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  • Ubuntu 13.10 Installer freezes on partition recognition on Alienware M17x

    - by Mutewinter
    I'm trying to install Ubuntu 13.10 on an Alienware M17x, after bypassing the graphical problems with "nomodeset" (annoying!), I'm facing a rather different problem: I click on "Install Ubuntu", the guided installation progress reach the point where the partition, mount point etc. needs to be selected and...nothing. In the partition selection menu it sees only dev/sda, but in the window where the actual way the disk is partitioned should appear nothing shows, it's blank. I've tried to click on "change..." to try to force it to read something, but the installer simply quits. The button "change partition table" etcetera are greyed out (well, obviously, since no partition table has been read). What's that? The Alienware has Windows 7 and legacy BIOS (so no UEFI here). Anyone has an idea? Thanks for your time and help!

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  • Librated error when creating partition table

    - by Marko
    I bought a Dell Inspiron 5521 laptop a few days ago that came with Ubuntu preinstalled. I haven't used Ubuntu yet, and I don't have any experience in using it. I wanted to install Windows 7 64-bit on my laptop alongside Ubuntu, and made two bootable USB drives with Gparted and Windows 7. There wasn't a suitable partition on my laptop in which I could install Windows 7. I've read the instructions for using Gparted to create or manage my hard drive. I inserted the USB, booted from BIOS, and followed the procedure in installing Gparted. Then I entered Gparted, and the following error occurred: Librated error when Creating partition table. It asked me to click on either OK or Cancel. Either way I had my hard disk shown to me in the user window, in partitions that were made by the manufacturer: Partition File sys Label Size Flags /dev/sda1 fat32 dellutility 300.00 Mib diag /dev/sda2 fat32 os 3.00 Gib lba /dev/sda3 ext4 912.46 Gib boot /dev/sda4 extended 15.75 Gib (had a subpart) /dev/sda5 linux-swap 15.75 Gib ...and a option to switch to dev/sdb that's unused and of capacity 3Gib. I've used the biggest partition 912.46 Gib, and tried to reduce its size, and clicked OK. Then when I tried to make a new partition, it said it can't make any more partitions, no more than a maximum of 5. I would like to keep Ubuntu and slowly learn, but I also need to use programs that work in Windows. Thank you for taking the time to answer my question.

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  • update-grub is setting a wrong linux root partition

    - by adrian m
    Initially, Ubuntu was installed on another partition (sda5 or sda4). At some point I did manually move the root partition to sda2. Now, the problem is that whenever a new kernel is installed, the automatic regeneration of the menu.lst is using the OLD Linux partition. So I have to manually change in menu.lst the lines : root (hd0,5) into root (hd0,1) How can I configure update-grub to automatically generate menu.lst with the current Linux root partition? I assume that the (hd0,5) was written somewhere at installation, but I couldn't find it.

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  • How to I make my bootcamp partition bootable again?

    - by KJFMusic
    I'm having a similar problem as everyone else in this posting. I have 5 partitions. 3 of which I created for my Mac OS Lion installation, Windows 7 installation and a 3rd for storage. Everything was running fine for quite sometime until recently. My Windows 7 installation has suddenly stopped booting. Instead of a start up screen I get: Windows failed to start. A recent hardware or software change might be the cause. File: \BOOT\BCD Status: 0xc000000d Info: An error occurred while attempting to read the boot configuration data Mac OS Lion starts up fine. I'm unable to mount my "Bootcamp" partition nor the "Storage" partition. On top of that "Storage" has been renamed to "disk0s5". When I installed Windows 7 it didn't recognize the "Storage" partition that was created in Lion so it merged what it thought was free diskspace (I'm assuming the same space that Mac OS recognized as Storage) to the Root Drive of Windows 7 (Bootcamp). Are you able to assist?

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  • How to resize(increase) a root+home cloned partition from a smaller hdd?

    - by saulo
    Here is my story: I had a hd failing so I cloned it with dd to larger 500gb hd. It seemed to have worked well, but I have all this unallocated free space at the end of the disk (230gb). I tried to used gparted from a liveUSB to allocate all this space to my root+home partition (other than that I only have a small extended partition with the swap). It won't let me do this. I can only create another partition, or allocate the free space to the extend partition. I can however reduce the size of the root+home partition since I wasn't using all the space Is there a way to put this free space to my root+home partition after all? Or do I have to go with another partition. If so I'd like to separate my home from the root, reduce the root partition and create a ext4 logical home partition. What would be the best safest way to do this? Thanks so much in advance, aloha

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  • Heaps of Trouble?

    - by Paul White NZ
    If you’re not already a regular reader of Brad Schulz’s blog, you’re missing out on some great material.  In his latest entry, he is tasked with optimizing a query run against tables that have no indexes at all.  The problem is, predictably, that performance is not very good.  The catch is that we are not allowed to create any indexes (or even new statistics) as part of our optimization efforts. In this post, I’m going to look at the problem from a slightly different angle, and present an alternative solution to the one Brad found.  Inevitably, there’s going to be some overlap between our entries, and while you don’t necessarily need to read Brad’s post before this one, I do strongly recommend that you read it at some stage; he covers some important points that I won’t cover again here. The Example We’ll use data from the AdventureWorks database, copied to temporary unindexed tables.  A script to create these structures is shown below: CREATE TABLE #Custs ( CustomerID INTEGER NOT NULL, TerritoryID INTEGER NULL, CustomerType NCHAR(1) COLLATE SQL_Latin1_General_CP1_CI_AI NOT NULL, ); GO CREATE TABLE #Prods ( ProductMainID INTEGER NOT NULL, ProductSubID INTEGER NOT NULL, ProductSubSubID INTEGER NOT NULL, Name NVARCHAR(50) COLLATE SQL_Latin1_General_CP1_CI_AI NOT NULL, ); GO CREATE TABLE #OrdHeader ( SalesOrderID INTEGER NOT NULL, OrderDate DATETIME NOT NULL, SalesOrderNumber NVARCHAR(25) COLLATE SQL_Latin1_General_CP1_CI_AI NOT NULL, CustomerID INTEGER NOT NULL, ); GO CREATE TABLE #OrdDetail ( SalesOrderID INTEGER NOT NULL, OrderQty SMALLINT NOT NULL, LineTotal NUMERIC(38,6) NOT NULL, ProductMainID INTEGER NOT NULL, ProductSubID INTEGER NOT NULL, ProductSubSubID INTEGER NOT NULL, ); GO INSERT #Custs ( CustomerID, TerritoryID, CustomerType ) SELECT C.CustomerID, C.TerritoryID, C.CustomerType FROM AdventureWorks.Sales.Customer C WITH (TABLOCK); GO INSERT #Prods ( ProductMainID, ProductSubID, ProductSubSubID, Name ) SELECT P.ProductID, P.ProductID, P.ProductID, P.Name FROM AdventureWorks.Production.Product P WITH (TABLOCK); GO INSERT #OrdHeader ( SalesOrderID, OrderDate, SalesOrderNumber, CustomerID ) SELECT H.SalesOrderID, H.OrderDate, H.SalesOrderNumber, H.CustomerID FROM AdventureWorks.Sales.SalesOrderHeader H WITH (TABLOCK); GO INSERT #OrdDetail ( SalesOrderID, OrderQty, LineTotal, ProductMainID, ProductSubID, ProductSubSubID ) SELECT D.SalesOrderID, D.OrderQty, D.LineTotal, D.ProductID, D.ProductID, D.ProductID FROM AdventureWorks.Sales.SalesOrderDetail D WITH (TABLOCK); The query itself is a simple join of the four tables: SELECT P.ProductMainID AS PID, P.Name, D.OrderQty, H.SalesOrderNumber, H.OrderDate, C.TerritoryID FROM #Prods P JOIN #OrdDetail D ON P.ProductMainID = D.ProductMainID AND P.ProductSubID = D.ProductSubID AND P.ProductSubSubID = D.ProductSubSubID JOIN #OrdHeader H ON D.SalesOrderID = H.SalesOrderID JOIN #Custs C ON H.CustomerID = C.CustomerID ORDER BY P.ProductMainID ASC OPTION (RECOMPILE, MAXDOP 1); Remember that these tables have no indexes at all, and only the single-column sampled statistics SQL Server automatically creates (assuming default settings).  The estimated query plan produced for the test query looks like this (click to enlarge): The Problem The problem here is one of cardinality estimation – the number of rows SQL Server expects to find at each step of the plan.  The lack of indexes and useful statistical information means that SQL Server does not have the information it needs to make a good estimate.  Every join in the plan shown above estimates that it will produce just a single row as output.  Brad covers the factors that lead to the low estimates in his post. In reality, the join between the #Prods and #OrdDetail tables will produce 121,317 rows.  It should not surprise you that this has rather dire consequences for the remainder of the query plan.  In particular, it makes a nonsense of the optimizer’s decision to use Nested Loops to join to the two remaining tables.  Instead of scanning the #OrdHeader and #Custs tables once (as it expected), it has to perform 121,317 full scans of each.  The query takes somewhere in the region of twenty minutes to run to completion on my development machine. A Solution At this point, you may be thinking the same thing I was: if we really are stuck with no indexes, the best we can do is to use hash joins everywhere. We can force the exclusive use of hash joins in several ways, the two most common being join and query hints.  A join hint means writing the query using the INNER HASH JOIN syntax; using a query hint involves adding OPTION (HASH JOIN) at the bottom of the query.  The difference is that using join hints also forces the order of the join, whereas the query hint gives the optimizer freedom to reorder the joins at its discretion. Adding the OPTION (HASH JOIN) hint results in this estimated plan: That produces the correct output in around seven seconds, which is quite an improvement!  As a purely practical matter, and given the rigid rules of the environment we find ourselves in, we might leave things there.  (We can improve the hashing solution a bit – I’ll come back to that later on). Faster Nested Loops It might surprise you to hear that we can beat the performance of the hash join solution shown above using nested loops joins exclusively, and without breaking the rules we have been set. The key to this part is to realize that a condition like (A = B) can be expressed as (A <= B) AND (A >= B).  Armed with this tremendous new insight, we can rewrite the join predicates like so: SELECT P.ProductMainID AS PID, P.Name, D.OrderQty, H.SalesOrderNumber, H.OrderDate, C.TerritoryID FROM #OrdDetail D JOIN #OrdHeader H ON D.SalesOrderID >= H.SalesOrderID AND D.SalesOrderID <= H.SalesOrderID JOIN #Custs C ON H.CustomerID >= C.CustomerID AND H.CustomerID <= C.CustomerID JOIN #Prods P ON P.ProductMainID >= D.ProductMainID AND P.ProductMainID <= D.ProductMainID AND P.ProductSubID = D.ProductSubID AND P.ProductSubSubID = D.ProductSubSubID ORDER BY D.ProductMainID OPTION (RECOMPILE, LOOP JOIN, MAXDOP 1, FORCE ORDER); I’ve also added LOOP JOIN and FORCE ORDER query hints to ensure that only nested loops joins are used, and that the tables are joined in the order they appear.  The new estimated execution plan is: This new query runs in under 2 seconds. Why Is It Faster? The main reason for the improvement is the appearance of the eager Index Spools, which are also known as index-on-the-fly spools.  If you read my Inside The Optimiser series you might be interested to know that the rule responsible is called JoinToIndexOnTheFly. An eager index spool consumes all rows from the table it sits above, and builds a index suitable for the join to seek on.  Taking the index spool above the #Custs table as an example, it reads all the CustomerID and TerritoryID values with a single scan of the table, and builds an index keyed on CustomerID.  The term ‘eager’ means that the spool consumes all of its input rows when it starts up.  The index is built in a work table in tempdb, has no associated statistics, and only exists until the query finishes executing. The result is that each unindexed table is only scanned once, and just for the columns necessary to build the temporary index.  From that point on, every execution of the inner side of the join is answered by a seek on the temporary index – not the base table. A second optimization is that the sort on ProductMainID (required by the ORDER BY clause) is performed early, on just the rows coming from the #OrdDetail table.  The optimizer has a good estimate for the number of rows it needs to sort at that stage – it is just the cardinality of the table itself.  The accuracy of the estimate there is important because it helps determine the memory grant given to the sort operation.  Nested loops join preserves the order of rows on its outer input, so sorting early is safe.  (Hash joins do not preserve order in this way, of course). The extra lazy spool on the #Prods branch is a further optimization that avoids executing the seek on the temporary index if the value being joined (the ‘outer reference’) hasn’t changed from the last row received on the outer input.  It takes advantage of the fact that rows are still sorted on ProductMainID, so if duplicates exist, they will arrive at the join operator one after the other. The optimizer is quite conservative about introducing index spools into a plan, because creating and dropping a temporary index is a relatively expensive operation.  It’s presence in a plan is often an indication that a useful index is missing. I want to stress that I rewrote the query in this way primarily as an educational exercise – I can’t imagine having to do something so horrible to a production system. Improving the Hash Join I promised I would return to the solution that uses hash joins.  You might be puzzled that SQL Server can create three new indexes (and perform all those nested loops iterations) faster than it can perform three hash joins.  The answer, again, is down to the poor information available to the optimizer.  Let’s look at the hash join plan again: Two of the hash joins have single-row estimates on their build inputs.  SQL Server fixes the amount of memory available for the hash table based on this cardinality estimate, so at run time the hash join very quickly runs out of memory. This results in the join spilling hash buckets to disk, and any rows from the probe input that hash to the spilled buckets also get written to disk.  The join process then continues, and may again run out of memory.  This is a recursive process, which may eventually result in SQL Server resorting to a bailout join algorithm, which is guaranteed to complete eventually, but may be very slow.  The data sizes in the example tables are not large enough to force a hash bailout, but it does result in multiple levels of hash recursion.  You can see this for yourself by tracing the Hash Warning event using the Profiler tool. The final sort in the plan also suffers from a similar problem: it receives very little memory and has to perform multiple sort passes, saving intermediate runs to disk (the Sort Warnings Profiler event can be used to confirm this).  Notice also that because hash joins don’t preserve sort order, the sort cannot be pushed down the plan toward the #OrdDetail table, as in the nested loops plan. Ok, so now we understand the problems, what can we do to fix it?  We can address the hash spilling by forcing a different order for the joins: SELECT P.ProductMainID AS PID, P.Name, D.OrderQty, H.SalesOrderNumber, H.OrderDate, C.TerritoryID FROM #Prods P JOIN #Custs C JOIN #OrdHeader H ON H.CustomerID = C.CustomerID JOIN #OrdDetail D ON D.SalesOrderID = H.SalesOrderID ON P.ProductMainID = D.ProductMainID AND P.ProductSubID = D.ProductSubID AND P.ProductSubSubID = D.ProductSubSubID ORDER BY D.ProductMainID OPTION (MAXDOP 1, HASH JOIN, FORCE ORDER); With this plan, each of the inputs to the hash joins has a good estimate, and no hash recursion occurs.  The final sort still suffers from the one-row estimate problem, and we get a single-pass sort warning as it writes rows to disk.  Even so, the query runs to completion in three or four seconds.  That’s around half the time of the previous hashing solution, but still not as fast as the nested loops trickery. Final Thoughts SQL Server’s optimizer makes cost-based decisions, so it is vital to provide it with accurate information.  We can’t really blame the performance problems highlighted here on anything other than the decision to use completely unindexed tables, and not to allow the creation of additional statistics. I should probably stress that the nested loops solution shown above is not one I would normally contemplate in the real world.  It’s there primarily for its educational and entertainment value.  I might perhaps use it to demonstrate to the sceptical that SQL Server itself is crying out for an index. Be sure to read Brad’s original post for more details.  My grateful thanks to him for granting permission to reuse some of his material. Paul White Email: [email protected] Twitter: @PaulWhiteNZ

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  • How to recover my invisible HD again?

    - by pattulus
    I made this several times now, but this time something bad happened. What I did: I installed Windows 7 at a 32GB partition on my slot 2 HD in my MacPro. Windows 7 made a 105MB partition… I knew this before, but what I didn’t know was that this partition is now on my slot 4 HD. My home folder, my private videos and some other stuff are on this 1TB drive. What I found out so far: I’m currently logged in as another admin since my OS partition as well as the two other HD's aren't harmed. Disk Utility: … only shows the 105MB NTSF partition on this 1TB volume. It isn’t showing my old 1TB partition/ex-HD named "storehouse". Only the partition tab is telling me that there now is a 1TB empty free unpartitioned space. Data Rescue II: … is showing the Volume as it used to be with it's old Name "storehouse". A quick scan and a thorough scan both were done in 1 second which leds me to the conclusion that there's isn’t something deleted at all (» hope!). Data Rescue doesn’t even mention the damn "system reserved" partition. Drive Genius: … also shows the old partition and doesn’t mention the new one. But looking at the info it tells me under "content": FDisk_partition_scheme (instead of Apple_partition_scheme). Well D'oh…. Tech Tools: … doesn’t show the volume, otherwise I'd might have been tempted to press rebuild/repair. What to do next?? I think the best approach is to buy another 1TB HD and let Disk Warrior Clone my old one to it… just to be on the safe side. But what is the best thing to do after this… ???

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  • Fixing damaged partition table

    - by dr4cul4
    This is continuation of Recover Extended Partition , but this time I have different problem related partition table it self. I managed to restore partition that I needed and backed up files that were crucial to me (at least those that I had space to store somewhere) OK now get to the problem. My partition table is corrupted, booting RIP Linux I can mount it in truecrypt (and other ones that recovered), but that's basically it. When I launch GParted I have unallocated drive. GParted Dev info: Device Information Model: ATA ST2000DL003-9VT1 Size: 1.82TiB Path: /dev/sda Partition table: unrecognized Heads: 255 Sectors/track: 63 Cylinders: 243201 Total Sectors: 3907029168 Sector size: 512 When I check information on unallocated space I get: File system: unallocated Size: 1.82TiB First sector: 0 Last sector: 3907029167 Total sectors: 3907029168 Warning: Can't have a partition outside the disk! Now the output of testdisc (Analyze): TestDisk 6.13, Data Recovery Utility, November 2011 Christophe GRENIER <[email protected]> http://www.cgsecurity.org Disk /dev/sda - 2000 GB / 1863 GiB - CHS 243201 255 63 Current partition structure: Partition Start End Size in sectors > 1 P Linux 13132 242 39 16353 233 8 51744768 2 E extended LBA 16807 223 1 243201 254 63 3637021626 No partition is bootable 5 L Linux 16807 223 57 20430 39 25 58191872 X extended 20430 70 1 243201 78 13 3578816632 Invalid NTFS or EXFAT boot 6 L HPFS - NTFS 20430 71 58 243201 78 13 3578816512 6 LNext Now fdisk: # fdisk -l /dev/sda Disk /dev/sda: 2000.4 GB, 2000398934016 bytes 255 heads, 63 sectors/track, 243201 cylinders, total 3907029168 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: 0x00039cd0 Device Boot Start End Blocks Id System /dev/sda1 210980864 262725631 25872384 83 Linux /dev/sda2 270018504 3907040129 1818510813 f W95 Ext'd (LBA) /dev/sda5 270018560 328210431 29095936 83 Linux /dev/sda6 328212480 3907028991 1789408256 7 HPFS/NTFS/exFAT Now I would like to fix that to arrange partitions correctly, but I have no idea which tool is capable of fixing that (tried, a few, some of them offered fixing, but it was to risky at the moment - still backing up data).

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  • Multi-partition USB stick

    - by nightcracker
    In my freelance job as "the dude that fixes your computer" I have an extremely handy tool, a bootable USB stick with Ubuntu LiveCD that allows me to recover and investigate in a known, working environment. Now, I want to reformat this USB stick and reinstall with Casper-RW persistance. I did this a few times before with a FAT-formatted USB stick. It was a horror. The USB drive corrupted constantly, by people accidently removing the USB stick, the computer not properly shutting down, ETC. Now what I want to create a multi-partition USB stick so I can put Ubuntu on a ext partition, but still be able to store some Windows stuff in it, by having a secondary FAT partition. However I read somewhere that Windows will only check the first partition on USB sticks, giving a problem with the first bootable linux partition. Is this possible on some way? EDIT Perhaps it wasn't clear what the problem is. The problem is that I read somewhere that Windows will only recognize the first partition on a USB stick. But I want two partitions, a ext partition and a FAT partition. No issues so far, but in order to be bootable the ext partition must be the first one!

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  • Can I delete EFI System Partition without harming other data on drive?

    - by Andy
    I have three external HDD's in a USB enclosure. After a recent upgrade to Win7, during which these three drives were actually installed inside the PC tower, two of the three drives now have a 200MB EFI partition, and the two drives do not show up as usable drives under either Win7 or Snow Leopard. One of the drives is empty; the other one, however, has a bunch of stuff on it that I want to save if possible. My question is, how can I get back to this data? Can I simply delete the EFI partition, and all will be well? Or do I have to do something trickier? Or am I just hosed?

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  • Partition and mount my secondary hard drive on CentOS 5.5 64bit?

    - by Andrew Fashion
    I am trying to prepare my second hard drive for user image uploads. Here is the current layout: # sudo parted /dev/sda print Model: ATA WDC WD2500KS-00M (scsi) Disk /dev/sda: 250GB Sector size (logical/physical): 512B/512B Partition Table: msdos Number Start End Size Type File system Flags 1 32.3kB 107MB 107MB primary ext3 boot 2 107MB 8595MB 8488MB primary linux-swap 3 8595MB 10.7GB 2147MB primary ext3 4 10.7GB 250GB 239GB extended 5 10.7GB 250GB 239GB logical ext3 Information: Don't forget to update /etc/fstab, if necessary. I am assuming #4 is my secondary drive? How do I partition and mount it so I can begin using it? And how do I add to fstab? I understand if it's to many questions in one, just help me with whatever you can I guess :) Thank you for any help!

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  • Free space not reclaimed after online resizing ext4 in Ubuntu 9.10

    - by TiansHUo
    My root partition was filling up, with only 500 mbs left, I wanted to resize my root partition from 20 Gb to 40Gb So I resized my partition by using these steps: Using Gparted to resize another partition to give space for the EXT4 Using fdisk, deleting the root partition (on /dev/sda2), and creating it again using the new size resize2fs /dev/sda2 Updating grub2 But now the problem is that although I can boot in my new partition and the new partition shows it is 40Gb, but the free size was still 500mb. So I booted from a LiveCD and checked with e2fsck -p /dev/sda2, it reported clean. So I added the -f flag (force check), still, the drive is full.

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  • Cannot resize OS X partition

    - by David Pearce
    I am trying to resize my existing Mac OS Extended partition on my Macbook to install Windows 7 (using steps similar to these), but when ever I go to apply the changes, I get this error: Partition failed Partition failed with the error: The partition cannot be resized. Try reducing the amount of change in the size of the partition. The total capacity of the hard drive in question is 260GB, with the entirety being taken up by the OS X boot partition. There is I am aiming to shrink that partition down to 60GB. How can I fix this problem? I have been reducing the amount of change by 10GB each attempt, but it still is not working. I assume the problem is that there is not a large amount of continuous space on the device. Is there some way to can do a manual defrag that would rectify this problem?

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  • Cannot resize OS X partition

    - by joshhunt
    I am trying to resize my existing Mac OS Extended partition on my Macbook to install Windows 7 (using steps similar to these), but when ever I go to apply the changes, I get this error: Partition failed Partition failed with the error: The partition cannot be resized. Try reducing the amount of change in the size of the partition. The total capacity of the hard drive in question is 260GB, with the entirety being taken up by the OS X boot partition. There is I am aiming to shrink that partition down to 60GB. How can I fix this problem? I have been reducing the amount of change by 10GB each attempt, but it still is not working. I assume the problem is that there is not a large amount of continuous space on the device. Is there some way to can do a manual defrag that would rectify this problem?

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  • Partitioning a bootable Flash drive

    - by mmc
    Is it possible to have a 2 partition Flash drive that looks like the following: A partition that is bootable to OS X (this will require a GUID partition table) A second partition formatted either FAT32 or NTFS that is readable on both OS X and various flavors of Windows I have set up a disk using Disk Utility on the Mac, and it boots fine with a second FAT32 partition... but Windows does not see it. Any flavor of Windows wants to format the entire drive. Has anyone done this, and if so, can you explain the steps you followed? EDIT: Making it bootable is no problem. I have that. I'm wondering how to make the second partition on a Flash drive visible to Windows. It's possible that the "second partition" is the problem, and I need Windows to be first, and HFS to be second. I'll try that tonight.

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  • HELP! Free space not reclaimed after online resizing ext4 in Ubuntu 9.10

    - by TiansHUo
    My root partition was filling up, with only 500 mbs left, I wanted to resize my root partition from 20 Gb to 40Gb So I resized my partition by using these steps: Using Gparted to resize another partition to give space for the EXT4 Using fdisk, deleting the root partition (on /dev/sda2), and creating it again using the new size resize2fs /dev/sda2 Updating grub2 But now the problem is that although I can boot in my new partition and the new partition shows it is 40Gb, but the free size was still 500mb. So I booted from a LiveCD and checked with e2fsck -p /dev/sda2, it reported clean. So I added the -f flag (force check), still, the drive is full.

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  • Can a virus on a windows 7 partition make its way into the OS X partition?

    - by hatorade
    I have a Windows 7 partition on my MBP that I installed with Boot Camp. I have reason to believe that there was a virus on my Windows 7 partition (did some scans, got some sketchy results from Avira). I decided to just wipe the entire partition using Boot Camp Restore to reformat the old partition and add it back to my OS X partition. I'm wondering however if in the time period I had the two partitions up a virus could have jumped from the Windows 7 partition onto the OS X partition, in which case I now need to worry about a virus on my OS X installation?

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  • Create new partition on ssd hard (remove hibernate)

    - by Aleks
    I bought Dell Vostro 3360 notebook with Windows 7 It has 128 GB SSD hard disk, it is already has 4 partitions: Dell partition, Recovery partition, OS partition, Hibernate partition. Here is screenshot (Russian language, but I made some marks): I want to split OS partition, because I need c:\ and d:\ . Reason is that I have a lot of difficulties with administration mode on c:. So I tried to split OS partition but I already have 4 partitions. Can I remove hibernate partition without consequences? I have Hibernate disabled in settings, but I have hiberfil.sys file on c:\ If I can remove it, how can I do this, I can't do this with standard GUI disk managment tool

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  • Combine OS partion with data partition on NAS4Free/FreeNAS

    - by Pak
    I recently built a NAS4Free (formerly FreeNAS) machine using a 256MB (yes, MB) USB drive for the OS. When I did the original install, I had the bright idea of making the OS partition just big enough for the OS and a then creating a second partition using the remainder of the drive to store stuff pertaining to the OS. I never really found a use for the data partition and I ended up running out of space on the OS partition, so now I'd like to combine the partitions into a single partition. Is this something that is possible to do while everything is up and running? If it comes down to it, I can take down the machine and do a fresh install of the OS using the entire space of the USB drive, but I'd like to use this as an opportunity to better familiarize myself with FreeBSD/UNIX type systems. If this is possible, will it interfere with the NAS4Free things? The data partition shows up in the web interface under the disks section. If I end up manually changing the partitions, I'd be concerned with NAS4Free getting confused by the missing partition.

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  • In SQL, a Join is actually an Intersection? And it is also a linkage or a "Sideway Union"?

    - by Jian Lin
    I always thought of a Join in SQL as some kind of linkage between two tables. For example, select e.name, d.name from employees e, departments d where employees.deptID = departments.deptID In this case, it is linking two tables, to show each employee with a department name instead of a department ID. And kind of like a "linkage" or "Union" sideway". But, after learning about inner join vs outer join, it shows that a Join (Inner join) is actually an intersection. For example, when one table has the ID 1, 2, 7, 8, while another table has the ID 7 and 8 only, the way we get the intersection is: select * from t1, t2 where t1.ID = t2.ID to get the two records of "7 and 8". So it is actually an intersection. So we have the "Intersection" of 2 tables. Compare this with the "Union" operation on 2 tables. Can a Join be thought of as an "Intersection"? But what about the "linking" or "sideway union" aspect of it?

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  • Issues with LVM partition size in Server 13.04

    - by Michael
    I am new to ubuntu and a little confused about how hard drive partitions and LVM works. I remember setting up Ubuntu server 13.04 and telling to to use 1TB of a 3TB server. Well I have maxed that out with blu-ray rips and want the rest of the drive for space. On log-in it says: System load: 2.24 Processes: 179 Usage of /: 88.7% of 912.89GB Users logged in: 0 Memory usage: 6% IP address for p5p1: 192.168.0.100 Swap usage: 0% => / is using 88.7% of 912.89GB lvdisplay outputs: --- Logical volume --- LV Path /dev/DeathStar-vg/root LV Name root VG Name DeathStar-vg LV Write Access read/write LV Creation host, time DeathStar, 2013-05-18 22:21:11 -0400 LV Status available # open 1 LV Size 2.70 TiB Current LE 707789 Segments 2 Allocation inherit Read ahead sectors auto - currently set to 256 Block device 252:0 --- Logical volume --- LV Path /dev/DeathStar-vg/swap_1 LV Name swap_1 VG Name DeathStar-vg LV Write Access read/write LV Creation host, time DeathStar, 2013-05-18 22:21:11 -0400 LV Status available # open 2 LV Size 3.75 GiB Current LE 959 Segments 1 Allocation inherit Read ahead sectors auto - currently set to 256 Block device 252:1 vgdisplay outputs: VG Name DeathStar-vg System ID Format lvm2 Metadata Areas 1 Metadata Sequence No 4 VG Access read/write VG Status resizable MAX LV 0 Cur LV 2 Open LV 2 Max PV 0 Cur PV 1 Act PV 1 VG Size 2.73 TiB PE Size 4.00 MiB Total PE 715335 Alloc PE / Size 708748 / 2.70 TiB Free PE / Size 6587 / 25.73 GiB df outputs: Filesystem 1K-blocks Used Available Use% Mounted on /dev/mapper/DeathStar--vg-root 957238932 848972636 59634696 94% / none 4 0 4 0% /sys/fs/cgroup udev 1864716 4 1864712 1% /dev tmpfs 374968 1060 373908 1% /run none 5120 4 5116 1% /run/lock none 1874824 148 1874676 1% /run/shm none 102400 24 102376 1% /run/user /dev/sda2 234153 56477 165184 26% /boot And fdisk /dev/sda -l outputs: Disk /dev/sda: 3000.6 GB, 3000592982016 bytes 255 heads, 63 sectors/track, 364801 cylinders, total 5860533168 sectors Units = sectors of 1 * 512 = 512 bytes Sector size (logical/physical): 512 bytes / 4096 bytes I/O size (minimum/optimal): 4096 bytes / 4096 bytes Disk identifier: 0x00000000 Device Boot Start End Blocks Id System /dev/sda1 1 4294967295 2147483647+ ee GPT Partition 1 does not start on physical sector boundary. I just don't know what to make of all this and am not sure how I can make it use all 2.73TBs. Thanks in advance for any help. EDIT-- Yes I did make changes to the LVM Config, but it didnt do anything. As requested, output of parted -l /dev/sda Model: ATA WDC WD30EFRX-68A (scsi) Disk /dev/sda: 3001GB Sector size (logical/physical): 512B/4096B Partition Table: gpt Number Start End Size File system Name Flags 1 1049kB 2097kB 1049kB bios_grub 2 2097kB 258MB 256MB ext2 3 258MB 3001GB 3000GB lvm Model: ATA WDC WD30EFRX-68A (scsi) Disk /dev/sdb: 3001GB Sector size (logical/physical): 512B/4096B Partition Table: msdos Number Start End Size Type File system Flags Model: Linux device-mapper (linear) (dm) Disk /dev/mapper/DeathStar--vg-swap_1: 4022MB Sector size (logical/physical): 512B/4096B Partition Table: loop Number Start End Size File system Flags 1 0.00B 4022MB 4022MB linux-swap(v1) Model: Linux device-mapper (linear) (dm) Disk /dev/mapper/DeathStar--vg-root: 2969GB Sector size (logical/physical): 512B/4096B Partition Table: loop Number Start End Size File system Flags 1 0.00B 2969GB 2969GB ext4

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  • What's the different between these 2 mysql queries? one using left join

    - by Lyon
    Hi, I see people using LEFT JOIN in their mysql queries to fetch data from two tables. But I normally do it without left join. Is there any differences besides the syntax, e.g. performance? Here's my normal query style: SELECT * FROM table1 as tbl1, table2 as tbl2 WHERE tbl1.id=tbl2.table_id as compared to SELECT * FROM table1 as tbl1 LEFT JOIN table2 as tbl2 on tbl1.id=tbl2.id Personally I prefer the first style...hmm..

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  • In SQL / MySQL, can a Left Outer Join be used to find out the duplicates when there is no Primary ID

    - by Jian Lin
    I would like to try using Outer Join to find out duplicates in a table: If a table has Primary Index ID, then the following outer join can find out the duplicate names: mysql> select * from gifts; +--------+------------+-----------------+---------------------+ | giftID | name | filename | effectiveTime | +--------+------------+-----------------+---------------------+ | 2 | teddy bear | bear.jpg | 2010-04-24 04:36:03 | | 3 | coffee | coffee123.jpg | 2010-04-24 05:10:43 | | 6 | beer | beer_glass.png | 2010-04-24 05:18:12 | | 10 | heart | heart_shape.jpg | 2010-04-24 05:11:29 | | 11 | ice tea | icetea.jpg | 2010-04-24 05:19:53 | | 12 | cash | cash.png | 2010-04-24 05:27:44 | | 13 | chocolate | choco.jpg | 2010-04-25 04:04:31 | | 14 | coffee | latte.jpg | 2010-04-27 05:49:52 | | 15 | coffee | espresso.jpg | 2010-04-27 06:03:03 | +--------+------------+-----------------+---------------------+ 9 rows in set (0.00 sec) mysql> select * from gifts g1 LEFT JOIN (select * from gifts group by name) g2 on g1.giftID = g2.giftID where g2.giftID IS NULL; +--------+--------+--------------+---------------------+--------+------+----------+---------------+ | giftID | name | filename | effectiveTime | giftID | name | filename | effectiveTime | +--------+--------+--------------+---------------------+--------+------+----------+---------------+ | 14 | coffee | latte.jpg | 2010-04-27 05:49:52 | NULL | NULL | NULL | NULL | | 15 | coffee | espresso.jpg | 2010-04-27 06:03:03 | NULL | NULL | NULL | NULL | +--------+--------+--------------+---------------------+--------+------+----------+---------------+ 2 rows in set (0.00 sec) But what if the table doesn't have a Primary Index ID, then can an outer join still be used to find out duplicates?

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  • Why would I do an inner join on a non-distinct field?

    - by froadie
    I just came across a query that does an inner join on a non-distinct field. I've never seen this before and I'm a little confused about this usage. Something like: SELECT distinct all, my, stuff FROM myTable INNER JOIN myOtherTable ON myTable.nonDistinctField = myOtherTable.nonDistinctField (WHERE some filters here...) I'm not quite sure what my question is or how to phrase it, or why exactly this confuses me, but I was wondering if anyone could explain why someone would need to do an inner join on a non-distinct field and then select only distinct values...? Is there ever a legitimate use of an inner join on a non-distinct field? What would be the purpose? And if there's is a legitimate reason for such a query, can you give examples of where it would be used?

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