Join the content of a bash array with commas

(IFS=,; echo "${array[*]}")
This type of join is clearly documented in the bash manual. Only the first character of IFS is used for the delimiter.
Sample Output
$ array=(a b c)
$ (IFS=,; echo "${array[*]}")
a,b,c

2 Alternatives + Submit Alt

  • DOCKER_APP_VARS=(DATABASE_USER=dbuserro, DATABASE_PASSWORD=maipass) [jeff@omniscience container] (master)$ echo docker run $(printf -- " -e %s" ${DOCKER_APP_VARS[*]}) -name 12factorapp mattdm/fedora-small docker run -e DATABASE_USER=dbuserro, -e DATABASE_PASSWORD=maipass -name 12factorapp mattdm/fedora-small Note that the printf method by itsself doesn't include a newline (\n), so you'll need to embed it into an echo statement or something that does. Show Sample Output


    1
    printf -- " -e %s" ${ARRAY[*]}
    SEJeff · 2014-02-25 03:34:12 0
  • printf reapeats the format as longer as it has arguments. Then the idea is to make cut retain as much fields as we have elements in the array. As usual with such join/split string manipulation, you have to make sure you don't have conflicts between your separator and your array content.


    0
    printf "%s," "${LIST[@]}" | cut -d "," -f 1-${#LIST[@]}
    Valise · 2012-06-04 14:56:12 0

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