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  • unavailable disc after installing windows7

    - by niao
    Greetings, I know this is a site about programming but I don't know where to put this question. The problem is that after installing Windows7 one partition is unavailable. The error is: "E:\ is unavailable" Access denied. There is no information about available space etc. How can I resolve this problem because there are some very important files on that partition.

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  • gunzip: invalid compressed data--format violated

    - by Arunjith
    Problem definition: I transferred a tar.gz file from a Linux machine to a Windows partition.The Windows partition has mounted with the Linux server as cifs. OS : Red Hat Enterprise Linux Server release 5 Symptom: After the copy process is successful, doing an integrity check with gunzip -t and the process get the following error: gunzip -t Backup-28--Jun--2011--Tuesday.tar.gz gunzip: Backup-28--Jun--2011--Tuesday.tar.gz: invalid compressed data--format violated And further tried to untar (tar -xvzf) and the process as well is failed.

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  • Easy way to deploy the recovery parition in Windows 7?

    - by Jesse K
    We're using ImageX to deploy Windows 7 Professional. We've gotten the Windows partition to work, but the recovery partition (100-200MB at the front of the drive in a standard install) isn't as simple. Here's a Technet guide I found: http://technet.microsoft.com/en-us/library/dd744280%28WS.10%29.aspx That looks like it could work, but would take alot of time if we need to do that for every single machine we deploy. Is there a faster/automated way?

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  • I've two swap partitions, how can i delete one?

    - by Gp2mv3
    I've installed Ubuntu on my computer but I did a mistake during the installation and it created two swap's. In fact I had tree partitions (system, home, swap) but the installator crashed so I restarted the installation and it has installed everything in the system partition. So now I separated the home in the appropriate partition but I've two swap partitions. How can I delete one ? If I delete one, how will it go ?

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  • Installing/dual-booting Fedora 17 on existing Windows 7 HDD

    - by Moose4
    I have a 64-bit Windows 7 install as the only partition on a 1 TB HDD, with about 350 GB free. I would like to install Fedora 17 as a dual-boot option on this system and give it about 100 GB to play with. If in the Fedora install utility I choose to shrink the W7 partition by 100 GB to give it space, will that cause me to lose my existing W7 data? And how do I go about setting up dual-boot (with Windows 7 as the default)?

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  • How to image an fresh hard drive

    - by JoelHess
    I'm looking for something that can drop an image of Windows onto an uninitialized Hard Drive. I have new units that haven't even had the partition table created on them, and I'm trying to come up with a one (or two) step process for getting a disk image onto them. I've tried clonezilla, but that doesn't seem to like the lack of partition table, and as far as I can tell, Acronis doesn't work either.

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  • Get extra hard drive space from windows 7

    - by abhinole
    I am using Ubuntu 12.04 and Windows 7 (dual-boot) on my laptop.For some reasons I want some more space in my Ubuntu partition.I have installed gParted in Ubuntu.Now is it recommended to get this required extra space from Windows 7 drive (*where my linux is installed *) directly using gParted? Will it cause damage to my boot loader or my data on the partition from where I wish to grab some space? Any help will be appreciated. Thanks in advance.

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  • Does Windows 7 deleted files generated during hibernation?

    - by Koffeehaus
    Somebody was using my Windows 7 and she hibernated it instead of shutting down. Later, I booted up Ubuntu and moved about 2GB worth of files from the Ubuntu partition to the Windows partition. After booting up Windows (from hibernation), I couldn't find any of the files. Then I restarted the PC, and the files showed for a second or two and then disappeared. Did Windows delete all the files I put on it while it was hibernating?

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  • Windows 7 startup MUCH slower after reinstall (on an SSD)

    - by user326639
    I installed Windows 7 Prof 64 bits OEM (Spanish) on my new machine. As I wanted my Windows to be in English, the web shop where I bought the DVD recomended me to download an ISO file with the same Windows version (but in English), burn it on a DVD and install it. And that I should be able to use my registration code. Location ISO: http://msft-dnl.digitalrivercontent.net/msvista/pub/X15-65805/X15-65805.iso I've done this and everything works (I have not activated my Windows yet but I expect no problem there). Just one thing: its startup is MUCH slower now! Have a look at my PC specs (bottom). On my first install (Spanish), it was like: - motherboard splash screen -- shows for a second or two - list of found drives -- a few seconds - the text "Windows starting" -- about a second before the dots appear - four collored dots form the Windows logo -- a few seconds after the logo is fully formed it moves on to the login screen. On my second install (English): - motherboard splash screen -- shows for 15 seconds - list of found drives -- a few seconds - the text "Windows starting" -- shows for 40 seconds before the dots appear - four collored dots form the Windows logo -- now it moves on to the login screen about equally fast as before. Ones it's up and running it seems to be as responsive as before, although it's possible that I'm not noticing the difference. I did the first install on the virgin SSD drive straight from the box. The second time I let the Windows installation program format the drive first to get rid of the old installation. I noticed that there were two partitions on my SSD: partition 1, 100 Mb, "reserved for the system" and partition 2, 111.7 Gb. I only formated the big partition, and I left the system partition untouched. Between the two installs, I didn't open the computer so everything is connected to the same port. I did not change anything in BIOS. Has Windows not recognized my SSD as an SSD but as a normal HDD. I suspect that Windows has not done the neccesary automatic configuration settings that it should do for SSD's (but that's just a hunch). How do I get my SSD back into its virgin state, as if it came right from the box, so I can go for a 3rd attempt to install windows. Should I use DISKPART? Other ideas are welcome. Specifications: mobo: Gigabyte GA-Z68X-UD3H-B3 CPU: i7-2600K SSD: OCZ Agility3 2,5" HDD: Samsung Spinpoint F4 mem: Kingston HyperX DIMM 8 Gb DDR3-1600

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  • Python: Slicing a list into n nearly-equal-length partitions

    - by Drew
    I'm looking for a fast, clean, pythonic way to divide a list into exactly n nearly-equal partitions. partition([1,2,3,4,5],5)->[[1],[2],[3],[4],[5]] partition([1,2,3,4,5],2)->[[1,2],[3,4,5]] (or [[1,2,3],[4,5]]) partition([1,2,3,4,5],3)->[[1,2],[3,4],[5]] (there are other ways to slice this one too) There are several answers in here http://stackoverflow.com/questions/1335392/iteration-over-list-slices that run very close to what I want, except they are focused on the size of the list, and I care about the number of the lists (some of them also pad with None). These are trivially converted, obviously, but I'm looking for a best practice. Similarly, people have pointed out great solutions here http://stackoverflow.com/questions/312443/how-do-you-split-a-list-into-evenly-sized-chunks-in-python for a very similar problem, but I'm more interested in the number of partitions than the specific size, as long as it's within 1. Again, this is trivially convertible, but I'm looking for a best practice.

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  • mysql partitioning

    - by Yang
    just want to verify that database partition is implemented only at the database level, when we query a partitioned table, we still do our normal query, nothing special with our queries, the optimization is performed automatically when parsing the query, is that correct? e.g. we have a table called 'address' with a column called 'country_code' and 'city'. so if i want to get all the addresses in New York, US, normally i wound do something like this: select * from address where country_code = 'US' and city = 'New York' if now the table is partitioned by 'country_code', and i know that now the query will only be executed on the partition which contains country_code = US. My question is do I need to explicitly specify the partition to query in my sql statement? or i still use the previous statement and the db server will optimize it automatically? Thanks in advance!

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  • iTextSharp renders image with poor quality in PDF

    - by Sebastian
    Hello, I'm using iTextSharp to print a PDF document. Everything goes ok until I have to print the company logo in it. First I noticed that the logo had poor quality, but after testing with several images, I realize that was the iTextSharp rendering it poorly. The test I did to say this was to print the PDF using my code and then edit the document with Acrobat 8.0 and I drew an image. Then printed the two documents and saw the noticeable difference. My question is that if anyone know if this can be due to a scaling problem where I'm failing to tell iTextSharp how it must render the image or is an iTextSharp limitation. The code to render the image is the following: Dim para As Paragraph = New Paragraph para.Alignment = Image.RIGHT_ALIGN para.Add(text) Dim imageFile As String = String.Format("{0}{1}", GetAppSetting("UploadDirectory"), myCompany.LogoUrl) Dim thisImage As Image = Image.GetInstance(imageFile) thisImage.Alignment = Image.LEFT_ALIGN para.Add(thisImage) The printed images are the following: Image printed directly with iTextSharp Image edited and printed with Acrobat 8 Please let me know if I need to provide more info. Thanks

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  • Installed Ubuntu 12.04.01 with Windows XP but lost access to Windows XP

    - by Bob D
    The First time I tried to install Ubuntu the installer installed it on my D drive. This resulted in only booting to Windows XP with no access to Ubuntu. I had to download a disk partitioning program to undo all of this. A tip from the Internet said to create a partition on the C drive for Ubuntu, so I did along with a Swap Partition. I did this manually because the installer on the CD would not do so and would not let me do so from within the installer program. With the fresh partitions created for Ubuntu I let the installer do its thing. The computer rebooted and came up in Ubuntu. I then installed WINE and all was well. Then I shut the computer down for the night. The next day I turned on the computer and it booted directly into Ubuntu. I can see the Windows partition and all the files but it will not allow me to switch to the Windows XP OS. Does not even give me a choice to do so. I have reinstalled Ubuntu several times and each time is the same, I cannot access Windows XP anymore. Right now I am in a fresh install with only whatever the installer installed. How do I fix this?! I have tried the hold the shift key to see if something called GRUB shows up, but no. I tried shifting the order of boot in GRUB but that did not work either. I tried using EasyBCD but that will not run. One symptom I do not understand, my monitor will post a graphic when the computer reboots that the cable is disconnected, this is normal. Then when the computer gets to the actual boot process it will display the splash screens etc and it did this for Windows XP as well. But now something new has popped up, while booting Ubuntu after where it probably should be showing me a menu to pick what OS I want to boot, the monitor posts "Input Unsupported" until Ubuntu loads. I have never seen it post this before, maybe a clue to someone.

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  • Where to install Ubuntu bootloader on a Mac?

    - by ylhtravis
    I'm trying to install Ubuntu on a Mac, but I can't get it to boot with default mac bootloader or rEFI. It shows up as "Windows" On the mac bootloader and shows up as "Linux" on rEFI. I'm not sure where should I install the bootloader to during installation. Here's my partition table: sda : Hard drive manufacturer name sda0: Mac partition sda1: ... sda4: SWAP sda5: Ubuntu And I have installed the bootloader on sda5. Should I install it on sda instead? I'm afraid this will replace the default Mac OS X bootloader and I won't be able to go into the Mac side again. Any help will be greatly appreciated. Thank you.

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  • Apache DS fails to list users

    - by CuriousMind
    Apache ds fails to list the users INFO | jvm 1 | 2012/03/28 15:54:04 | java.lang.Error: ERR_546 CRITICAL: page header magic for block 59 not OK 0 INFO | jvm 1 | 2012/03/28 15:54:04 | at jdbm.recman.PageHeader.(PageHeader.java:95) INFO | jvm 1 | 2012/03/28 15:54:04 | at jdbm.recman.PageHeader.getView(PageHeader.java:124) INFO | jvm 1 | 2012/03/28 15:54:04 | at jdbm.recman.PageManager.getNext(PageManager.java:234) INFO | jvm 1 | 2012/03/28 15:54:04 | at jdbm.recman.PageCursor.next(PageCursor.java:104) INFO | jvm 1 | 2012/03/28 15:54:04 | at jdbm.recman.PhysicalRowIdManager.fetch(PhysicalRowIdManager.java:158) INFO | jvm 1 | 2012/03/28 15:54:04 | at jdbm.recman.BaseRecordManager.fetch(BaseRecordManager.java:324) INFO | jvm 1 | 2012/03/28 15:54:04 | at jdbm.recman.CacheRecordManager.fetch(CacheRecordManager.java:262) INFO | jvm 1 | 2012/03/28 15:54:04 | at jdbm.btree.BPage.loadBPage(BPage.java:899) INFO | jvm 1 | 2012/03/28 15:54:04 | at jdbm.btree.BPage.childBPage(BPage.java:890) INFO | jvm 1 | 2012/03/28 15:54:04 | at jdbm.btree.BPage.find(BPage.java:284) INFO | jvm 1 | 2012/03/28 15:54:04 | at jdbm.btree.BPage.find(BPage.java:285) INFO | jvm 1 | 2012/03/28 15:54:04 | at jdbm.btree.BTree.find(BTree.java:408) INFO | jvm 1 | 2012/03/28 15:54:04 | at org.apache.directory.server.core.partition.impl.btree.jdbm.JdbmTable.get(JdbmTable.java:395) INFO | jvm 1 | 2012/03/28 15:54:04 | at org.apache.directory.server.core.partition.impl.btree.jdbm.JdbmMasterTable.get(JdbmMasterTable.java:155) INFO | jvm 1 | 2012/03/28 15:54:04 | at org.apache.directory.server.core.partition.impl.btree.jdbm.JdbmStore.lookup(JdbmStore.java:1332) INFO | jvm 1 | 2012/03/28 15:54:04 | at org.apache.directory.server.core.partition.impl.btree.jdbm.JdbmStore.lookup(JdbmStore.java:70) INFO | jvm 1 | 2012/03/28 15:54:04 | at org.apache.directory.server.xdbm.search.impl.EqualityEvaluator.evaluate(EqualityEvaluator.java:126) INFO | jvm 1 | 2012/03/28 15:54:04 | at org.apache.directory.server.xdbm.search.impl.AndCursor.matches(AndCursor.java:234) INFO | jvm 1 | 2012/03/28 15:54:04 | at org.apache.directory.server.xdbm.search.impl.AndCursor.next(AndCursor.java:143) INFO | jvm 1 | 2012/03/28 15:54:04 | at org.apache.directory.server.xdbm.search.impl.AndCursor.next(AndCursor.java:139) INFO | jvm 1 | 2012/03/28 15:54:04 | at org.apache.directory.server.core.partition.impl.btree.ServerEntryCursorAdaptor.next(ServerEntryCursorAdaptor.java:178) INFO | jvm 1 | 2012/03/28 15:54:04 | at org.apache.directory.server.core.filtering.BaseEntryFilteringCursor.next(BaseEntryFilteringCursor.java:499) INFO | jvm 1 | 2012/03/28 15:54:04 | at org.apache.directory.server.ldap.handlers.SearchHandler.readResults(SearchHandler.java:314) INFO | jvm 1 | 2012/03/28 15:54:04 | at org.apache.directory.server.ldap.handlers.SearchHandler.doSimpleSearch(SearchHandler.java:749) INFO | jvm 1 | 2012/03/28 15:54:04 | at org.apache.directory.server.ldap.handlers.SearchHandler.handleIgnoringReferrals(SearchHandler.java:978) INFO | jvm 1 | 2012/03/28 15:54:04 | at org.apache.directory.server.ldap.handlers.SearchHandler.handleIgnoringReferrals(SearchHandler.java:78) INFO | jvm 1 | 2012/03/28 15:54:04 | at org.apache.directory.server.ldap.handlers.ReferralAwareRequestHandler.handle(ReferralAwareRequestHandler.java:83) INFO | jvm 1 | 2012/03/28 15:54:04 | at org.apache.directory.server.ldap.handlers.ReferralAwareRequestHandler.handle(ReferralAwareRequestHandler.java:57) INFO | jvm 1 | 2012/03/28 15:54:04 | at org.apache.directory.server.ldap.handlers.LdapRequestHandler.handleMessage(LdapRequestHandler.java:208) INFO | jvm 1 | 2012/03/28 15:54:04 | at org.apache.directory.server.ldap.handlers.LdapRequestHandler.handleMessage(LdapRequestHandler.java:58) INFO | jvm 1 | 2012/03/28 15:54:04 | at org.apache.mina.handler.demux.DemuxingIoHandler.messageReceived(DemuxingIoHandler.java:232) INFO | jvm 1 | 2012/03/28 15:54:04 | at org.apache.directory.server.ldap.LdapProtocolHandler.messageReceived(LdapProtocolHandler.java:193) INFO | jvm 1 | 2012/03/28 15:54:04 | at org.apache.mina.core.filterchain.DefaultIoFilterChain$TailFilter.messageReceived(DefaultIoFilterChain.java:713) INFO | jvm 1 | 2012/03/28 15:54:04 | at org.apache.mina.core.filterchain.DefaultIoFilterChain.callNextMessageReceived(DefaultIoFilterChain.java:434) INFO | jvm 1 | 2012/03/28 15:54:04 | at org.apache.mina.core.filterchain.DefaultIoFilterChain.access$1200(DefaultIoFilterChain.java:46) INFO | jvm 1 | 2012/03/28 15:54:04 | at org.apache.mina.core.filterchain.DefaultIoFilterChain$EntryImpl$1.messageReceived(DefaultIoFilterChain.java:793) INFO | jvm 1 | 2012/03/28 15:54:04 | at org.apache.mina.core.filterchain.IoFilterEvent.fire(IoFilterEvent.java:71) INFO | jvm 1 | 2012/03/28 15:54:04 | at org.apache.mina.core.session.IoEvent.run(IoEvent.java:63) INFO | jvm 1 | 2012/03/28 15:54:04 | at org.apache.mina.filter.executor.UnorderedThreadPoolExecutor$Worker.runTask(UnorderedThreadPoolExecutor.java:480) INFO | jvm 1 | 2012/03/28 15:54:04 | at org.apache.mina.filter.executor.UnorderedThreadPoolExecutor$Worker.run(UnorderedThreadPoolExecutor.java:434) INFO | jvm 1 | 2012/03/28 15:54:04 | at java.lang.Thread.run(Thread.java:619) INFO | jvm 1 | 2012/03/28 15:54:04 | [15:54:04] WARN [org.apache.directory.server.ldap.LdapProtocolHandler] - Null LdapSession given to cleanUpSession. INFO | jvm 1 | 2012/03/28 15:55:20 | [15:55:20] WARN [org.apache.directory.server.ldap.LdapProtocolHandler] - Unexpected exception forcing session to close: sending disconnect notice to client.

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  • Draw underlined / strikethrough text ( MULTILINE STRING ) ?

    - by Madhup
    Hi, I have to draw underlined-multiline text with all types of text alignment. I have searched on forums and got some results like: http://davidjhinson.wordpress.com/2009/11/26/underline-text-on-the-iphone/ http://forums.macrumors.com/showthread.php?t=561572 But all draw text for single line only. while i have multi-line text. The situation even become worse when the text alignment is centered. I searched and found that in iphone-sdk-3.2 there are some core-text attributes for underlining a text but no idea how to use that. Besides if I use these my problem would not be solved fully. As I have to draw strikethrough text also. Anybody having idea about this please help.

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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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  • 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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  • Is it possible to share a C struct in shared memory between apps compiled with different compilers?

    - by Joseph Garvin
    I realize that in general the C and C++ standards gives compiler writers a lot of latitude. But in particular it guarantees that POD types like C struct members have to be laid out in memory the same order that they're listed in the structs definition, and most compilers provide extensions letting you fix the alignment of members. So if you had a header that defined a struct and manually specified the alignment of its members, then compiled two apps with different compilers using the header, shouldn't one app be able to write an instance of the struct into shared memory and the other app be able to read it without errors? I am assuming though that the size of the types contained is consistent across two compilers on the same architecture (it has to be the same platform already since we're talking about shared memory). I realize that this is not always true for some types (e.g. long vs. long long in GCC and MSVC 64-bit) but nowadays there are uint16_t, uint32_t, etc. types, and float and double are specified by IEEE standards.

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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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