Shell scripting to find length of filenames

ridertech · 08-25-2004, 10:30 AM

I'm trying to copy files from one hard drive to another. The problem is that the destination drive does not accept files over 30 characters (very weird!). I already hava a script (below) that recursively searches the current directory and outputs the total number of files in each folder.

How can I recursively search through the current directory and output the complete path to only filenames that are over 30 characters? I assume it would be a simple modification to the script below, but my shell scripting is still limited...

Code:

#!/bin/sh

IFS='
'
for i in `find . -type d`; do
    DIRNAME=`echo $i | sed "s/'/\'/g"`
    echo "Directory $i has [`ls $DIRNAME | wc -l`] files"
done

ridertech · 08-25-2004, 12:01 PM

Sorry to post twice, but here is the solution is below...

Code:

#!/bin/sh

IFS='
'
for i in `find . -type d`; do
    DIRNAME=`echo $i | sed "s/'/\'/g"`
    
    for f in `ls $DIRNAME`; do
        LENGTH=`echo $f | wc -m`
        #echo "[$LENGTH] $DIRNAME/$f"
        
        if [ $LENGTH -gt 31 ]; then
            echo "[$LENGTH] $DIRNAME/$f"
        fi
    done
done

Now I just need to figure out how to cut-down the long filenames. I know moving it to a new name is the process, but I'm not sure how to get only the first 26 characters of the long name.

For example...
"long-ass-file-names-are-not-my-friend.mp3" would be moved to "long-ass-file-names-are-no.mp3"

Demonbane · 08-25-2004, 12:07 PM

you can pipe it to "cut"

Code:

echo "long-ass-file-names-are-not-my-friend.mp3" | cut -c 1-26