Ygtjki

So my task is to find the file with the most hardlinks in a directory.
So far i have :

find . -name "file*" | xargs -I -n 1 find . -samefile 

which gives me:

./hardlinkFIle245
./hardlinkFIle23
./hardlinkFIle2
./file2.txt
./hardlinkFIle1234
./hardlinkFIle123
./hardlinkFIle12
./hardlinkFIle1
./file1.txt

Now when I pipe it in with |wc -l, I get the total number of lines 9:

find . -name "file*" | xargs -I -n 1 find . -samefile | wc -l

What I want is for each xargs batch -n 1 to give me the count :
so i want:

4
5

If using GNU find, you can have it report the number of links with -printf %n. So you can get the maximum value with:

find . -name 'file*' -printf '%nn' | sort -rn | head -n1

Note however that that number may include links that are not found under . or for entries that don't match the file* pattern.

If you only want to count hardlinks named file* found under ., and see the paths for those, you could do:

find . -name 'file*' -printf '%i%p' | gawk -v RS='' '
 
 inode = $0
 getline file
 
 ++count[inode] >= max 
 files[inode] = files[inode] " - " file ORS
 max = count[inode]
 max_inode = inode
 
 END 
 printf "%s", "File with most links ("max"):n" files[max_inode]
 '

Which would still run just one find invocation instead of one per file.

You can just spawn a new shell on each xargs:

find . -name "file*" | xargs -n 1 sh -c 'echo "$1"; find . -samefile "$1" | wc -l' xargs-sh

Though using xargs is a bad idea here as it would break if file paths contain whitespace of quoting characters.

Here, using wc -l is also brittle as it breaks if file paths contain newline characters.

You could use the standard find -exec cmd + syntax and save having to run one sh per file by using a loop:

find . -name "file*" -exec sh -c '
 for file do
 printf "%sn" "$file"
 find .//. -samefile "$file" | grep -c //
 done' find-sh +