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up vote 1 down vote favorite 
arrFormat=( jpg jpeg bmp tiff png )
varExtension="jpg"

for elem in "${arrFormat[@]}"
do
  echo "${elem}"
  # do something on $elem #
done


#for i in $( find -E . -iregex '.*\.($arrFormat)' )  ; do   #problem
#for i in $( find -E . -iregex '.*\.("$arrFormat")' )  ; do #problem
#for i in $( find -E . -iregex '.*\.($varExtension)' )  ; do        #problem
#for i in $( find -E . -iregex '.*\.("$varExtension")' )  ; do  #problem

for  i in $( find -E . -iregex '.*\.(jpg|png)' )  ; do # this works fine

  echo "${i}"
  # do something on $i #
done

So it seems that the regular expression has a problem with variable(s) How can I give an array or just one variable as argument to a regular expression

Shell info, MACINTOSH :

Darwin Kernel Version 15.3.0: Thu Dec 10 18:40:58 PST 2015; root:xnu-3248.30.4~1/RELEASE_X86_64 x86_64
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up vote 5 down vote accepted

The problem are the single quotes. Variables aren't expanded in single quotes (from man bash):

Enclosing characters in single quotes preserves the literal value of each character within the quotes.

If you want to use a variable, you need double quotes. To illustrate:

$ foo="bar"
$ echo '$foo'
$foo
$ echo "$foo"
bar

So, in your case, you want something like:

for i in $( find -E . -iregex ".*\.$varExtension")  ; do

Or, better since the above can't deal with strange file names:

find -E . -iregex ".*\.$varExtension" -print0 | while IFS= read -r -d '' i; do

Or, forget the loop and have find do whatever the job is for you:

find -E . -iregex ".*\.$varExtension" -exec something {} \;
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