Longest filename in a directory

dresch · 01-04-2007, 09:00 AM

Can anyone help me to write a shell script that when given a directory as an argument, finds the longest filename in that directory and any subdirectories?

ronkymac · 01-04-2007, 10:21 AM

Hi,

After having searched the net for about 2 minutes, I came up with this, which should be of help:

http://www.linuxforums.org/forum/lin...postcount47550

Regards,
Ronald.

colucix · 01-04-2007, 10:33 AM

Try this:

Code:

#!/bin/bash
maxlen=0

for file in `find $1 -type f` ; do

    len=`echo "$file" | gawk -F\/ '{printf "%s", $NF}' | wc -c`
    [ $len -gt $maxlen ] && maxlen=$len && maxfile=$file
    
done

echo $len $maxfile

dresch · 01-04-2007, 07:48 PM

Thanks for the help.

However my code below does not quite work. It is looking in current directory and sub-directories but when looking for the longest filename is is counting the path to that file aswell.

Is there something I am leaving out of the code?

#!/bin/bash
# Given a directory name as argument
# print the longest filename in that directory.

name=${0##*/}
FILES=${1-'pwd'}
length=0
filename=''

if [ ! -d $directory ]; then
echo "'$directory' is not a directory"
echo "Usage: $name [directory]"
exit 1
fi

for file in `find $FILES`; do
len=${#file}
if [ $len -gt $length ]; then
length=$len
filename=$file
fi
done

echo "The longest filename in directory is $filename [/SIZE][/SIZE][/FONT]with $length characters."[/SIZE]

colucix · 01-05-2007, 01:46 AM

Quote:

Originally Posted by dresch

However my code below does not quite work. It is looking in current directory and sub-directories but when looking for the longest filename is is counting the path to that file aswell.

for file in `find $FILES`; do
len=${#file}
if [ $len -gt $length ]; then
length=$len
filename=$file
fi
done

This is beacuse the find command give the full path of the files it finds. This is the reason why I parsed the output with gawk in the script above.

Code:

len=`echo "$file" | gawk -F\/ '{printf "%s", $NF}' | wc -c`

This echoes the full-path, extract the last field with gawk considering "/" as the field separator (and without the newline control-character, that would give a wrong result) and then count the characters in it.

Hope this will help...

muha · 01-05-2007, 02:56 PM

Quote:

Originally Posted by colucix

Try this:

Code:

#!/bin/bash
maxlen=0

for file in `find $1 -type f` ; do

    len=`echo "$file" | gawk -F\/ '{printf "%s", $NF}' | wc -c`
    [ $len -gt $maxlen ] && maxlen=$len && maxfile=$file
    
done

echo $len $maxfile

Shouldn't the last line be:
echo $maxlen $maxfile
Otherwise it will just spew out the last length ...

anomie · 01-05-2007, 03:29 PM

My lazy way of doing this:

Code:

ls | gawk '{ print length, $0 }' | sort -rn | head -1

edit -- sorry, I missed the "must recurse" requirement at first read.

edit2 -- Here's my next stab at it.

Code:

find . -exec basename {} \; | gawk '{ print length, $0 }' | sort -rn | head -1

(note sure if -type f is another requirement for you; if so you can add it.)