/GPT: MBR ¦ ¦ Size: 1.50GB CHS: 196 255 63 ¦ ¦ Sectors: 3149824 Disk Signature: 0xEE3EEE3E ¦ ¦ Partitions: 1 Partition Order: 1 ¦ ¦ Media Type: Fixed Interface: SCSI ¦ ¦ Description: Msft Virtual Disk ¦ +-----------------------------------------------------------------------------¦ ¦Pos Idx Type/Name Size Boot Hide Start Sector Total Sectors DL Vol Label ¦ +--- --- --------- ---- ---- ---- -------------- -------------- -- -----------¦ ¦ 1 1 07-NTFS 1.5G Yes No 63 3,148,677 F: <None> ¦ +-----------------------------------------------------------------------------+

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  • Portable USB drives hidden pertition - New request

    - by ZXC
    This question was made by Francesco on Jul 29 '11 at 17:14. and the replies were not satisfactory due they not point to an important problem that´s: Why could anyone want to make certain data only accesible for a program but not to the users?. For example: If I want to do a safe distribution of original music for demostration purposes I will need several requisites: 1) The music should be heard using a simple procedure like selecting the name of each song on a playlist of a mediaplayer. 2) The portable media, ussually a portable USB drive, must hide for complete and should make unaccesible the files that contain the audio data to anything but the mediaplayer, that must be in the first partition, the one that is visible. 3) Considering that´s impossible to really hide files in a non-hidden partition, a second hidden partition should be created in the USB drive and the audio data will be stored there. 4) The trick is to read the audio data files stored in the hidden partition with a mediaplayer stored in the visible partition, the media player also should be a complete standalone program and independent from any library of the operating system except of the OS audio system. 5) The hidden partition should have a copy protection scheme that could impede to do copies of the data or create working ISO images of it. I know that this description could not be technically accurate but it has a complete logic from the needs of a music producer against the problem of piracy. The philosophy that surrounds the concept is to transform a virtual object like a digital string of audio in a solid object like the analog vinyl discs are.

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  • How do I remove a USB drive's write protection?

    - by nate
    I have a SanDisk Cruser Blade USB stick that suddenly seems to be write protected. I tried running DiskPart but after I write the command "attributes disk clear readonly" it displays this: Microsoft DiskPart version 5.1.3565 ADD - Add a mirror to a simple volume. ACTIVE - Marks the current basic partition as an active boot partition. ASSIGN - Assign a drive letter or mount point to the selected volume. BREAK - Break a mirror set. CLEAN - Clear the configuration information, or all information, off the disk. CONVERT - Converts between different disk formats. CREATE - Create a volume or partition. DELETE - Delete an object. DETAIL - Provide details about an object. EXIT - Exit DiskPart EXTEND - Extend a volume. HELP - Prints a list of commands. IMPORT - Imports a disk group. LIST - Prints out a list of objects. INACTIVE - Marks the current basic partition as an inactive partition. ONLINE - Online a disk that is currently marked as offline. REM - Does nothing. Used to comment scripts. REMOVE - Remove a drive letter or mount point assignment. REPAIR - Repair a RAID-5 volume. RESCAN - Rescan the computer looking for disks and volumes. RETAIN - Place a retainer partition under a simple volume. SELECT - Move the focus to an object. It's like when you type help at the DiskPart prompt, so how do I get past this? This problem started when I plugged the stick into a laptop which had viruses, if that's any help.

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  • Corrupted NTFS Drive showing multiple unallocated partitions

    - by volting
    My external hdd with a single NTFS partition was accidentaly plugged out (kids!)... and is now corrupted. Iv tried running ntfsfix - with no luck - output below.. When I look at the disk under disk management in Windows 7 it shows up as having 5 partitions 2 of which are unallocated - none have drive letters and it is not possible to set any (that option and most others are greyed out) - so I can't run chkdsk /f Iv tried using Minitool partition wizard which was mentioned as a solution to another similar question here. It showed the whole drive as one partition, but as unallocated, and the option -- "Check File System" was greyout. Is there anything else I could try ? Output of fdisk -l Disk /dev/sdb: 1500.3 GB, 1500299395072 bytes 255 heads, 63 sectors/track, 182401 cylinders, total 2930272256 sectors Units = sectors of 1 * 512 = 512 bytes Sector size (logical/physical): 512 bytes / 512 bytest I/O size (minimum/optimal): 512 bytes / 512 bytes Disk identifier: 0x69205244 This doesn't look like a partition table Probably you selected the wrong device. Device Boot Start End Blocks Id System /dev/sdb1 ? 218129509 1920119918 850995205 72 Unknown /dev/sdb2 ? 729050177 1273024900 271987362 74 Unknown /dev/sdb3 ? 168653938 168653938 0 65 Novell Netware 386 /dev/sdb4 2692939776 2692991410 25817+ 0 Empty Partition table entries are not in disk order Output of ntfsfix me@vaio:/dev$ sudo ntfsfix /dev/sdb Mounting volume... ntfs_mst_post_read_fixup_warn: magic: 0xffffffff size: 1024 usa_ofs: 65535 usa_count: 65534: Invalid argument Record 0 has no FILE magic (0xffffffff) Failed to load $MFT: Input/output error FAILED Attempting to correct errors... ntfs_mst_post_read_fixup_warn: magic: 0xffffffff size: 1024 usa_ofs: 65535 usa_count: 65534: Invalid argument Record 0 has no FILE magic (0xffffffff) Failed to load $MFT: Input/output error FAILED Failed to startup volume: Input/output error Checking for self-located MFT segment... ntfs_mst_post_read_fixup_warn: magic: 0xffffffff size: 1024 usa_ofs: 65535 usa_count: 65534: Invalid argument OK ntfs_mst_post_read_fixup_warn: magic: 0xffffffff size: 1024 usa_ofs: 65535 usa_count: 65534: Invalid argument Record 0 has no FILE magic (0xffffffff) Failed to load $MFT: Input/output error Volume is corrupt. You should run chkdsk. Options available with MiniTool: Related questions: How to fix a damaged/corrupted NTFS filesystem/partition without losing the data on it? Repair corrupted NTFS File System

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  • Changing the default boot option without losing the boot menu

    - by hvd
    I've had a working multi-boot setup with the Windows boot loader, containing menu items for two Windows 7 systems, and one for Grub. Grub in turn contains multiple menu items, but I think that's not relevant here. I've upgraded one system to Windows 8. When I now set a different system as the default, I lose the boot menu, and I lose the possibility of booting into the other systems. I've set Windows 7 as the default, rebooted, and get Windows 7, but I don't get to choose which system to boot into. I can run its own bcdedit to change the default back to Windows 8, and another reboot shows the boot menu again, but how can I avoid defaulting to Windows 8? Here are my current boot settings, is there anything that is misconfigured? C:\WINDOWS\system32>bcdedit Windows Boot Manager -------------------- identifier {bootmgr} device partition=F: description Windows Boot Manager locale nl-NL inherit {globalsettings} integrityservices Enable default {current} resumeobject {2f8b77f0-a30b-11e1-a9c6-a4bd8d37f662} displayorder {current} {2f8b77e3-a30b-11e1-a9c6-a4bd8d37f662} {2f8b77ee-a30b-11e1-a9c6-a4bd8d37f662} toolsdisplayorder {memdiag} timeout 30 Windows Boot Loader ------------------- identifier {current} device partition=C: path \WINDOWS\system32\winload.exe description Windows 8 locale nl-NL inherit {bootloadersettings} integrityservices Enable recoveryenabled No allowedinmemorysettings 0x15000075 osdevice partition=C: systemroot \WINDOWS resumeobject {2f8b77f0-a30b-11e1-a9c6-a4bd8d37f662} nx OptIn bootmenupolicy Standard Windows Boot Loader ------------------- identifier {2f8b77e3-a30b-11e1-a9c6-a4bd8d37f662} device partition=D: path \Windows\system32\winload.exe description Windows 7 locale nl-NL osdevice partition=D: systemroot \Windows resumeobject {59616f59-a2ba-11e1-b73a-806e6f6e6963} nx OptIn pae Default bootmenupolicy Standard hypervisorlaunchtype Auto detecthal Yes sos No debug No Real-mode Boot Sector --------------------- identifier {2f8b77ee-a30b-11e1-a9c6-a4bd8d37f662} device partition=C: path \grub\winloader\grub.boot description Grub 2

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  • How to verify TRIM/discard on encrypted swap?

    - by svarni
    I am using an encrypted swap partition via ecryptfs-setup-swap on my Ubuntu 13.04 computer using a SSD. I have manually set up trim for my ext4 root partition (simply by adding the "discard" option in /etc/fstab). I also manually ran fstrim on the root partition prior to booting and using dstat I saw that for a few seconds several GB/s of data have been written to the disk. That was presumably the effect of the trim command. These high writerates are reproducable by deleting huge files and have not occured before setting up trim, so I take them as evidence for working trim/discard. Manually enabling trim on my root partition has stopped the wearout of my precious new disk from 365 used reserved blocks (out of 6176 total) within three months down to 0 additional used reserved blocks within three additional months (data from SMART attributes). Because I want to minimize the wearout of my SSD I now would like to know whether my swap partition (which is encrypted using ecryptfs-setup-swap) also makes use of the trim/discard option. I tried sudo swapon -d -v /dev/mapper/cryptswap1 but did not receive particular information ("-v") about whether trim/discard ("-d") was applied. If unsupported, i would expect a message. Then I tried sudo dd if=/dev/sda6 count=1 BS=1M | xxd | less directly after booting and when no swapspace was used but I saw not only zeroes. I assume, when looking at freshly trimmed regions, the disk would send zeroes instead of reading random sectors (and according to some forums, (unencrypted) swap space is trimmed once upon boot). Long story short: Are there any ideas on how to test if trim is effectively used for my encrypted swap? And if not, any ideas on how to - at least manually, for once - trim the whole swap space? I wouldn't want to tinker with the partition itself, because I dont know if it needs to be reinitialized as (encrypted) swap - I dont want to be left with an unbootable system :)

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  • Is there a way to do a sector level copy/clone from one hard drive to another?

    - by irrational John
    Without going into distracting details, I'm attempting to duplicate the contents of the 500GB drive in my MacBook to another 500GB drive. But this is turning out to be an unexpected hassle because the drive contains both the OS X partition and an NTFS partition with Win 7 via Apple's Boot Camp. With the exception of Clonezilla, the tools I have looked at so far all have some limitation. The Mac tools don't want to deal with the NTFS partition. The Windows tools are totally clueless about either the HFS+ partition and/or the hybrid MBR/GPT Boot Camp partitioning. Clonezilla looked like it would do what I want but apparently I can't figure out how to use it. After doing what I thought was a sector to sector copy I found that only the NTFS partition had been migrated. The others were apparently empty. (And frankly, I'm not positive Clonezilla migrated the partition table correctly either). Note: It takes over 2 hours using SATA to read/write all sectors with these drives. So I'm not up for using trial & error to narrow in on the right combination of Clonezilla options to use. I'm beginning to think that maybe the answer is to boot Linux (probably Ubuntu) and then use some ancient BSD command. Trouble is I don't know what command (or parameters to use) in order to do a sector level copy from one drive to another. As far as I know the drives have the same number of sectors so this should be trivial. Sigh.

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  • Howto align partitions in Linux + NetApp

    - by santisaez
    NetApp support has suggested us aligning partitions to improve performance, in short: starting sector must be divisible by 8. How can I move the start point in a misaligned partition -in production, with ext3- under Linux? A screenshot with a misaligned (start=63s) and aligned (start=64s) partition is available at: http://filesocial.com/lkwvvn2 (If anyone is interested in this topic, NetApp has a good document explaining performance issues in misaligned partitions, search for "tr-3747": Best Practices for File System Alignment in Virtual Environments.) I have tried using parted "resize + move" commands, but when moving start point a get this error: (parted) resize Partition number? 1 Start? [64s]? End? [419425019s]? 419425018 (parted) move Partition number? 1 Start? 65 End? [419425019s]? 419425019 Error: Can't move a partition onto itself. Try using resize, perhaps? Using fdisk 'b' command in expert mode ('move beginning of data in a partition') works, but it doesn't move the file system.. thanks!!

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  • After installing Windows 7, I can no longer get to Debian 6 (dual boot)

    - by Jeremy
    I had Debian 6 on my machine (Dell Vostro 260) and used GParted to shrink the partition. I then tried installing Windows 7 on that partition. After Windows 7 installed, I could not choose which OS to run. It would just boot into Windows. I ran GParted again, and saw that Windows created another partition, labeled "System Reserved". That partition had the boot flag set. I tried moving the boot flag to other partitions, including my Debian partition and one with a file system "linux-swap". No option would actually load an OS or anything except for the Windows partition, which is not what I want. Is it worth it to try to fix this installation, or should I start over. I have all my data backed up, so I can easily install from scratch if I need to. If I do start from scratch, which OS should I install first? And then, how do I set up the partitions to install the other OS? Let me know if you have any questions. Thanks for your help.

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  • Recovering a broken NTFS filesystem?

    - by OverTheRainbow
    A much-needed Windows Update broke a Vista laptop that was running fine until then: After booting up, Windows displays "Please wait..." but it never goes anywhere. I waited for a couple of hours, there is a bit of disk activity, but it didn't work out in the end. I booted with the Vista DVD, chose "Repair your computer" which said that there was nothing wrong :-/ Next, I booted it up with a Linux USB keydrive, and ran Gparted 0.8.1 (which includes ntfsresize v2011.4.12AR.4 libntfs-3g) which displays a bunch of warnings for the NTFS partition where the Vista system is located such as: ntfs_mst_post_read_fixup: magic: 0x00000000 size: 1024 usa_ofs: 0 usa_count: 65535: Invalid argument Record 16 has no FILE magic (0x0) Next, I ran ntfsfix /dev/sda2, which said: Mounting volume... OK Processing of $MFT and $MFTMirr completed successfully. NTFS volume version is 3.1. NTFS partition /dev/sda2 was processed successfully. Next, I rebooted Vista, which did a CHKDSK, before rebooting. But I'm still getting nowhere with "Please wait..." Before I copy the user's data to another host and reinstall Vista from a DVD, does someone know what I could try? Thank you. Edit: In case someone else has the same issue... After the BIOS, hit F8 and choose "Repair your computer", followed by "Toshiba HDD Recovery". In addition to a 1,5GB partition labelled "WinRE", the hard disk contains a second partition labeled "Data" from which the application will fetch a system image and reinstall it in the "Vista" partition. Make sure you copy your data out of the system partition before doing this.

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  • Shrinking windows and recovery partitions on the samsung new series 9

    - by bobbaluba
    I just bought a samsung NP900X3C, and as I was going to install linux, I noticed the windows partitions and recovery partitions occupied a major portion of the disk. The disk is a 128 GB SSD, and I want to keep the windows partition in order to play some games once in a while, but the windows disk is already 45GB full (with no installed programs) and the recovery partition is 20GB. That leaves under 60 GB for linux, which is not optimal, since that is what I'm going to be using most of the time, and there would be no room for games on the windows partition. There are also two small partitions that I don't know what are doing, one 100mb at the start of the disk that I'm guessing is some kind of boot partition, and one 5GB, that is described as an OS/2 hidden C: drive What I'm wondering is: can i delete the recovery partition? What about the mystical 5gb partition? Here is what fdisk reports: ubuntu@ubuntu:~$ sudo fdisk -l Disk /dev/sda: 128.0 GB, 128035676160 bytes 255 heads, 63 sectors/track, 15566 cylinders, total 250069680 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: 0x83953ffc Device Boot Start End Blocks Id System /dev/sda1 * 2048 206847 102400 7 HPFS/NTFS/exFAT /dev/sda2 206848 198273023 99033088 7 HPFS/NTFS/exFAT /dev/sda3 198273024 207276031 4501504 84 OS/2 hidden C: drive /dev/sda4 207276032 250068991 21396480 27 Hidden NTFS WinRE

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