find the length of file names in a directory?

koti_rama · #1 (**permalink**) 06-02-2008

Hi,
how can find length of file names in a directory.

Examp:

I have a directory with name "d1".
directory: d1

files:
aaa.log
bbb.log
abcd.log
abcdef.log

I wold like out put like:
file name legnth
aaa.log 3
bbb.log 3
abcd.log 4
abcdef.log 5

Thanks

Yogesh Sawant · #2 (**permalink**) 06-02-2008

using bash:

Code:

for file in *
do
    ff=`echo $file | cut -d '.' -f1`
    echo -n "$ff "
    echo ${#ff}
done

robotronic · #3 (**permalink**) 06-02-2008

Code:

ls | awk '$(NF+1)=length'

linuxbynight · #4 (**permalink**) 06-02-2008

If I understand your question, the following should work or you can modify it to do what you need.

for i in `ls -1`
do
val=`echo $i | cut -d '.' -f1 | wc | tr '\t' ' ' | tr -s ' ' | sed 's/^ //' | cut -d' ' -f 3`
val=`/usr/bin/expr $val - 1`
echo $i $val
done

robotronic · #5 (**permalink**) 06-02-2008

If you want to exclude the file extension from the character count, try this instead:

Code:

ls | awk -F'.' '{ print($0, NF>1?length-length($NF)-1:length) }'

linuxbynight · #6 (**permalink**) 06-04-2008

is more elegant. In my case, I just never took the time to learn awk. Thanks for the alternative.

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