I have partition structure like :
$ df
Filesystem 1K-blocks Used Available Use% Mounted on
/dev/sda6 51606140 16939248 34142692 34% /
/dev/sda5 495844 72969 397275 16% /boot
/dev/sda7 113022648 57515608 49765728 50% /home
/dev/sda8 113022648 57515608 49765728 4% /mnt
while parsing directories content using readdir() - how to find out which file resides on what device?
readdir() invoked from root directory and parses the file name and prints its size.
like from device : /dev/sda6 and list the filenames under that partition.
When it reads contents from /home - it should display reading content from /dev/sda7 and list filenames
Please let me know,if you need more details/info
you can just do
df <file_name>
that will give you the device and partition for the particuar file
There is a st_dev member in struct stat, it should uniquely identify one partition.
Example in bash:
stat ~/.vimrc
File: `/home2//leonard/.vimrc' -> `local-priv/vimrc'
Size: 16 Blocks: 0 IO Block: 4096 symbolic link
Device: 802h/2050d Inode: 6818899 Links: 1
Access: (0777/lrwxrwxrwx) Uid: ( 1024/ leonard) Gid: ( 1024/ leonard)
Access: 2012-06-22 16:36:45.341371003 +0300
Modify: 2012-06-22 16:36:45.341371003 +0300
Change: 2012-06-22 16:36:45.341371003 +0300
The stat utility does no additional magic. Here is strace -vvv output:
lstat64("/home2//leonard/.vimrc", {st_dev=makedev(8, 2), st_ino=6818899, st_mode=S_IFLNK|0777, st_nlink=1, st_uid=1024, st_gid=1024, st_blksize=4096, st_blocks=0, st_size=16, st_atime=2012/06/22-16:36:45, st_mtime=2012/06/22-16:36:45, st_ctime=2012/06/22-16:36:45}) = 0
0x0802 is major 8(sd) partition 2, so /dev/sda2
In order to map this to actual partitions you can iterate /proc/mounts and stat all the devices (first column). The contents of /proc/mounts is just like the output of mount(1) except it comes directly from the kernel. Some distros symlink /etc/mtab to /proc/mounts.
Or you can parse /proc/partitions:
$ cat /proc/partitions
major minor #blocks name
8 0 976762584 sda
8 1 3998720 sda1
8 2 972762112 sda2
Of course /dev/sda might not actually exist, the device could be using a long udev name like /dev/disk/by-uuid/c4181217-a753-4cf3-b61d-190ee3981a3f. Major/Minor numbers should be a reliable unique identifier of a partition.
Related
For those reaching here; Unfortunately I could not recover the data, after various tries and reproducing the problem it was too costy to keep trying, so we just used a past backup to recreate the information needed
A human error broke an 150G UFS filesystem (Solaris).
When trying to do a backup of the filesytem (c0t0d0s3) the ufsdump(1M) hasn't been correctly used.
I will explain the background that led to this ...
The admin used:
# ufsdump 0f /dev/dsk/c0t0d0s3 > output_1
root#ats-br000432 # ufsdump 0f /dev/dsk/c0t0d0s3 > output_1
Usage: ufsdump [0123456789fustdWwnNDCcbavloS [argument]] filesystem
This is a bad usage, so it created a file called output_1 with 0 bytes:
# ls -la output_1
-rw-r--r-- 1 root root 0 abr 12 14:12 output_1
Then, the syntax used was:
# ufsdump 0f /dev/rdsk/c0t0d0s3 output_1
Which wrote that 0 bytes file output_1 to /dev/rdsk/c0t0d0s3 - which was the partition slice
Now, interestingly, due to being a 0 bytes file, we thought that this would cause no harm to the filesystem, but it did.
When trying to ls in the mountpoint, the partition claimed there was an I/O error, when umounting and mounting again, the filesystem showed no contents, but the disk space was still showing as used just like it was previously.
I assume, at some point, the filesystem 'header' was affected, right? Or was it the slice information?
A small fsck try brings up this:
** /dev/rdsk/c0t0d0s3
** Last Mounted on /database
** Phase 1 - Check Blocks and Sizes
INCORRECT DISK BLOCK COUNT I=11 (400 should be 208)
CORRECT?
Disk block count / I=11
this seems that the command broke filesystem information regarding its own contents, right?
When we tried to fsck -y -F ufs /dev/dsk.. various files have been recovered, but not the dbf files we are after (which are GB sized)
What can be done now? Should I try every superblock information from newfs -N ?
EDIT: new information regarding partition
newfs output showing superblock information
# newfs -N /dev/rdsk/c0t0d0s3
Warning: 2826 sector(s) in last cylinder unallocated
/dev/rdsk/c0t0d0s3: 265104630 sectors in 43149 cylinders of 48 tracks, 128 sectors
129445,6MB in 2697 cyl groups (16 c/g, 48,00MB/g, 5824 i/g)
super-block backups (for fsck -F ufs -o b=#) at:
98464, 196896, 295328, 393760, 492192, 590624, 689056, 787488, 885920,
Initializing cylinder groups:
.....................................................
super-block backups for last 10 cylinder groups at:
264150944, 264241184, 264339616, 264438048, 264536480, 264634912, 264733344,
264831776, 264930208, 265028640
I accidentally removed a wrong folder using rm -rf, every tool I've tried tell me either I have no hard disk or that the filesystem is not found.
When I type df I get :
Filesystem Size Used Avail Use% Mounted on
/dev/simfs 25G 7.6G 18G 31% /
none 256M 12K 256M 1% /dev
But if I type debugfs -w /dev/simfs I get the error :
/dev/simfs: No such file or directory while opening filesystem
I'm using a VPS hosted by OVH, what's wrong with that filesystem ?
I have the same scenario/issue. To take the conversation a step further, I've also tried
tune2fs -l /dev/simfs
which outputs
tune2fs 1.42.9 (4-Feb-2014)
tune2fs: No such file or directory while trying to open /dev/simfs
Couldn't find valid filesystem superblock.
You'll get the same invalid superblock error if you try to scan an entire disk eg /dev/sda
For example, a sample df command output is
Filesystem MB blocks Free %Used Iused %Iused Mounted on
/dev/hd4 512.00 322.96 37% 4842 7% /
/dev/hd2 4096.00 717.96 83% 68173 29% /usr
/dev/hd9var 1024.00 670.96 35% 6385 4% /var
/dev/hd3 5120.00 0.39 100% 158 10% /tmp
Now if I specify something like /tmp/dummy.txt I should be able to get /dev/hd3 or just hd3.
EDIT : Thanks torek for the answer. But probing the /proc would become very tedious. Can anyone suggest me some system calls which can do the same internally?
df `pwd`
...Super simple, works, and also tells you how much space is there...
[stackuser#rhel62 ~]$ pwd
/home/stackuser
[stackuser#rhel62 ~]$ df `pwd`
Filesystem 1K-blocks Used Available Use% Mounted on
/dev/sda7 250056240 196130640 41223408 83% /
[stackuser#rhel62 ~]$ cd isos
[stackuser#rhel62 isos]$ pwd
/home/stackuser/isos
[stackuser#rhel62 isos]$ df `pwd`
Filesystem 1K-blocks Used Available Use% Mounted on
/dev/sda5 103216920 90417960 11750704 89% /mnt/sda5
[stackuser#rhel62 isos]$ df $(pwd)
Filesystem 1K-blocks Used Available Use% Mounted on
/dev/sda5 103216920 90417960 11750704 89% /mnt/sda5
...which is the likely cause of the mount point query in the first place.
Note those are backticks, and the alternate (modern) method, providing further control over slashes and expansion is df $(pwd). Tested and traverses symlinks correctly on bash, dash, busybox, zsh. Note that tcsh won't like the $(...), so stick to the older backtick style in csh-variants.
There are also extra switches in pwd and df for further enjoyment.
On Linux, use /proc/<pid>/mounts to access a list of mount points for a given pid, or /proc/self/mounts (with the literal word self) to refer to yourself. (cat the /proc/self/mount* files to see what they look like.)
Then, for each file system, you can do a statfs() call and compare f_fsid the f_fsid field to the result from an earlier statfs() on the path in question. Once the fsid's match, you have found the appropriate mounted file system and can use the other data from /proc/self/mounts. (However, see statfs(2) for restrictions on doing anything useful with f_fsid.)
I need to create a very high number of files which are not very large (like 4kb,8kb).
It's not possible on my computer cause it takes all inodes up to 100% and I cannot create more files :
$ df -i /dev/sda5
Filesystem Inodes IUsed IFree IUse% Mounted on
/dev/sda5 54362112 36381206 17980906 67% /scratch
(I started deleting files, it's why it's now 67%)
The bytes-per-nodes are of 256 on my filesystem (ext4)
$ sudo tune2fs -l /dev/sda5 | grep Inode
Inode count: 54362112
Inodes per group: 8192
Inode blocks per group: 512
Inode size: 256
I wonder if it's possible to set this value very low even below 128(during reformating). If yes,what value should I use?
Thx
The default bytes per inode is usually 16384, which is the default inode_ratio in /etc/mke2fs.conf (it's read prior to filesystem creation). If you're running out of inodes, you might try for example:
mkfs.ext4 -i 8192 /dev/mapper/main-var2
Another option that affects this is -T, typically -T news which further reduces it to 4096.
Also, you can not change the number of inodes in a ext3 or ext4 filesystem without re-creating or hex-editing it. Reiser filesystems are dynamic so you'll never have an issue with them.
You can find out the approximate inode ratio by dividing the size of available space by the number of available inodes. For example:
$ sudo tune2fs -l /dev/sda1 | awk -F: ' \
/^Block count:/ { blocks = $2 } \
/^Inode count:/ { inodes = $2 } \
/^Block size:/ { block_size = $2 } \
END { blocks_per_inode = blocks/inodes; \
print "blocks per inode:\t", blocks_per_inode, \
"\nbytes per inode:\t", blocks_per_inode * block_size }'
blocks per inode: 3.99759
bytes per inode: 16374.1
I have found solution to my problem on the mke2fs man page :
-I inode-size
Specify the size of each inode in bytes. mke2fs creates 256-byte inodes by default. In kernels after 2.6.10 and some earlier vendor kernels it is possible to utilize
inodes larger than 128 bytes to store extended attributes for improved performance. The inode-size value must be a power of 2 larger or equal to 128. The larger the
inode-size the more space the inode table will consume, and this reduces the usable space in the filesystem and can also negatively impact performance. Extended
attributes stored in large inodes are not visible with older kernels, and such filesystems will not be mountable with 2.4 kernels at all. It is not possible to change
this value after the filesystem is created.
The maximun you will be able to set is given by your block-size.
sudo tune2fs -l /dev/sda5 | grep "Block size"
Block size: 4096
Hope this can help....
I have a directory I’m archiving:
$ du -sh oldcode
1400848
$ tar cf oldcode.tar oldcode
So the directory is 1.4gb. The file is significantly smaller, though:
$ ls -l oldcode.tar
-rw-r--r-- 1 ieure ieure 940339200 2002-01-30 10:33 oldcode.tar
Only 897mb. It’s not compressed in any way:
$ file oldcode.tar
oldcode.tar: POSIX tar archive
Why is the tar file smaller than its contents?
You get a difference because of the way the filesystem works.
In a nutshell your disk is made out of clusters. Each cluster has a fixed size of - let's say - 4 kilobytes. If you store a 1kb file in such a cluster 3kb will be unused. The exact details vary with the kind of file-system that you use, but most file-systems work that way.
3kb wasted space is not much for a single file, but if you have lots of very small files the waste can become a significant part of the disk usage.
Inside the tar-archive the files are not stored in clusters but one after another. That's where the difference comes from.
Having no knowledge of what tar you're using or what sort of Unix system you're using, here's my guess: oldcode contains numerous smaller files, which when by themselves use disk space inefficiently, since disk space is allocated by some sort of block, rather than byte by byte. In the tar file, they're concatenated, and make maximum use of the disk space they're assigned.
This has something to do with the blocksize of your filesystem. man 1 du on MacOSX 10.5.6 states:
The du utility displays the file system block usage for each file argument and for each directory in the file hierarchy rooted in each directory argument. If no file is specified, the block usage of the hierarchy rooted in the current directory is displayed.
[mirko#borg foo]$ ls -la
total 0
drwxr-xr-x 2 mirko wheel 68 Jan 30 21:20 .
drwxrwxrwt 10 root wheel 340 Jan 30 21:16 ..
[mirko#borg foo]$ du -sh
0B .
[mirko#borg foo]$ touch foo
[mirko#borg foo]$ ls -la
total 0
drwxr-xr-x 3 mirko wheel 102 Jan 30 21:20 .
drwxrwxrwt 10 root wheel 340 Jan 30 21:16 ..
-rw-r--r-- 1 mirko wheel 0 Jan 30 21:20 foo
[mirko#borg foo]$ du -sh
0B .
[mirko#borg foo]$ echo 1 > foo
[mirko#borg foo]$ ls -la
total 8
drwxr-xr-x 3 mirko wheel 102 Jan 30 21:20 .
drwxrwxrwt 10 root wheel 340 Jan 30 21:16 ..
-rw-r--r-- 1 mirko wheel 2 Jan 30 21:20 foo
[mirko#borg foo]$ du -sh
4.0K .
As you see even a file of 2 bytes takes a whole block of 4kb. There are some filesystems which avoid this waste of space by block suballocation.
There are 2 possibilities.
Small files
Most likely, it isn't smaller than its contents. As Nils Pipenbrinck wrote, du displays the amount of space the filesystem allocates, which since files are stored in filesystem blocks is more than the logical size of the file.
To view the logical size of the file, use du --apparent-size. In this case, the result should be smaller than the tar file.
Sparse files
Tar files can store sparse files. If the tarball was created using --sparse, the holes in the sparse files will be recorded, so the tarball could be smaller than the logical size of the files.
If the sparseness information in your extracted copy was somehow lost (e.g. if you extracted the tarball onto a filesystem that doesn't support sparse files, or if it was zipped and then unzipped, etc.), then df will report the expanded size.
du counts disk blocks, not file size duder.