A few years ago I found an interesting code snippet that prints each type of file in its corresponding color according to the colors set up in LS_COLORS. Unfortunately, I can't remember the link anymore.

Here is the snippet test_colors.sh in question

eval $(echo "no:global default;fi:normal file;di:directory;ln:symbolic link;pi:named pipe;so:socket;do:door;bd:block device;cd:character device;or:orphan symlink;mi:missing file;su:set uid;sg:set gid;tw:sticky other writable;ow:other w\
ritable;st:sticky;ex:executable;"|sed -e 's/:/="/g; s/\;/"\n/g')                                                                                                                                                                            
  for i in $LS_COLORS                                                                                                                                                                                                                       
    echo -e "\e[${i#*=}m$( x=${i%=*}; [ "${!x}" ] && echo "${!x}" || echo "$x" )\e[m"                                                                                                                                                       

The snippet works great in bash, but not in zsh, and I can't tell why. When I run it in zsh I get the following error:

> sh .test_colors.sh
.eval_colors:1: * not found

Update (Nov. 1, 2011)

I tested the script by @Stéphane Gimenez below. I noticed that some characters seem to not escape correctly. Any thoughts why?

Answer: See comments on @Stéphane Gimenez's answer.

                                                      enter image description here


2 Answers 2


The same written for zsh in a much cleaner way:


typeset -A names
names[no]="global default"
names[fi]="normal file"
names[ln]="symbolic link"
names[pi]="named pipe"
names[bd]="block device"
names[cd]="character device"
names[or]="orphan symlink"
names[mi]="missing file"
names[su]="set uid"
names[sg]="set gid"
names[tw]="sticky other writable"
names[ow]="other writable"

for i in ${(s.:.)LS_COLORS}
    printf '\e[%sm%s\e[m\n' $color $name
  • You might want to replace \n by a space at the end of the printf for compactness maybe. Sep 15, 2011 at 0:27
  • Thanks @Stéphane Gimenez. I have updated my OP with a problem I am getting when printing some characters using your script. Not sure if this is strictly related to your script though (it may be my own terminal?) Nov 1, 2011 at 18:09
  • 1
    @intrpc: running zsh as sh your are using some compatibility mode. Either call your script as zsh ./test_color_scheme or add double quotes around $color and $name. Nov 2, 2011 at 12:35
  • @Stéphane Gimenez: I was directed here from my related question here: unix.stackexchange.com/questions/52659/… . Your expansion of abbreviations is very helpful. I have three abbreviations, rs, ca and mh that are not included above. Can you please tell me where to find their expansions? Thanks.
    – chandra
    Oct 24, 2012 at 5:13
  • Interpreted from dircolors -p rs=reset, ca=capability, mh=multi-hard_link
    – weldabar
    Nov 3, 2013 at 20:16

You need to escape the = in ${i%=*} because otherwise the suffix pattern =* undergoes = expansion, so = is interpreted as a command name. This is the cause of the * not found error.

Zsh doesn't split words on variable substitutions by default, so $LS_COLORS expands to a single word. To have the for loop operate on the colon-separated parts of $LS_COLORS, use for i in $=LS_COLORS. Or more idiomatically in zsh, don't use IFS but instead specify explicitly how to split: for i in ${(s.:.)LS_COLORS}.

The syntax ${!x} to mean “the value of the variable whose name is $x” is specific to bash. Zsh has an equivalent construct, the P parameter expansion flag: ${(P)x}.

  • 1
    There are two more reasons for this script is not working in zsh. No automatic word-splitting for LS_COLORS and = needs to be escaped in substitution patterns. Sep 15, 2011 at 0:54
  • @StéphaneGimenez You're right, thanks, I only hit on where bash is non-standard but these are two non-standard zsh features that need to be addressed as well. You should expand your answer to have all the explanations, and then I can delete mine. Sep 15, 2011 at 7:18

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