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Old 10-12-2012, 10:43 PM   #1
NickPats
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# of files.


how can i find the directory with max files. i listed the #files and name of directory in a file.
 
Old 10-13-2012, 12:22 AM   #2
Elv13
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not optimal at all, but easy to read all in one
Code:
IFS=`echo -en "\n\b"`;for dir in $(find ./ -type d -maxdepth 1); do ls -a $dir | wc -l | xargs -i echo {} $dir;done | sort -n | tail -n1
as for using your file, if the number is the first column, then just use 'sort -n'. If it is the last one, then use
Code:
 awk '{printf $NF;for (i=1;i<=NF;i++){printf $i};printf "\n"}' < myfile.txt | sort -n | tail -n1 #untested
to put the number column back in the first position. You can also use ark or perl for the whole thing too. There is a million way to do this, thinking longer about it, I probably could come up with something a lot better

EDIT: ls -> ls -a

Last edited by Elv13; 10-13-2012 at 10:17 PM.
 
Old 10-13-2012, 12:57 AM   #3
shivaa
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What exactly your requirement is? Do you want to find a directory with maximum number of files it contains or just want to check the size of the directory for space checking purpose?
For checking the number of files, simply use wc -l option while listing the content i.e. ls -la <directory> | wc -l
For space checking purpose, use du -Sh command
 
Old 10-13-2012, 10:06 AM   #4
NickPats
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though how can i print the directory name with max files in it?is this correct?

for dir in *; do
( echo $dir: $(ls "$dir" | wc -l)) > /dev/null
done

awk '{for (i=1;i<=NF;i++){printf $i};printf "\n"}' < /dev/null | sort -n | tail -n1
 
Old 10-13-2012, 10:16 PM   #5
Elv13
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Thinking about it, I made a big mistake in the first one, use "ls -a", ls without -a wont tell you the number of file, only the number of file not starting with ".". Of course, there will be 2 additional "files", '.' and '..'. So, if you want the accurate number, better use "find $dir -maxdepth 1 -type f | wc -l" this will count only file, not directories, remove the "-type f" if you want dirs too.
 
  


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