3

Is there a way I can do this? For example if a gets a long name like:

i-have-names-that-are-too-long-to-describe/
i-have-names-that-are-too-long-to-describe-2/
i-have-names-that-are-too-long-to-descri-3/

Can I "upgrade" over from ls to ls -l given that I have a name of a file or directory that is longer than than say, 20 characters?

Is there a way to set up a bash function in my .bashrc to do this? I'll call the resulting function lls().

@tripleee asked:

Do you want ls -l when the input file name is long? Why? It will make the output longer, not shorter. What if you receive a mix of long and short filenames?

I want it more so that reading the long filenames are systematized to a list (and easier for me to digest and read going down a fixed column); for a mix of long and short filenames, I would default to the list format.

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3 Answers 3

Reset to default
3

There's no builtin option in ls that does what you want. You'd have to parse the output then restart if "long" filenames are found, or do something like:

$ ls ??????????* >& /dev/null && ls -l || ls

(Put as many ? as your length limit. You can set that up as an alias.)

Why don't you simply use ls -1? (That's a one, not a lowercase L.) It always lists files in a single column. (Or pipe ls to more or less, which also goes to single column display.) Or use find with -maxdepth 1.

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1
  • I did not know about ls -1, thanks! Also, great answer, that's a neat way of going about doing what I wanted
    – Kevin Lee
    Jul 17, 2012 at 4:07
2 
if [ $(ls "$@" | ( max=0; while read l ; do len=${#l} ; [[ $max -lt $len ]] && max=$len; done; echo $max )) -gt 20 ]
then
    ls "$@"
else
    ls -l "$@"
fi

or, thanks to manatwork suggestion, this much simpler way which assumes GNU wc is available:

[[ $(ls "$@" | wc -L) -gt 20 ]] && ls "$@" || ls -l "$@"
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3
  • 1
    While assuming GNU tools, would not be easier using wc's -L or --max-line-length switch?
    – manatwork
    Jul 16, 2012 at 9:47
  • Even if wc's -L switch is not available, using wc for counting is still shorter and faster: ls | tr -c 'a\n' 'a' | sort -r | head -1 | wc -m. But there is a small inconvenience: this will output actual max length+1.
    – manatwork
    Jul 16, 2012 at 9:59
  • @manatwork Thanks for the comments, reply updated.
    – jlliagre
    Jul 16, 2012 at 10:26
0

This is my independent attempt of what I wanted:

lls(){
  opt="   "      
  for i in $(ls -l | tr -s " " | cut -d' ' -f9)  
  do   
    count=$(echo $i | wc -m)  
     if [ $count -gt 20 ] ; then    
       opt=" -l"; break;           
     fi    
  done  
  ls $opt
}

I placed this in my .bashrc. This does require the use of tr, cut, and wc though.

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