Hard Links Help

Unknown50862 · October 4, 2010, 7:55pm

Ksh newbie here, so please bear with me.

I'm currently writing a script that searches through a directory and displays files with multiple hard links. The way I have it set up, is that it displays the i-node number and then each of the link names. In addition to this, I need to know if there are any hard links outside of the searched directory and if so, how many. Which I'm not completely sure how to do.

find /etc ! -type d -links +1 -ls
718093    8 -rw-r--r--   2 root     root          223 Aug 22  2009 /etc/hosts
718091    8 -rw-r--r--   3 root     root          204 Aug 22  2009 /etc/sysconfig/network-scripts/ifcfg-eth0
718093    8 -rw-r--r--   2 root     root          223 Aug 22  2009 /etc/sysconfig/networking/profiles/default/hosts
718094    8 -rw-r--r--   2 root     root           68 Aug 22  2009 /etc/sysconfig/networking/profiles/default/resolv.conf
718091    8 -rw-r--r--   3 root     root          204 Aug 22  2009 /etc/sysconfig/networking/profiles/default/ifcfg-eth0
718091    8 -rw-r--r--   3 root     root          204 Aug 22  2009 /etc/sysconfig/networking/devices/ifcfg-eth0
718094    8 -rw-r--r--   2 root     root           68 Aug 22  2009 /etc/resolv.conf

I understand the $4 is the number of hard links, however does that include any links that are found outside of the current directory? If not, then what exactly do I need to do to get a total number of hard links in and out of the current directory?

Also, on a side note what is $2?

agama · October 4, 2010, 9:15pm

The second column is the number of disk blocks required to house the file. Generally the value is 512 byte blocks, though some versions of find might default to 1024 byte blocks.

The fourth column is the number of links, but only if the file is a regular file. If it is a directory it is the number of files contained/referenced in the directory.

Because links (old timers call them links as symbolic links were introduced later) cannot span filesystems, I'd approach the problem by executing a find on the whole file system and matching inodes for files that are listed within the target directory. The following is a way to do this and assumes that the target directory is the current working directory.

#!/usr/bin/env ksh

p=$(df -h . )
find ${p##* } -type file -ls 2>/dev/null | awk -v cwd="$(pwd)/" '
        {
                if( $4 > 1  )                           # only care about files with multiple hardlinks
                {
                        dir = $NF;
                        sub( "[^/]+$", "", dir );       # lop off the filename
                        if( dir == cwd )                # collect for printing only files in the target directory
                                listit[$1]++;

                        paths[$1] = paths[$1] $NF " ";  # collect paths associated with the inode (regardless of directory)
                }
        }

        END {
                for( f in listit )
                {
                        x = split( paths[f], a, " " );  # split paths for a "vertical" listing
                        printf( "%s\n", f );            # print the inode number
                        for( i = 1; i <= x; i++ )       # list each path
                                printf( "\t%s\n", a );
                        printf( "\n" );
                }
        }
'

[/size]

Hope this gets you started.

Chubler_XL · October 4, 2010, 9:35pm

Nice work agama,

Can I suggest 1 slight change: add a -mount flag on the find, to keep it to the 1 filesystem.

Unknown50862 · October 4, 2010, 10:09pm

Thanks for the replies. Just one last question though. In the fourth column above, is that the total number of hard links for that file or just the number of links in that directory?

agama · October 4, 2010, 10:15pm

Total number of links to the real file.

@Chubler_XL -- yes, good call. Thanks.

michaelrozar17 · October 5, 2010, 2:10am

can you please tell wot does ##* mean in the below piece of your code.

find ${p##* } -type file -ls 2>/dev/null | awk -v cwd="$(pwd)/" '

Aia · October 5, 2010, 3:14am

An excerpt from here

Chopping strings like a pro

While basename and dirname are great tools, there are times where we may need to perform more advanced string "chopping" operations than just standard pathname manipulations. When we need more punch, we can take advantage of bash's advanced built-in variable expansion functionality. We've already used the standard kind of variable expansion, which looks like this: ${MYVAR}. But bash can also perform some handy string chopping on its own. Take a look at these examples:

$ MYVAR=foodforthought.jpg
$ echo ${MYVAR##*fo}
rthought.jpg
$ echo ${MYVAR#*fo}
odforthought.jpg

In the first example, we typed ${MYVAR##*fo}. What exactly does this mean? Basically, inside the ${ }, we typed the name of the environment variable, two ##s, and a wildcard ("*fo"). Then, bash took MYVAR, found the longest substring from the beginning of the string "foodforthought.jpg" that matched the wildcard "*fo", and chopped it off the beginning of the string. That's a bit hard to grasp at first, so to get a feel for how this special "##" option works, let's step through how bash completed this expansion. First, it began searching for substrings at the beginning of "foodforthought.jpg" that matched the "*fo" wildcard. Here are the substrings that it checked:

f
fo MATCHES *fo
foo
food
foodf
foodfo MATCHES *fo
foodfor
foodfort
foodforth
foodfortho
foodforthou
foodforthoug
foodforthought
foodforthought.j
foodforthought.jp
foodforthought.jpg

After searching the string for matches, you can see that bash found two. It selects the longest match, removes it from the beginning of the original string, and returns the result.

The second form of variable expansion shown above appears identical to the first, except it uses only one "#" -- and bash performs an almost identical process. It checks the same set of substrings as our first example did, except that bash removes the shortest match from our original string, and returns the result. So, as soon as it checks the "fo" substring, it removes "fo" from our string and returns "odforthought.jpg".

After reading that put it to the test.

p=$(df -h .) # free hard disk space in the current directory in a human readable format
echo ${p##* } # there's a space after the *

michaelrozar17 · October 5, 2010, 4:20am

Thank you.. It was really helpful:b: