How to print input arguments in bash script as is












1















Could you please help, does exist a command in bash to quote input argument?



script ./test.sh:



#!/bin/bash



echo ${1}


./test.sh "It costs $1"
This prints It costs, but how to print it as is It costs $1.



Of course it is possible to quote the argument directly in the command:
./test.sh "It costs $1"
and it prints It costs $1. But how to quote it in the script?



UPDATED: It is possible with single quotes ./test.sh 'It costs $1'






bash quote







share|improve this question




















  • 1





    Exactly as you just did: echo "It costs ${1}", or with single quotes: echo 'It costs ${1}'.

    – Amadan
    Nov 26 '18 at 3:19






  • 1





    When your script executes, it's already too late. There is nothing your script can do to get the original misquoted $1

    – that other guy
    Nov 26 '18 at 3:20











  • Did you mean to print the an argument, or that the cost is a single dollar? $1 will do the latter. If you run as-is with just ./test.sh you'll get what you're seeing, but try ./test.sh "a buck". For real clarity, read this manual, especially here & here.

    – Paul Hodges
    Nov 26 '18 at 15:25
















1















Could you please help, does exist a command in bash to quote input argument?



script ./test.sh:



#!/bin/bash



echo ${1}


./test.sh "It costs $1"
This prints It costs, but how to print it as is It costs $1.



Of course it is possible to quote the argument directly in the command:
./test.sh "It costs $1"
and it prints It costs $1. But how to quote it in the script?



UPDATED: It is possible with single quotes ./test.sh 'It costs $1'






bash quote







share|improve this question




















  • 1





    Exactly as you just did: echo "It costs ${1}", or with single quotes: echo 'It costs ${1}'.

    – Amadan
    Nov 26 '18 at 3:19






  • 1





    When your script executes, it's already too late. There is nothing your script can do to get the original misquoted $1

    – that other guy
    Nov 26 '18 at 3:20











  • Did you mean to print the an argument, or that the cost is a single dollar? $1 will do the latter. If you run as-is with just ./test.sh you'll get what you're seeing, but try ./test.sh "a buck". For real clarity, read this manual, especially here & here.

    – Paul Hodges
    Nov 26 '18 at 15:25














1












1








1








Could you please help, does exist a command in bash to quote input argument?



script ./test.sh:



#!/bin/bash



echo ${1}


./test.sh "It costs $1"
This prints It costs, but how to print it as is It costs $1.



Of course it is possible to quote the argument directly in the command:
./test.sh "It costs $1"
and it prints It costs $1. But how to quote it in the script?



UPDATED: It is possible with single quotes ./test.sh 'It costs $1'






bash quote







share|improve this question
















Could you please help, does exist a command in bash to quote input argument?



script ./test.sh:



#!/bin/bash



echo ${1}


./test.sh "It costs $1"
This prints It costs, but how to print it as is It costs $1.



Of course it is possible to quote the argument directly in the command:
./test.sh "It costs $1"
and it prints It costs $1. But how to quote it in the script?



UPDATED: It is possible with single quotes ./test.sh 'It costs $1'






bash quote




bash quote






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Nov 26 '18 at 3:53







Walrus

















asked Nov 26 '18 at 3:17









WalrusWalrus

7418




7418








  • 1





    Exactly as you just did: echo "It costs ${1}", or with single quotes: echo 'It costs ${1}'.

    – Amadan
    Nov 26 '18 at 3:19






  • 1





    When your script executes, it's already too late. There is nothing your script can do to get the original misquoted $1

    – that other guy
    Nov 26 '18 at 3:20











  • Did you mean to print the an argument, or that the cost is a single dollar? $1 will do the latter. If you run as-is with just ./test.sh you'll get what you're seeing, but try ./test.sh "a buck". For real clarity, read this manual, especially here & here.

    – Paul Hodges
    Nov 26 '18 at 15:25














  • 1





    Exactly as you just did: echo "It costs ${1}", or with single quotes: echo 'It costs ${1}'.

    – Amadan
    Nov 26 '18 at 3:19






  • 1





    When your script executes, it's already too late. There is nothing your script can do to get the original misquoted $1

    – that other guy
    Nov 26 '18 at 3:20











  • Did you mean to print the an argument, or that the cost is a single dollar? $1 will do the latter. If you run as-is with just ./test.sh you'll get what you're seeing, but try ./test.sh "a buck". For real clarity, read this manual, especially here & here.

    – Paul Hodges
    Nov 26 '18 at 15:25








1




1





Exactly as you just did: echo "It costs ${1}", or with single quotes: echo 'It costs ${1}'.

– Amadan
Nov 26 '18 at 3:19





Exactly as you just did: echo "It costs ${1}", or with single quotes: echo 'It costs ${1}'.

– Amadan
Nov 26 '18 at 3:19




1




1





When your script executes, it's already too late. There is nothing your script can do to get the original misquoted $1

– that other guy
Nov 26 '18 at 3:20





When your script executes, it's already too late. There is nothing your script can do to get the original misquoted $1

– that other guy
Nov 26 '18 at 3:20













Did you mean to print the an argument, or that the cost is a single dollar? $1 will do the latter. If you run as-is with just ./test.sh you'll get what you're seeing, but try ./test.sh "a buck". For real clarity, read this manual, especially here & here.

– Paul Hodges
Nov 26 '18 at 15:25





Did you mean to print the an argument, or that the cost is a single dollar? $1 will do the latter. If you run as-is with just ./test.sh you'll get what you're seeing, but try ./test.sh "a buck". For real clarity, read this manual, especially here & here.

– Paul Hodges
Nov 26 '18 at 15:25












2 Answers
2






active

oldest

votes


















2














You may update your program like below to print/display all arguments.



#!/bin/bash



echo "$@"


This will print all the arguments passed while running test.sh. $1 is a variable which substituted with empty string while calling your program test.sh. Inside test.sh, first argument you passed in command line becomes $1, second becomes $2 and so forth.



$@ prints all arguments.






share|improve this answer



















  • 2





    Indeed, it is useful to do man bash and read everything under EXPANSION heading. There is so much good info there that many people never see.

    – Amadan
    Nov 26 '18 at 3:57





















0














I often use:



printf "'%s' "  "$@"


This is an alternative to echo when you want each argument quoted.






share|improve this answer























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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2














    You may update your program like below to print/display all arguments.



    #!/bin/bash
    

    
echo "$@"


    This will print all the arguments passed while running test.sh. $1 is a variable which substituted with empty string while calling your program test.sh. Inside test.sh, first argument you passed in command line becomes $1, second becomes $2 and so forth.



    $@ prints all arguments.






    share|improve this answer



















    • 2





      Indeed, it is useful to do man bash and read everything under EXPANSION heading. There is so much good info there that many people never see.

      – Amadan
      Nov 26 '18 at 3:57


















    2














    You may update your program like below to print/display all arguments.



    #!/bin/bash
    

    
echo "$@"


    This will print all the arguments passed while running test.sh. $1 is a variable which substituted with empty string while calling your program test.sh. Inside test.sh, first argument you passed in command line becomes $1, second becomes $2 and so forth.



    $@ prints all arguments.






    share|improve this answer



















    • 2





      Indeed, it is useful to do man bash and read everything under EXPANSION heading. There is so much good info there that many people never see.

      – Amadan
      Nov 26 '18 at 3:57
















    2












    2








    2







    You may update your program like below to print/display all arguments.



    #!/bin/bash
    

    
echo "$@"


    This will print all the arguments passed while running test.sh. $1 is a variable which substituted with empty string while calling your program test.sh. Inside test.sh, first argument you passed in command line becomes $1, second becomes $2 and so forth.



    $@ prints all arguments.






    share|improve this answer













    You may update your program like below to print/display all arguments.



    #!/bin/bash
    

    
echo "$@"


    This will print all the arguments passed while running test.sh. $1 is a variable which substituted with empty string while calling your program test.sh. Inside test.sh, first argument you passed in command line becomes $1, second becomes $2 and so forth.



    $@ prints all arguments.







    share|improve this answer












    share|improve this answer



    share|improve this answer










    answered Nov 26 '18 at 3:28









    Robert RanjanRobert Ranjan

    363410




    363410








    • 2





      Indeed, it is useful to do man bash and read everything under EXPANSION heading. There is so much good info there that many people never see.

      – Amadan
      Nov 26 '18 at 3:57
















    • 2





      Indeed, it is useful to do man bash and read everything under EXPANSION heading. There is so much good info there that many people never see.

      – Amadan
      Nov 26 '18 at 3:57










    2




    2





    Indeed, it is useful to do man bash and read everything under EXPANSION heading. There is so much good info there that many people never see.

    – Amadan
    Nov 26 '18 at 3:57







    Indeed, it is useful to do man bash and read everything under EXPANSION heading. There is so much good info there that many people never see.

    – Amadan
    Nov 26 '18 at 3:57















    0














    I often use:



    printf "'%s' "  "$@"


    This is an alternative to echo when you want each argument quoted.






    share|improve this answer




























      0














      I often use:



      printf "'%s' "  "$@"


      This is an alternative to echo when you want each argument quoted.






      share|improve this answer


























        0












        0








        0







        I often use:



        printf "'%s' "  "$@"


        This is an alternative to echo when you want each argument quoted.






        share|improve this answer













        I often use:



        printf "'%s' "  "$@"


        This is an alternative to echo when you want each argument quoted.







        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered Nov 26 '18 at 4:19









        CraigCraig

        5368




        5368






























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