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How to prevent command substitution on the command line?


How to use a special character as a normal one?How can we prevent parameter expansion?Number of backslashes needed for escaping regex backslash on the command-lineHow do I echo an expression with both single and double quotes?Single quote within double quotes and the Bash reference manualWhat can single quote mean specially within double quotes?command substitution within single quotes for aliasWhat does POSIX require for quoted here documents inside command substitution?shell expanding rename command backreferencesBad substitution: no closing “`” in a heredoc / EOFCommand substitution, assignment, tests and spacesWhy does single quotes around arguments fail to prevent “syntax error near unexpected token '('”?






.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;








0















I find that when writing text as input to another program, any command substitutions in double quotes within the intended text are interpreted and expanded by the shell



The links in the answer here states that single quotes can be used to prevent parameter expansion or command substitution. However I'm finding that enclosing a command substitution in single-quotes also fails to stop the shell from expanding the command substitution



How do you prevent the shell from interpreting command substitutions that are intended as text rather than a command to be executed?



A demonstration



$ echo "`wc -l *`"


attempts to count lines in all files in the current directory



$ echo "'`wc -l *`'"


Same result, i.e. counts lines in all files in the current directory



update From this demonstration I've spotted that the problem seems to be that I am quoting the single quotes. I think enclosing single quotes and ` (backtick) in double quotes preserves the literal meaning of (i.e. suppresses) the single quotes but does not preserve the literal meaning of the backquote (i.e. backtick) that introduces the command substitution.



In my use case the input for another command needs to be quoted. With this document saying that:




A single-quote cannot occur within single quotes




How do you prevent a single-quoted command substitution from being expanded when the single-quoted command substitution is within a (double) quoted string? There should be a way to do it other than using backslash escapes










share|improve this question





















  • 3





    You mean echo '$(ls)' does not produce $(ls) as its output? Can you add the example that is not working for you in the question?

    – NickD
    8 hours ago












  • Note that adding single quotes within double quotes won't help.

    – Andy Dalton
    7 hours ago











  • @NickD good suggestion, I've updated the question

    – MyWrathAcademia
    7 hours ago












  • Can you not just use single quotes instead of double quotes?

    – Andy Dalton
    7 hours ago






  • 1





    You know, instead of us guessing, you should put your actual example in the question.

    – NickD
    4 hours ago

















0















I find that when writing text as input to another program, any command substitutions in double quotes within the intended text are interpreted and expanded by the shell



The links in the answer here states that single quotes can be used to prevent parameter expansion or command substitution. However I'm finding that enclosing a command substitution in single-quotes also fails to stop the shell from expanding the command substitution



How do you prevent the shell from interpreting command substitutions that are intended as text rather than a command to be executed?



A demonstration



$ echo "`wc -l *`"


attempts to count lines in all files in the current directory



$ echo "'`wc -l *`'"


Same result, i.e. counts lines in all files in the current directory



update From this demonstration I've spotted that the problem seems to be that I am quoting the single quotes. I think enclosing single quotes and ` (backtick) in double quotes preserves the literal meaning of (i.e. suppresses) the single quotes but does not preserve the literal meaning of the backquote (i.e. backtick) that introduces the command substitution.



In my use case the input for another command needs to be quoted. With this document saying that:




A single-quote cannot occur within single quotes




How do you prevent a single-quoted command substitution from being expanded when the single-quoted command substitution is within a (double) quoted string? There should be a way to do it other than using backslash escapes










share|improve this question





















  • 3





    You mean echo '$(ls)' does not produce $(ls) as its output? Can you add the example that is not working for you in the question?

    – NickD
    8 hours ago












  • Note that adding single quotes within double quotes won't help.

    – Andy Dalton
    7 hours ago











  • @NickD good suggestion, I've updated the question

    – MyWrathAcademia
    7 hours ago












  • Can you not just use single quotes instead of double quotes?

    – Andy Dalton
    7 hours ago






  • 1





    You know, instead of us guessing, you should put your actual example in the question.

    – NickD
    4 hours ago













0












0








0








I find that when writing text as input to another program, any command substitutions in double quotes within the intended text are interpreted and expanded by the shell



The links in the answer here states that single quotes can be used to prevent parameter expansion or command substitution. However I'm finding that enclosing a command substitution in single-quotes also fails to stop the shell from expanding the command substitution



How do you prevent the shell from interpreting command substitutions that are intended as text rather than a command to be executed?



A demonstration



$ echo "`wc -l *`"


attempts to count lines in all files in the current directory



$ echo "'`wc -l *`'"


Same result, i.e. counts lines in all files in the current directory



update From this demonstration I've spotted that the problem seems to be that I am quoting the single quotes. I think enclosing single quotes and ` (backtick) in double quotes preserves the literal meaning of (i.e. suppresses) the single quotes but does not preserve the literal meaning of the backquote (i.e. backtick) that introduces the command substitution.



In my use case the input for another command needs to be quoted. With this document saying that:




A single-quote cannot occur within single quotes




How do you prevent a single-quoted command substitution from being expanded when the single-quoted command substitution is within a (double) quoted string? There should be a way to do it other than using backslash escapes










share|improve this question
















I find that when writing text as input to another program, any command substitutions in double quotes within the intended text are interpreted and expanded by the shell



The links in the answer here states that single quotes can be used to prevent parameter expansion or command substitution. However I'm finding that enclosing a command substitution in single-quotes also fails to stop the shell from expanding the command substitution



How do you prevent the shell from interpreting command substitutions that are intended as text rather than a command to be executed?



A demonstration



$ echo "`wc -l *`"


attempts to count lines in all files in the current directory



$ echo "'`wc -l *`'"


Same result, i.e. counts lines in all files in the current directory



update From this demonstration I've spotted that the problem seems to be that I am quoting the single quotes. I think enclosing single quotes and ` (backtick) in double quotes preserves the literal meaning of (i.e. suppresses) the single quotes but does not preserve the literal meaning of the backquote (i.e. backtick) that introduces the command substitution.



In my use case the input for another command needs to be quoted. With this document saying that:




A single-quote cannot occur within single quotes




How do you prevent a single-quoted command substitution from being expanded when the single-quoted command substitution is within a (double) quoted string? There should be a way to do it other than using backslash escapes







bash shell command-line quoting command-substitution






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited 7 hours ago







MyWrathAcademia

















asked 8 hours ago









MyWrathAcademiaMyWrathAcademia

3462 silver badges12 bronze badges




3462 silver badges12 bronze badges










  • 3





    You mean echo '$(ls)' does not produce $(ls) as its output? Can you add the example that is not working for you in the question?

    – NickD
    8 hours ago












  • Note that adding single quotes within double quotes won't help.

    – Andy Dalton
    7 hours ago











  • @NickD good suggestion, I've updated the question

    – MyWrathAcademia
    7 hours ago












  • Can you not just use single quotes instead of double quotes?

    – Andy Dalton
    7 hours ago






  • 1





    You know, instead of us guessing, you should put your actual example in the question.

    – NickD
    4 hours ago












  • 3





    You mean echo '$(ls)' does not produce $(ls) as its output? Can you add the example that is not working for you in the question?

    – NickD
    8 hours ago












  • Note that adding single quotes within double quotes won't help.

    – Andy Dalton
    7 hours ago











  • @NickD good suggestion, I've updated the question

    – MyWrathAcademia
    7 hours ago












  • Can you not just use single quotes instead of double quotes?

    – Andy Dalton
    7 hours ago






  • 1





    You know, instead of us guessing, you should put your actual example in the question.

    – NickD
    4 hours ago







3




3





You mean echo '$(ls)' does not produce $(ls) as its output? Can you add the example that is not working for you in the question?

– NickD
8 hours ago






You mean echo '$(ls)' does not produce $(ls) as its output? Can you add the example that is not working for you in the question?

– NickD
8 hours ago














Note that adding single quotes within double quotes won't help.

– Andy Dalton
7 hours ago





Note that adding single quotes within double quotes won't help.

– Andy Dalton
7 hours ago













@NickD good suggestion, I've updated the question

– MyWrathAcademia
7 hours ago






@NickD good suggestion, I've updated the question

– MyWrathAcademia
7 hours ago














Can you not just use single quotes instead of double quotes?

– Andy Dalton
7 hours ago





Can you not just use single quotes instead of double quotes?

– Andy Dalton
7 hours ago




1




1





You know, instead of us guessing, you should put your actual example in the question.

– NickD
4 hours ago





You know, instead of us guessing, you should put your actual example in the question.

– NickD
4 hours ago










2 Answers
2






active

oldest

votes


















5














Use single-quote strong quoting:



printf '%sn' '`wc -l *`'


And if you want to also include single quotes in that argument passed to printf, you'd need to use different quotes for ' itself like:



printf '%sn' '`wc -l *` and a '"'"' character'


Or:



printf '%sn' '`wc -l *` and a ''' character'


Other alternatives include escaping the ` with backslash inside double quotes:



printf '%sn' "`wc -l *` and a ' character"


Or have ` be the result of some expansion:



backtick='`'
printf '%sn' "$backtickwc -l *$backtick and a ' character"


Also note that:



cat << 'EOF'
`wc -l *` and a ' character and a " character
EOF


to output arbitrary text without having to worry about quoting (note the quotes around the first EOF).



In the fish shell, you can embed ' within '...' with ':



printf '%sn' '`wc -l *` and a ' character'


but anyway ` is not special there, so:



printf '%sn' "`wc -l *` and a ' character"


would work as well.



In rc, es or zsh -o rcquotes, you can insert a ' within '...' with '':



printf '%sn' '`wc -l *` and a '' character'


See How to use a special character as a normal one? for more details.






share|improve this answer
































    1














    You can escape the backticks by using a backslash as shown below:



    echo "`wc -l *`"





    share|improve this answer








    New contributor



    user138278 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.





















    • I mentioned prefering solving the problem without using backticks because you would need to escape every character you don't want to be interpreted by the shell where as if quoting could solve the problem you could use one single-quote to quote a substring that contains special characters such as $, `` ` `` and ` ` that you don't want to be interpreted by the shell

      – MyWrathAcademia
      7 hours ago












    • ok, so the reason you need to use double quotes is because you have a variable that needs to be interpreted as well? So, as a result, echo 'wc -l *' wont work as solution?

      – user138278
      6 hours ago














    Your Answer








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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    5














    Use single-quote strong quoting:



    printf '%sn' '`wc -l *`'


    And if you want to also include single quotes in that argument passed to printf, you'd need to use different quotes for ' itself like:



    printf '%sn' '`wc -l *` and a '"'"' character'


    Or:



    printf '%sn' '`wc -l *` and a ''' character'


    Other alternatives include escaping the ` with backslash inside double quotes:



    printf '%sn' "`wc -l *` and a ' character"


    Or have ` be the result of some expansion:



    backtick='`'
    printf '%sn' "$backtickwc -l *$backtick and a ' character"


    Also note that:



    cat << 'EOF'
    `wc -l *` and a ' character and a " character
    EOF


    to output arbitrary text without having to worry about quoting (note the quotes around the first EOF).



    In the fish shell, you can embed ' within '...' with ':



    printf '%sn' '`wc -l *` and a ' character'


    but anyway ` is not special there, so:



    printf '%sn' "`wc -l *` and a ' character"


    would work as well.



    In rc, es or zsh -o rcquotes, you can insert a ' within '...' with '':



    printf '%sn' '`wc -l *` and a '' character'


    See How to use a special character as a normal one? for more details.






    share|improve this answer





























      5














      Use single-quote strong quoting:



      printf '%sn' '`wc -l *`'


      And if you want to also include single quotes in that argument passed to printf, you'd need to use different quotes for ' itself like:



      printf '%sn' '`wc -l *` and a '"'"' character'


      Or:



      printf '%sn' '`wc -l *` and a ''' character'


      Other alternatives include escaping the ` with backslash inside double quotes:



      printf '%sn' "`wc -l *` and a ' character"


      Or have ` be the result of some expansion:



      backtick='`'
      printf '%sn' "$backtickwc -l *$backtick and a ' character"


      Also note that:



      cat << 'EOF'
      `wc -l *` and a ' character and a " character
      EOF


      to output arbitrary text without having to worry about quoting (note the quotes around the first EOF).



      In the fish shell, you can embed ' within '...' with ':



      printf '%sn' '`wc -l *` and a ' character'


      but anyway ` is not special there, so:



      printf '%sn' "`wc -l *` and a ' character"


      would work as well.



      In rc, es or zsh -o rcquotes, you can insert a ' within '...' with '':



      printf '%sn' '`wc -l *` and a '' character'


      See How to use a special character as a normal one? for more details.






      share|improve this answer



























        5












        5








        5







        Use single-quote strong quoting:



        printf '%sn' '`wc -l *`'


        And if you want to also include single quotes in that argument passed to printf, you'd need to use different quotes for ' itself like:



        printf '%sn' '`wc -l *` and a '"'"' character'


        Or:



        printf '%sn' '`wc -l *` and a ''' character'


        Other alternatives include escaping the ` with backslash inside double quotes:



        printf '%sn' "`wc -l *` and a ' character"


        Or have ` be the result of some expansion:



        backtick='`'
        printf '%sn' "$backtickwc -l *$backtick and a ' character"


        Also note that:



        cat << 'EOF'
        `wc -l *` and a ' character and a " character
        EOF


        to output arbitrary text without having to worry about quoting (note the quotes around the first EOF).



        In the fish shell, you can embed ' within '...' with ':



        printf '%sn' '`wc -l *` and a ' character'


        but anyway ` is not special there, so:



        printf '%sn' "`wc -l *` and a ' character"


        would work as well.



        In rc, es or zsh -o rcquotes, you can insert a ' within '...' with '':



        printf '%sn' '`wc -l *` and a '' character'


        See How to use a special character as a normal one? for more details.






        share|improve this answer













        Use single-quote strong quoting:



        printf '%sn' '`wc -l *`'


        And if you want to also include single quotes in that argument passed to printf, you'd need to use different quotes for ' itself like:



        printf '%sn' '`wc -l *` and a '"'"' character'


        Or:



        printf '%sn' '`wc -l *` and a ''' character'


        Other alternatives include escaping the ` with backslash inside double quotes:



        printf '%sn' "`wc -l *` and a ' character"


        Or have ` be the result of some expansion:



        backtick='`'
        printf '%sn' "$backtickwc -l *$backtick and a ' character"


        Also note that:



        cat << 'EOF'
        `wc -l *` and a ' character and a " character
        EOF


        to output arbitrary text without having to worry about quoting (note the quotes around the first EOF).



        In the fish shell, you can embed ' within '...' with ':



        printf '%sn' '`wc -l *` and a ' character'


        but anyway ` is not special there, so:



        printf '%sn' "`wc -l *` and a ' character"


        would work as well.



        In rc, es or zsh -o rcquotes, you can insert a ' within '...' with '':



        printf '%sn' '`wc -l *` and a '' character'


        See How to use a special character as a normal one? for more details.







        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered 6 hours ago









        Stéphane ChazelasStéphane Chazelas

        327k57 gold badges636 silver badges1002 bronze badges




        327k57 gold badges636 silver badges1002 bronze badges


























            1














            You can escape the backticks by using a backslash as shown below:



            echo "`wc -l *`"





            share|improve this answer








            New contributor



            user138278 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
            Check out our Code of Conduct.





















            • I mentioned prefering solving the problem without using backticks because you would need to escape every character you don't want to be interpreted by the shell where as if quoting could solve the problem you could use one single-quote to quote a substring that contains special characters such as $, `` ` `` and ` ` that you don't want to be interpreted by the shell

              – MyWrathAcademia
              7 hours ago












            • ok, so the reason you need to use double quotes is because you have a variable that needs to be interpreted as well? So, as a result, echo 'wc -l *' wont work as solution?

              – user138278
              6 hours ago
















            1














            You can escape the backticks by using a backslash as shown below:



            echo "`wc -l *`"





            share|improve this answer








            New contributor



            user138278 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
            Check out our Code of Conduct.





















            • I mentioned prefering solving the problem without using backticks because you would need to escape every character you don't want to be interpreted by the shell where as if quoting could solve the problem you could use one single-quote to quote a substring that contains special characters such as $, `` ` `` and ` ` that you don't want to be interpreted by the shell

              – MyWrathAcademia
              7 hours ago












            • ok, so the reason you need to use double quotes is because you have a variable that needs to be interpreted as well? So, as a result, echo 'wc -l *' wont work as solution?

              – user138278
              6 hours ago














            1












            1








            1







            You can escape the backticks by using a backslash as shown below:



            echo "`wc -l *`"





            share|improve this answer








            New contributor



            user138278 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
            Check out our Code of Conduct.









            You can escape the backticks by using a backslash as shown below:



            echo "`wc -l *`"






            share|improve this answer








            New contributor



            user138278 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 answer



            share|improve this answer






            New contributor



            user138278 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
            Check out our Code of Conduct.








            answered 7 hours ago









            user138278user138278

            313 bronze badges




            313 bronze badges




            New contributor



            user138278 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
            Check out our Code of Conduct.




            New contributor




            user138278 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
            Check out our Code of Conduct.

















            • I mentioned prefering solving the problem without using backticks because you would need to escape every character you don't want to be interpreted by the shell where as if quoting could solve the problem you could use one single-quote to quote a substring that contains special characters such as $, `` ` `` and ` ` that you don't want to be interpreted by the shell

              – MyWrathAcademia
              7 hours ago












            • ok, so the reason you need to use double quotes is because you have a variable that needs to be interpreted as well? So, as a result, echo 'wc -l *' wont work as solution?

              – user138278
              6 hours ago


















            • I mentioned prefering solving the problem without using backticks because you would need to escape every character you don't want to be interpreted by the shell where as if quoting could solve the problem you could use one single-quote to quote a substring that contains special characters such as $, `` ` `` and ` ` that you don't want to be interpreted by the shell

              – MyWrathAcademia
              7 hours ago












            • ok, so the reason you need to use double quotes is because you have a variable that needs to be interpreted as well? So, as a result, echo 'wc -l *' wont work as solution?

              – user138278
              6 hours ago

















            I mentioned prefering solving the problem without using backticks because you would need to escape every character you don't want to be interpreted by the shell where as if quoting could solve the problem you could use one single-quote to quote a substring that contains special characters such as $, `` ` `` and ` ` that you don't want to be interpreted by the shell

            – MyWrathAcademia
            7 hours ago






            I mentioned prefering solving the problem without using backticks because you would need to escape every character you don't want to be interpreted by the shell where as if quoting could solve the problem you could use one single-quote to quote a substring that contains special characters such as $, `` ` `` and ` ` that you don't want to be interpreted by the shell

            – MyWrathAcademia
            7 hours ago














            ok, so the reason you need to use double quotes is because you have a variable that needs to be interpreted as well? So, as a result, echo 'wc -l *' wont work as solution?

            – user138278
            6 hours ago






            ok, so the reason you need to use double quotes is because you have a variable that needs to be interpreted as well? So, as a result, echo 'wc -l *' wont work as solution?

            – user138278
            6 hours ago


















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