How may I echo all but the last parameter in bash?











up vote
2
down vote

favorite












I have the following



#!/bin/bash

function f1 ()

{

  echo "${@:1:-2}"

}

f1 1 2 3 4 5 6


I need to echo 1 2 3 4 5
man bash tells me that when I use @ I can't use a negative length.



I resorted to using a calculating ("${@:1:$((${#@}-1))}") which is seeming unorthodox to me.



How do I exclude the last parameter from outputting?






bash parameter bash-functions







share|improve this question









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Bret Joseph is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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    up vote
    2
    down vote

    favorite












    I have the following



    #!/bin/bash
    
function f1 ()
    
{
    
  echo "${@:1:-2}"
    
}
    
f1 1 2 3 4 5 6


    I need to echo 1 2 3 4 5
    man bash tells me that when I use @ I can't use a negative length.



    I resorted to using a calculating ("${@:1:$((${#@}-1))}") which is seeming unorthodox to me.



    How do I exclude the last parameter from outputting?






    bash parameter bash-functions







    share|improve this question









    New contributor




    Bret Joseph is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.






















      up vote
      2
      down vote

      favorite









      up vote
      2
      down vote

      favorite











      I have the following



      #!/bin/bash
      
function f1 ()
      
{
      
  echo "${@:1:-2}"
      
}
      
f1 1 2 3 4 5 6


      I need to echo 1 2 3 4 5
      man bash tells me that when I use @ I can't use a negative length.



      I resorted to using a calculating ("${@:1:$((${#@}-1))}") which is seeming unorthodox to me.



      How do I exclude the last parameter from outputting?






      bash parameter bash-functions







      share|improve this question









      New contributor




      Bret Joseph is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











      I have the following



      #!/bin/bash
      
function f1 ()
      
{
      
  echo "${@:1:-2}"
      
}
      
f1 1 2 3 4 5 6


      I need to echo 1 2 3 4 5
      man bash tells me that when I use @ I can't use a negative length.



      I resorted to using a calculating ("${@:1:$((${#@}-1))}") which is seeming unorthodox to me.



      How do I exclude the last parameter from outputting?






      bash parameter bash-functions




      bash parameter bash-functions






      share|improve this question









      New contributor




      Bret Joseph is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











      share|improve this question









      New contributor




      Bret Joseph is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.









      share|improve this question




      share|improve this question








      edited 16 hours ago









      Kusalananda

      119k16223365




      119k16223365






      New contributor




      Bret Joseph is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.









      asked 17 hours ago









      Bret Joseph

      274




      274




      New contributor




      Bret Joseph is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.





      New contributor





      Bret Joseph is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.






      Bret Joseph is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.






















          1 Answer
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          echo "${@:1:$#-1}"


          The length argument is already in an arithmetic context, so there's no need for $(( ... )), and the number of arguments is given by $# so there's no need to try to use the equivalent of ${#...[@]} on $@.






          share|improve this answer





















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            1 Answer
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            active

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            1 Answer
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            active

            oldest

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            active

            oldest

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            active

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            up vote
            3
            down vote



            accepted










            echo "${@:1:$#-1}"


            The length argument is already in an arithmetic context, so there's no need for $(( ... )), and the number of arguments is given by $# so there's no need to try to use the equivalent of ${#...[@]} on $@.






            share|improve this answer

























              up vote
              3
              down vote



              accepted










              echo "${@:1:$#-1}"


              The length argument is already in an arithmetic context, so there's no need for $(( ... )), and the number of arguments is given by $# so there's no need to try to use the equivalent of ${#...[@]} on $@.






              share|improve this answer























                up vote
                3
                down vote



                accepted







                up vote
                3
                down vote



                accepted






                echo "${@:1:$#-1}"


                The length argument is already in an arithmetic context, so there's no need for $(( ... )), and the number of arguments is given by $# so there's no need to try to use the equivalent of ${#...[@]} on $@.






                share|improve this answer












                echo "${@:1:$#-1}"


                The length argument is already in an arithmetic context, so there's no need for $(( ... )), and the number of arguments is given by $# so there's no need to try to use the equivalent of ${#...[@]} on $@.







                share|improve this answer












                share|improve this answer



                share|improve this answer










                answered 16 hours ago









                Kusalananda

                119k16223365




                119k16223365






















                    Bret Joseph is a new contributor. Be nice, and check out our Code of Conduct.










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                    Bret Joseph is a new contributor. Be nice, and check out our Code of Conduct.
















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