Temporarily declare a variable in Bash











up vote
1
down vote

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To declare a variable in Bash, say in a Bash script-file (that doesn't include Bash function), I do for example x=y and when I finish using it inside that script-file I do unset x.



Is there a way (without using a function), to unset the variable after say 5 minutes, both in one line? A plausible approach might be x=y && echo "unset x" | at now + 5 minutes.



In this particular case I run the script-file directly in the terminal by copy-pasting its content from GitHub to terminal. This falls under sourcing I assume".





Given I use GitHub an alternative might be executing a raw version of the script-file directly from GitHub as follows but I don't like that way because it can't be user-agnostic or repo-agnostic (of course):



wget -O - https://raw.githubusercontent.com/<username>/<project>/<branch>/<path>/<file> | bash









share|improve this question




















  • 1




    You need to do unset x only if you are sourcing the script file. Otherwise the script runs in a subshell and does not affect your shell environment, so you don't need to unset it. I use parentheses to run one-liners with temporary variables on the command line all the time, like ( for i in {1..20}; do dosomething; done), and after I execute this command I don't have $i in my shell.
    – Weijun Zhou
    yesterday










  • If you're trying to do that for "security" purposes (ie. that var holds a password), then keep in mind that unsetting a variable does not guarantee in any way that its content will be scrubbed from memory. Simply put, keeping sensitive data in plain text in shell variables is not something that should be done for 5 secs, 5 minutes or 5 hours.
    – pizdelect
    yesterday








  • 1




    @WeijunZhou I updated the question due to your comment.
    – JohnDoea
    15 hours ago















up vote
1
down vote

favorite
2












To declare a variable in Bash, say in a Bash script-file (that doesn't include Bash function), I do for example x=y and when I finish using it inside that script-file I do unset x.



Is there a way (without using a function), to unset the variable after say 5 minutes, both in one line? A plausible approach might be x=y && echo "unset x" | at now + 5 minutes.



In this particular case I run the script-file directly in the terminal by copy-pasting its content from GitHub to terminal. This falls under sourcing I assume".





Given I use GitHub an alternative might be executing a raw version of the script-file directly from GitHub as follows but I don't like that way because it can't be user-agnostic or repo-agnostic (of course):



wget -O - https://raw.githubusercontent.com/<username>/<project>/<branch>/<path>/<file> | bash









share|improve this question




















  • 1




    You need to do unset x only if you are sourcing the script file. Otherwise the script runs in a subshell and does not affect your shell environment, so you don't need to unset it. I use parentheses to run one-liners with temporary variables on the command line all the time, like ( for i in {1..20}; do dosomething; done), and after I execute this command I don't have $i in my shell.
    – Weijun Zhou
    yesterday










  • If you're trying to do that for "security" purposes (ie. that var holds a password), then keep in mind that unsetting a variable does not guarantee in any way that its content will be scrubbed from memory. Simply put, keeping sensitive data in plain text in shell variables is not something that should be done for 5 secs, 5 minutes or 5 hours.
    – pizdelect
    yesterday








  • 1




    @WeijunZhou I updated the question due to your comment.
    – JohnDoea
    15 hours ago













up vote
1
down vote

favorite
2









up vote
1
down vote

favorite
2






2





To declare a variable in Bash, say in a Bash script-file (that doesn't include Bash function), I do for example x=y and when I finish using it inside that script-file I do unset x.



Is there a way (without using a function), to unset the variable after say 5 minutes, both in one line? A plausible approach might be x=y && echo "unset x" | at now + 5 minutes.



In this particular case I run the script-file directly in the terminal by copy-pasting its content from GitHub to terminal. This falls under sourcing I assume".





Given I use GitHub an alternative might be executing a raw version of the script-file directly from GitHub as follows but I don't like that way because it can't be user-agnostic or repo-agnostic (of course):



wget -O - https://raw.githubusercontent.com/<username>/<project>/<branch>/<path>/<file> | bash









share|improve this question















To declare a variable in Bash, say in a Bash script-file (that doesn't include Bash function), I do for example x=y and when I finish using it inside that script-file I do unset x.



Is there a way (without using a function), to unset the variable after say 5 minutes, both in one line? A plausible approach might be x=y && echo "unset x" | at now + 5 minutes.



In this particular case I run the script-file directly in the terminal by copy-pasting its content from GitHub to terminal. This falls under sourcing I assume".





Given I use GitHub an alternative might be executing a raw version of the script-file directly from GitHub as follows but I don't like that way because it can't be user-agnostic or repo-agnostic (of course):



wget -O - https://raw.githubusercontent.com/<username>/<project>/<branch>/<path>/<file> | bash






bash variable at






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited 13 hours ago

























asked yesterday









JohnDoea

821132




821132








  • 1




    You need to do unset x only if you are sourcing the script file. Otherwise the script runs in a subshell and does not affect your shell environment, so you don't need to unset it. I use parentheses to run one-liners with temporary variables on the command line all the time, like ( for i in {1..20}; do dosomething; done), and after I execute this command I don't have $i in my shell.
    – Weijun Zhou
    yesterday










  • If you're trying to do that for "security" purposes (ie. that var holds a password), then keep in mind that unsetting a variable does not guarantee in any way that its content will be scrubbed from memory. Simply put, keeping sensitive data in plain text in shell variables is not something that should be done for 5 secs, 5 minutes or 5 hours.
    – pizdelect
    yesterday








  • 1




    @WeijunZhou I updated the question due to your comment.
    – JohnDoea
    15 hours ago














  • 1




    You need to do unset x only if you are sourcing the script file. Otherwise the script runs in a subshell and does not affect your shell environment, so you don't need to unset it. I use parentheses to run one-liners with temporary variables on the command line all the time, like ( for i in {1..20}; do dosomething; done), and after I execute this command I don't have $i in my shell.
    – Weijun Zhou
    yesterday










  • If you're trying to do that for "security" purposes (ie. that var holds a password), then keep in mind that unsetting a variable does not guarantee in any way that its content will be scrubbed from memory. Simply put, keeping sensitive data in plain text in shell variables is not something that should be done for 5 secs, 5 minutes or 5 hours.
    – pizdelect
    yesterday








  • 1




    @WeijunZhou I updated the question due to your comment.
    – JohnDoea
    15 hours ago








1




1




You need to do unset x only if you are sourcing the script file. Otherwise the script runs in a subshell and does not affect your shell environment, so you don't need to unset it. I use parentheses to run one-liners with temporary variables on the command line all the time, like ( for i in {1..20}; do dosomething; done), and after I execute this command I don't have $i in my shell.
– Weijun Zhou
yesterday




You need to do unset x only if you are sourcing the script file. Otherwise the script runs in a subshell and does not affect your shell environment, so you don't need to unset it. I use parentheses to run one-liners with temporary variables on the command line all the time, like ( for i in {1..20}; do dosomething; done), and after I execute this command I don't have $i in my shell.
– Weijun Zhou
yesterday












If you're trying to do that for "security" purposes (ie. that var holds a password), then keep in mind that unsetting a variable does not guarantee in any way that its content will be scrubbed from memory. Simply put, keeping sensitive data in plain text in shell variables is not something that should be done for 5 secs, 5 minutes or 5 hours.
– pizdelect
yesterday






If you're trying to do that for "security" purposes (ie. that var holds a password), then keep in mind that unsetting a variable does not guarantee in any way that its content will be scrubbed from memory. Simply put, keeping sensitive data in plain text in shell variables is not something that should be done for 5 secs, 5 minutes or 5 hours.
– pizdelect
yesterday






1




1




@WeijunZhou I updated the question due to your comment.
– JohnDoea
15 hours ago




@WeijunZhou I updated the question due to your comment.
– JohnDoea
15 hours ago










4 Answers
4






active

oldest

votes

















up vote
6
down vote



accepted










You can't use at jobs because they run in a different context, and can't affect the current shell.



But we can do something similar. This code will trigger an alarm signal, which we can catch and perform action on



#!/bin/bash

x=100

trap 'unset x' SIGALRM

mypid=$$

( /bin/sleep 3 ; kill -ALRM $mypid) &

for a in 1 2 3 4 5 6
do
echo Now x=$x
sleep 1
done


This example is only 3 seconds long to demonstrate the solution; you can pick your delay as you need.



In action:



Now x=100
Now x=100
Now x=100
Now x=
Now x=
Now x=


You can easily make it one line with ;...






share|improve this answer























  • Wow - great minds think alike -- and within 30 seconds of each other! :)
    – Jeff Schaller
    yesterday










  • @JeffSchaller I noticed that :-)
    – Stephen Harris
    yesterday








  • 1




    @JeffSchaller I rolled back your edit; I specifically put a 3 second pause in as an example; a 300 second (5minute) sleep would mean the for loop won't demonstrate the solution working. I'll edit the answer to make that clear.
    – Stephen Harris
    yesterday




















up vote
6
down vote













Sure:



trap 'unset x; trap - USR1' USR1; { sleep 5m; kill -USR1 $$; } &


This sets a trap on the USR1 signal, then (cheating with a semicolon to put it on one line) groups together the sleep and kill commands into a background job. When that job completes, it will send the USR1 signal to the current shell, which will execute the trap. The trap unsets x and then clears the trap.



No functions!






share|improve this answer






























    up vote
    1
    down vote













    With your revised question, based around the command line and not a script, your simplest solution is to use a subshell to do your work.



    eg



    $ bash
    $ x=100
    $ echo $x
    100
    $ exit
    $ echo $x

    $


    All the variables you set between running bash and exiting that shell will be forgotten.



    This is very close to how the wget ... | bash part works as well, except you can cut'n'paste/type commands.






    share|improve this answer




























      up vote
      0
      down vote













      Actually the simplest way might be to just use a compound command with a variable declaration at the start.



      If you declare a variable at the start of a command line, the variable is temporarily defined only for the scope of that command.



      madumlao@lezard:~$ x=foo
      madumlao@lezard:~$ x=y bash -c 'echo $x'
      y
      madumlao@lezard:~$ echo $x
      foo


      As you can see, the line with a variable declaration only sets the variable in the environment for that line only.



      Note that the following will not work:



      madumlao@lezard:~$ x=foo
      madumlao@lezard:~$ x=y echo $x

      madumlao@lezard:~$ echo $x
      foo


      The $x in the second line is parsed during command-line parsing, before the value is actually assigned, so it is unexpectedly blank!






      share|improve this answer





















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        4 Answers
        4






        active

        oldest

        votes








        4 Answers
        4






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes








        up vote
        6
        down vote



        accepted










        You can't use at jobs because they run in a different context, and can't affect the current shell.



        But we can do something similar. This code will trigger an alarm signal, which we can catch and perform action on



        #!/bin/bash

        x=100

        trap 'unset x' SIGALRM

        mypid=$$

        ( /bin/sleep 3 ; kill -ALRM $mypid) &

        for a in 1 2 3 4 5 6
        do
        echo Now x=$x
        sleep 1
        done


        This example is only 3 seconds long to demonstrate the solution; you can pick your delay as you need.



        In action:



        Now x=100
        Now x=100
        Now x=100
        Now x=
        Now x=
        Now x=


        You can easily make it one line with ;...






        share|improve this answer























        • Wow - great minds think alike -- and within 30 seconds of each other! :)
          – Jeff Schaller
          yesterday










        • @JeffSchaller I noticed that :-)
          – Stephen Harris
          yesterday








        • 1




          @JeffSchaller I rolled back your edit; I specifically put a 3 second pause in as an example; a 300 second (5minute) sleep would mean the for loop won't demonstrate the solution working. I'll edit the answer to make that clear.
          – Stephen Harris
          yesterday

















        up vote
        6
        down vote



        accepted










        You can't use at jobs because they run in a different context, and can't affect the current shell.



        But we can do something similar. This code will trigger an alarm signal, which we can catch and perform action on



        #!/bin/bash

        x=100

        trap 'unset x' SIGALRM

        mypid=$$

        ( /bin/sleep 3 ; kill -ALRM $mypid) &

        for a in 1 2 3 4 5 6
        do
        echo Now x=$x
        sleep 1
        done


        This example is only 3 seconds long to demonstrate the solution; you can pick your delay as you need.



        In action:



        Now x=100
        Now x=100
        Now x=100
        Now x=
        Now x=
        Now x=


        You can easily make it one line with ;...






        share|improve this answer























        • Wow - great minds think alike -- and within 30 seconds of each other! :)
          – Jeff Schaller
          yesterday










        • @JeffSchaller I noticed that :-)
          – Stephen Harris
          yesterday








        • 1




          @JeffSchaller I rolled back your edit; I specifically put a 3 second pause in as an example; a 300 second (5minute) sleep would mean the for loop won't demonstrate the solution working. I'll edit the answer to make that clear.
          – Stephen Harris
          yesterday















        up vote
        6
        down vote



        accepted







        up vote
        6
        down vote



        accepted






        You can't use at jobs because they run in a different context, and can't affect the current shell.



        But we can do something similar. This code will trigger an alarm signal, which we can catch and perform action on



        #!/bin/bash

        x=100

        trap 'unset x' SIGALRM

        mypid=$$

        ( /bin/sleep 3 ; kill -ALRM $mypid) &

        for a in 1 2 3 4 5 6
        do
        echo Now x=$x
        sleep 1
        done


        This example is only 3 seconds long to demonstrate the solution; you can pick your delay as you need.



        In action:



        Now x=100
        Now x=100
        Now x=100
        Now x=
        Now x=
        Now x=


        You can easily make it one line with ;...






        share|improve this answer














        You can't use at jobs because they run in a different context, and can't affect the current shell.



        But we can do something similar. This code will trigger an alarm signal, which we can catch and perform action on



        #!/bin/bash

        x=100

        trap 'unset x' SIGALRM

        mypid=$$

        ( /bin/sleep 3 ; kill -ALRM $mypid) &

        for a in 1 2 3 4 5 6
        do
        echo Now x=$x
        sleep 1
        done


        This example is only 3 seconds long to demonstrate the solution; you can pick your delay as you need.



        In action:



        Now x=100
        Now x=100
        Now x=100
        Now x=
        Now x=
        Now x=


        You can easily make it one line with ;...







        share|improve this answer














        share|improve this answer



        share|improve this answer








        edited yesterday

























        answered yesterday









        Stephen Harris

        24k24477




        24k24477












        • Wow - great minds think alike -- and within 30 seconds of each other! :)
          – Jeff Schaller
          yesterday










        • @JeffSchaller I noticed that :-)
          – Stephen Harris
          yesterday








        • 1




          @JeffSchaller I rolled back your edit; I specifically put a 3 second pause in as an example; a 300 second (5minute) sleep would mean the for loop won't demonstrate the solution working. I'll edit the answer to make that clear.
          – Stephen Harris
          yesterday




















        • Wow - great minds think alike -- and within 30 seconds of each other! :)
          – Jeff Schaller
          yesterday










        • @JeffSchaller I noticed that :-)
          – Stephen Harris
          yesterday








        • 1




          @JeffSchaller I rolled back your edit; I specifically put a 3 second pause in as an example; a 300 second (5minute) sleep would mean the for loop won't demonstrate the solution working. I'll edit the answer to make that clear.
          – Stephen Harris
          yesterday


















        Wow - great minds think alike -- and within 30 seconds of each other! :)
        – Jeff Schaller
        yesterday




        Wow - great minds think alike -- and within 30 seconds of each other! :)
        – Jeff Schaller
        yesterday












        @JeffSchaller I noticed that :-)
        – Stephen Harris
        yesterday






        @JeffSchaller I noticed that :-)
        – Stephen Harris
        yesterday






        1




        1




        @JeffSchaller I rolled back your edit; I specifically put a 3 second pause in as an example; a 300 second (5minute) sleep would mean the for loop won't demonstrate the solution working. I'll edit the answer to make that clear.
        – Stephen Harris
        yesterday






        @JeffSchaller I rolled back your edit; I specifically put a 3 second pause in as an example; a 300 second (5minute) sleep would mean the for loop won't demonstrate the solution working. I'll edit the answer to make that clear.
        – Stephen Harris
        yesterday














        up vote
        6
        down vote













        Sure:



        trap 'unset x; trap - USR1' USR1; { sleep 5m; kill -USR1 $$; } &


        This sets a trap on the USR1 signal, then (cheating with a semicolon to put it on one line) groups together the sleep and kill commands into a background job. When that job completes, it will send the USR1 signal to the current shell, which will execute the trap. The trap unsets x and then clears the trap.



        No functions!






        share|improve this answer



























          up vote
          6
          down vote













          Sure:



          trap 'unset x; trap - USR1' USR1; { sleep 5m; kill -USR1 $$; } &


          This sets a trap on the USR1 signal, then (cheating with a semicolon to put it on one line) groups together the sleep and kill commands into a background job. When that job completes, it will send the USR1 signal to the current shell, which will execute the trap. The trap unsets x and then clears the trap.



          No functions!






          share|improve this answer

























            up vote
            6
            down vote










            up vote
            6
            down vote









            Sure:



            trap 'unset x; trap - USR1' USR1; { sleep 5m; kill -USR1 $$; } &


            This sets a trap on the USR1 signal, then (cheating with a semicolon to put it on one line) groups together the sleep and kill commands into a background job. When that job completes, it will send the USR1 signal to the current shell, which will execute the trap. The trap unsets x and then clears the trap.



            No functions!






            share|improve this answer














            Sure:



            trap 'unset x; trap - USR1' USR1; { sleep 5m; kill -USR1 $$; } &


            This sets a trap on the USR1 signal, then (cheating with a semicolon to put it on one line) groups together the sleep and kill commands into a background job. When that job completes, it will send the USR1 signal to the current shell, which will execute the trap. The trap unsets x and then clears the trap.



            No functions!







            share|improve this answer














            share|improve this answer



            share|improve this answer








            edited yesterday

























            answered yesterday









            Jeff Schaller

            37.8k1053122




            37.8k1053122






















                up vote
                1
                down vote













                With your revised question, based around the command line and not a script, your simplest solution is to use a subshell to do your work.



                eg



                $ bash
                $ x=100
                $ echo $x
                100
                $ exit
                $ echo $x

                $


                All the variables you set between running bash and exiting that shell will be forgotten.



                This is very close to how the wget ... | bash part works as well, except you can cut'n'paste/type commands.






                share|improve this answer

























                  up vote
                  1
                  down vote













                  With your revised question, based around the command line and not a script, your simplest solution is to use a subshell to do your work.



                  eg



                  $ bash
                  $ x=100
                  $ echo $x
                  100
                  $ exit
                  $ echo $x

                  $


                  All the variables you set between running bash and exiting that shell will be forgotten.



                  This is very close to how the wget ... | bash part works as well, except you can cut'n'paste/type commands.






                  share|improve this answer























                    up vote
                    1
                    down vote










                    up vote
                    1
                    down vote









                    With your revised question, based around the command line and not a script, your simplest solution is to use a subshell to do your work.



                    eg



                    $ bash
                    $ x=100
                    $ echo $x
                    100
                    $ exit
                    $ echo $x

                    $


                    All the variables you set between running bash and exiting that shell will be forgotten.



                    This is very close to how the wget ... | bash part works as well, except you can cut'n'paste/type commands.






                    share|improve this answer












                    With your revised question, based around the command line and not a script, your simplest solution is to use a subshell to do your work.



                    eg



                    $ bash
                    $ x=100
                    $ echo $x
                    100
                    $ exit
                    $ echo $x

                    $


                    All the variables you set between running bash and exiting that shell will be forgotten.



                    This is very close to how the wget ... | bash part works as well, except you can cut'n'paste/type commands.







                    share|improve this answer












                    share|improve this answer



                    share|improve this answer










                    answered 13 hours ago









                    Stephen Harris

                    24k24477




                    24k24477






















                        up vote
                        0
                        down vote













                        Actually the simplest way might be to just use a compound command with a variable declaration at the start.



                        If you declare a variable at the start of a command line, the variable is temporarily defined only for the scope of that command.



                        madumlao@lezard:~$ x=foo
                        madumlao@lezard:~$ x=y bash -c 'echo $x'
                        y
                        madumlao@lezard:~$ echo $x
                        foo


                        As you can see, the line with a variable declaration only sets the variable in the environment for that line only.



                        Note that the following will not work:



                        madumlao@lezard:~$ x=foo
                        madumlao@lezard:~$ x=y echo $x

                        madumlao@lezard:~$ echo $x
                        foo


                        The $x in the second line is parsed during command-line parsing, before the value is actually assigned, so it is unexpectedly blank!






                        share|improve this answer

























                          up vote
                          0
                          down vote













                          Actually the simplest way might be to just use a compound command with a variable declaration at the start.



                          If you declare a variable at the start of a command line, the variable is temporarily defined only for the scope of that command.



                          madumlao@lezard:~$ x=foo
                          madumlao@lezard:~$ x=y bash -c 'echo $x'
                          y
                          madumlao@lezard:~$ echo $x
                          foo


                          As you can see, the line with a variable declaration only sets the variable in the environment for that line only.



                          Note that the following will not work:



                          madumlao@lezard:~$ x=foo
                          madumlao@lezard:~$ x=y echo $x

                          madumlao@lezard:~$ echo $x
                          foo


                          The $x in the second line is parsed during command-line parsing, before the value is actually assigned, so it is unexpectedly blank!






                          share|improve this answer























                            up vote
                            0
                            down vote










                            up vote
                            0
                            down vote









                            Actually the simplest way might be to just use a compound command with a variable declaration at the start.



                            If you declare a variable at the start of a command line, the variable is temporarily defined only for the scope of that command.



                            madumlao@lezard:~$ x=foo
                            madumlao@lezard:~$ x=y bash -c 'echo $x'
                            y
                            madumlao@lezard:~$ echo $x
                            foo


                            As you can see, the line with a variable declaration only sets the variable in the environment for that line only.



                            Note that the following will not work:



                            madumlao@lezard:~$ x=foo
                            madumlao@lezard:~$ x=y echo $x

                            madumlao@lezard:~$ echo $x
                            foo


                            The $x in the second line is parsed during command-line parsing, before the value is actually assigned, so it is unexpectedly blank!






                            share|improve this answer












                            Actually the simplest way might be to just use a compound command with a variable declaration at the start.



                            If you declare a variable at the start of a command line, the variable is temporarily defined only for the scope of that command.



                            madumlao@lezard:~$ x=foo
                            madumlao@lezard:~$ x=y bash -c 'echo $x'
                            y
                            madumlao@lezard:~$ echo $x
                            foo


                            As you can see, the line with a variable declaration only sets the variable in the environment for that line only.



                            Note that the following will not work:



                            madumlao@lezard:~$ x=foo
                            madumlao@lezard:~$ x=y echo $x

                            madumlao@lezard:~$ echo $x
                            foo


                            The $x in the second line is parsed during command-line parsing, before the value is actually assigned, so it is unexpectedly blank!







                            share|improve this answer












                            share|improve this answer



                            share|improve this answer










                            answered 14 hours ago









                            madumlao

                            1,13166




                            1,13166






























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