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.
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      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
            1






            active

            oldest

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            active

            oldest

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            active

            oldest

            votes








            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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