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How to get an alternate separator for brace expansion?



2019 Community Moderator ElectionGet the source directory of a Bash script from within the script itselfHow do I parse command line arguments in Bash?How to check if a string contains a substring in BashHow to check if a program exists from a Bash script?How do I tell if a regular file does not exist in Bash?How do I split a string on a delimiter in Bash?How to count all the lines of code in a directory recursively?How to use double or single brackets, parentheses, curly bracesHow do I reload .bashrc without logging out and back in?How to concatenate string variables in Bash










2















I find myself often doing silly things like:



cmd $( echo foo1,2,3 | tr ' ' , )


when I want to invoke cmd foo1,foo2,foo3. I would like instead to be able to do something along the lines of:



OFS=, ; cmd foo1,2,3



to force the brace expansion to give me a comma separated list. Is there any such functionality available?










share|improve this question






















  • If cmd is always the same, a function might help.

    – Cyrus
    Mar 7 at 18:51











  • Interestingly, the documentation for neither brace expansion nor pathname expansion (to which brace expansion is compared) ever actually says what the result is. It just says a "list" of words.

    – chepner
    Mar 7 at 18:52












  • The closest thing I know of would be a clumsy use of an array. cmd "$(IFS=,; x=(foo1,2,3); printf '%s' "$x[*]")".

    – chepner
    Mar 7 at 18:55
















2















I find myself often doing silly things like:



cmd $( echo foo1,2,3 | tr ' ' , )


when I want to invoke cmd foo1,foo2,foo3. I would like instead to be able to do something along the lines of:



OFS=, ; cmd foo1,2,3



to force the brace expansion to give me a comma separated list. Is there any such functionality available?










share|improve this question






















  • If cmd is always the same, a function might help.

    – Cyrus
    Mar 7 at 18:51











  • Interestingly, the documentation for neither brace expansion nor pathname expansion (to which brace expansion is compared) ever actually says what the result is. It just says a "list" of words.

    – chepner
    Mar 7 at 18:52












  • The closest thing I know of would be a clumsy use of an array. cmd "$(IFS=,; x=(foo1,2,3); printf '%s' "$x[*]")".

    – chepner
    Mar 7 at 18:55














2












2








2


1






I find myself often doing silly things like:



cmd $( echo foo1,2,3 | tr ' ' , )


when I want to invoke cmd foo1,foo2,foo3. I would like instead to be able to do something along the lines of:



OFS=, ; cmd foo1,2,3



to force the brace expansion to give me a comma separated list. Is there any such functionality available?










share|improve this question














I find myself often doing silly things like:



cmd $( echo foo1,2,3 | tr ' ' , )


when I want to invoke cmd foo1,foo2,foo3. I would like instead to be able to do something along the lines of:



OFS=, ; cmd foo1,2,3



to force the brace expansion to give me a comma separated list. Is there any such functionality available?







bash






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked Mar 7 at 18:43









William PursellWilliam Pursell

133k32206240




133k32206240












  • If cmd is always the same, a function might help.

    – Cyrus
    Mar 7 at 18:51











  • Interestingly, the documentation for neither brace expansion nor pathname expansion (to which brace expansion is compared) ever actually says what the result is. It just says a "list" of words.

    – chepner
    Mar 7 at 18:52












  • The closest thing I know of would be a clumsy use of an array. cmd "$(IFS=,; x=(foo1,2,3); printf '%s' "$x[*]")".

    – chepner
    Mar 7 at 18:55


















  • If cmd is always the same, a function might help.

    – Cyrus
    Mar 7 at 18:51











  • Interestingly, the documentation for neither brace expansion nor pathname expansion (to which brace expansion is compared) ever actually says what the result is. It just says a "list" of words.

    – chepner
    Mar 7 at 18:52












  • The closest thing I know of would be a clumsy use of an array. cmd "$(IFS=,; x=(foo1,2,3); printf '%s' "$x[*]")".

    – chepner
    Mar 7 at 18:55

















If cmd is always the same, a function might help.

– Cyrus
Mar 7 at 18:51





If cmd is always the same, a function might help.

– Cyrus
Mar 7 at 18:51













Interestingly, the documentation for neither brace expansion nor pathname expansion (to which brace expansion is compared) ever actually says what the result is. It just says a "list" of words.

– chepner
Mar 7 at 18:52






Interestingly, the documentation for neither brace expansion nor pathname expansion (to which brace expansion is compared) ever actually says what the result is. It just says a "list" of words.

– chepner
Mar 7 at 18:52














The closest thing I know of would be a clumsy use of an array. cmd "$(IFS=,; x=(foo1,2,3); printf '%s' "$x[*]")".

– chepner
Mar 7 at 18:55






The closest thing I know of would be a clumsy use of an array. cmd "$(IFS=,; x=(foo1,2,3); printf '%s' "$x[*]")".

– chepner
Mar 7 at 18:55













2 Answers
2






active

oldest

votes


















4














I don't think it's possible with a simple variable or setting.



However, you can capture the brace expansion in an array, and then use IFS and parameter substitution to achieve that effect:



$ ( a=( foo1,2,3 ); IFS=,; printf -- "%sn" "$a[*]" )
foo1,foo2,foo3


There, I'm using a subshell so I don't alter IFS in the current shell.




Here's a short
function that's much tidier:



$ join() local IFS=$1; shift; printf -- "%sn" "$*"; 
$ join , foo1,2,3
foo1,foo2,foo3





share|improve this answer






























    4














    You can create a function that will echo a comma-separated list of its args:



    commas() local IFS=,; echo "$*"; 


    Then you can use it in your calls to cmd:



    cmd "$(commas foo1,2,3)"


    Edit: Note that echo is somewhat broken, in that if you pass a single arg that looks like a valid echo options (e.g., -e, -n, etc), echo will consume it and interpret it as an option instead of printing it.



    $ commas -e #prints nothing
    $ commas -n #prints nothing
    $ commas -e -n
    -e,-n


    The last one works because "$*" expands to -e,-n, which echo sees as a single argument.



    And echo doesn't respect the -- argument as an "end of options" indicator like printf does. So you may want to use printf -- "%sn" "$*", as in @glennjackman's answer. Then again, the whole point of the commas function is to combine multiple args into a comma-separated list, as a single string. So it's probably unlikely that you'd run into any issues with echo. But hey, it only costs you a few more characters to type printf -- "%s" instead of echo.






    share|improve this answer

























    • The unquoted command substitution can cause problems, depending on what the arguments to commas end up being.

      – chepner
      Mar 7 at 18:59











    • @Aaron Fixed. Thanks.

      – Mike Holt
      Mar 7 at 18:59











    • @chepner Well spotted. Thanks.

      – Mike Holt
      Mar 7 at 19:00










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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    4














    I don't think it's possible with a simple variable or setting.



    However, you can capture the brace expansion in an array, and then use IFS and parameter substitution to achieve that effect:



    $ ( a=( foo1,2,3 ); IFS=,; printf -- "%sn" "$a[*]" )
    foo1,foo2,foo3


    There, I'm using a subshell so I don't alter IFS in the current shell.




    Here's a short
    function that's much tidier:



    $ join() local IFS=$1; shift; printf -- "%sn" "$*"; 
    $ join , foo1,2,3
    foo1,foo2,foo3





    share|improve this answer



























      4














      I don't think it's possible with a simple variable or setting.



      However, you can capture the brace expansion in an array, and then use IFS and parameter substitution to achieve that effect:



      $ ( a=( foo1,2,3 ); IFS=,; printf -- "%sn" "$a[*]" )
      foo1,foo2,foo3


      There, I'm using a subshell so I don't alter IFS in the current shell.




      Here's a short
      function that's much tidier:



      $ join() local IFS=$1; shift; printf -- "%sn" "$*"; 
      $ join , foo1,2,3
      foo1,foo2,foo3





      share|improve this answer

























        4












        4








        4







        I don't think it's possible with a simple variable or setting.



        However, you can capture the brace expansion in an array, and then use IFS and parameter substitution to achieve that effect:



        $ ( a=( foo1,2,3 ); IFS=,; printf -- "%sn" "$a[*]" )
        foo1,foo2,foo3


        There, I'm using a subshell so I don't alter IFS in the current shell.




        Here's a short
        function that's much tidier:



        $ join() local IFS=$1; shift; printf -- "%sn" "$*"; 
        $ join , foo1,2,3
        foo1,foo2,foo3





        share|improve this answer













        I don't think it's possible with a simple variable or setting.



        However, you can capture the brace expansion in an array, and then use IFS and parameter substitution to achieve that effect:



        $ ( a=( foo1,2,3 ); IFS=,; printf -- "%sn" "$a[*]" )
        foo1,foo2,foo3


        There, I'm using a subshell so I don't alter IFS in the current shell.




        Here's a short
        function that's much tidier:



        $ join() local IFS=$1; shift; printf -- "%sn" "$*"; 
        $ join , foo1,2,3
        foo1,foo2,foo3






        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered Mar 7 at 18:55









        glenn jackmanglenn jackman

        170k26147240




        170k26147240























            4














            You can create a function that will echo a comma-separated list of its args:



            commas() local IFS=,; echo "$*"; 


            Then you can use it in your calls to cmd:



            cmd "$(commas foo1,2,3)"


            Edit: Note that echo is somewhat broken, in that if you pass a single arg that looks like a valid echo options (e.g., -e, -n, etc), echo will consume it and interpret it as an option instead of printing it.



            $ commas -e #prints nothing
            $ commas -n #prints nothing
            $ commas -e -n
            -e,-n


            The last one works because "$*" expands to -e,-n, which echo sees as a single argument.



            And echo doesn't respect the -- argument as an "end of options" indicator like printf does. So you may want to use printf -- "%sn" "$*", as in @glennjackman's answer. Then again, the whole point of the commas function is to combine multiple args into a comma-separated list, as a single string. So it's probably unlikely that you'd run into any issues with echo. But hey, it only costs you a few more characters to type printf -- "%s" instead of echo.






            share|improve this answer

























            • The unquoted command substitution can cause problems, depending on what the arguments to commas end up being.

              – chepner
              Mar 7 at 18:59











            • @Aaron Fixed. Thanks.

              – Mike Holt
              Mar 7 at 18:59











            • @chepner Well spotted. Thanks.

              – Mike Holt
              Mar 7 at 19:00















            4














            You can create a function that will echo a comma-separated list of its args:



            commas() local IFS=,; echo "$*"; 


            Then you can use it in your calls to cmd:



            cmd "$(commas foo1,2,3)"


            Edit: Note that echo is somewhat broken, in that if you pass a single arg that looks like a valid echo options (e.g., -e, -n, etc), echo will consume it and interpret it as an option instead of printing it.



            $ commas -e #prints nothing
            $ commas -n #prints nothing
            $ commas -e -n
            -e,-n


            The last one works because "$*" expands to -e,-n, which echo sees as a single argument.



            And echo doesn't respect the -- argument as an "end of options" indicator like printf does. So you may want to use printf -- "%sn" "$*", as in @glennjackman's answer. Then again, the whole point of the commas function is to combine multiple args into a comma-separated list, as a single string. So it's probably unlikely that you'd run into any issues with echo. But hey, it only costs you a few more characters to type printf -- "%s" instead of echo.






            share|improve this answer

























            • The unquoted command substitution can cause problems, depending on what the arguments to commas end up being.

              – chepner
              Mar 7 at 18:59











            • @Aaron Fixed. Thanks.

              – Mike Holt
              Mar 7 at 18:59











            • @chepner Well spotted. Thanks.

              – Mike Holt
              Mar 7 at 19:00













            4












            4








            4







            You can create a function that will echo a comma-separated list of its args:



            commas() local IFS=,; echo "$*"; 


            Then you can use it in your calls to cmd:



            cmd "$(commas foo1,2,3)"


            Edit: Note that echo is somewhat broken, in that if you pass a single arg that looks like a valid echo options (e.g., -e, -n, etc), echo will consume it and interpret it as an option instead of printing it.



            $ commas -e #prints nothing
            $ commas -n #prints nothing
            $ commas -e -n
            -e,-n


            The last one works because "$*" expands to -e,-n, which echo sees as a single argument.



            And echo doesn't respect the -- argument as an "end of options" indicator like printf does. So you may want to use printf -- "%sn" "$*", as in @glennjackman's answer. Then again, the whole point of the commas function is to combine multiple args into a comma-separated list, as a single string. So it's probably unlikely that you'd run into any issues with echo. But hey, it only costs you a few more characters to type printf -- "%s" instead of echo.






            share|improve this answer















            You can create a function that will echo a comma-separated list of its args:



            commas() local IFS=,; echo "$*"; 


            Then you can use it in your calls to cmd:



            cmd "$(commas foo1,2,3)"


            Edit: Note that echo is somewhat broken, in that if you pass a single arg that looks like a valid echo options (e.g., -e, -n, etc), echo will consume it and interpret it as an option instead of printing it.



            $ commas -e #prints nothing
            $ commas -n #prints nothing
            $ commas -e -n
            -e,-n


            The last one works because "$*" expands to -e,-n, which echo sees as a single argument.



            And echo doesn't respect the -- argument as an "end of options" indicator like printf does. So you may want to use printf -- "%sn" "$*", as in @glennjackman's answer. Then again, the whole point of the commas function is to combine multiple args into a comma-separated list, as a single string. So it's probably unlikely that you'd run into any issues with echo. But hey, it only costs you a few more characters to type printf -- "%s" instead of echo.







            share|improve this answer














            share|improve this answer



            share|improve this answer








            edited Mar 7 at 19:23

























            answered Mar 7 at 18:56









            Mike HoltMike Holt

            2,9021920




            2,9021920












            • The unquoted command substitution can cause problems, depending on what the arguments to commas end up being.

              – chepner
              Mar 7 at 18:59











            • @Aaron Fixed. Thanks.

              – Mike Holt
              Mar 7 at 18:59











            • @chepner Well spotted. Thanks.

              – Mike Holt
              Mar 7 at 19:00

















            • The unquoted command substitution can cause problems, depending on what the arguments to commas end up being.

              – chepner
              Mar 7 at 18:59











            • @Aaron Fixed. Thanks.

              – Mike Holt
              Mar 7 at 18:59











            • @chepner Well spotted. Thanks.

              – Mike Holt
              Mar 7 at 19:00
















            The unquoted command substitution can cause problems, depending on what the arguments to commas end up being.

            – chepner
            Mar 7 at 18:59





            The unquoted command substitution can cause problems, depending on what the arguments to commas end up being.

            – chepner
            Mar 7 at 18:59













            @Aaron Fixed. Thanks.

            – Mike Holt
            Mar 7 at 18:59





            @Aaron Fixed. Thanks.

            – Mike Holt
            Mar 7 at 18:59













            @chepner Well spotted. Thanks.

            – Mike Holt
            Mar 7 at 19:00





            @chepner Well spotted. Thanks.

            – Mike Holt
            Mar 7 at 19:00

















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