Why is my function not re-evaluated in PS1?











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8
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I'm trying to have a part of my prompt set dynamically by a function, so in my .bashrc I have:



asdf ()
{
echo -n $(pwd)
}
PS1="u@h:w $(asdf)$ "


Opening a shell gives me what I expect at first:



$ bash
darthbith@server:~/test /home/darthbith/test$


However, when I change directory, the part defined by the function doesn't change:



darthbith@server:~/test /home/darthbith/test$ cd ~/test2
darthbith@server:~/test2 /home/darthbith/test$


My actual goal is to use the git-prompt.sh script to show the branch of my git repository when I'm in one with pretty colors and everything, but the problem is that it never updates the branch name when I change repositories. The trivial example above is the simplest reproduction I could come up with for my question.



The .bashrc lines that I have to integrate the git-prompt script:



source ~/.git-prompt.sh
PS1="[33[01;32m]u@h[33[00m]:[33[01;34m]w[33[00m]$(__git_ps1)$ "









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  • On a side note, if you want a git prompt I would recommend github.com/magicmonty/bash-git-prompt/blob/master/README.md
    – mgor
    Jul 23 '15 at 18:38















up vote
8
down vote

favorite












I'm trying to have a part of my prompt set dynamically by a function, so in my .bashrc I have:



asdf ()
{
echo -n $(pwd)
}
PS1="u@h:w $(asdf)$ "


Opening a shell gives me what I expect at first:



$ bash
darthbith@server:~/test /home/darthbith/test$


However, when I change directory, the part defined by the function doesn't change:



darthbith@server:~/test /home/darthbith/test$ cd ~/test2
darthbith@server:~/test2 /home/darthbith/test$


My actual goal is to use the git-prompt.sh script to show the branch of my git repository when I'm in one with pretty colors and everything, but the problem is that it never updates the branch name when I change repositories. The trivial example above is the simplest reproduction I could come up with for my question.



The .bashrc lines that I have to integrate the git-prompt script:



source ~/.git-prompt.sh
PS1="[33[01;32m]u@h[33[00m]:[33[01;34m]w[33[00m]$(__git_ps1)$ "









share|improve this question
























  • On a side note, if you want a git prompt I would recommend github.com/magicmonty/bash-git-prompt/blob/master/README.md
    – mgor
    Jul 23 '15 at 18:38













up vote
8
down vote

favorite









up vote
8
down vote

favorite











I'm trying to have a part of my prompt set dynamically by a function, so in my .bashrc I have:



asdf ()
{
echo -n $(pwd)
}
PS1="u@h:w $(asdf)$ "


Opening a shell gives me what I expect at first:



$ bash
darthbith@server:~/test /home/darthbith/test$


However, when I change directory, the part defined by the function doesn't change:



darthbith@server:~/test /home/darthbith/test$ cd ~/test2
darthbith@server:~/test2 /home/darthbith/test$


My actual goal is to use the git-prompt.sh script to show the branch of my git repository when I'm in one with pretty colors and everything, but the problem is that it never updates the branch name when I change repositories. The trivial example above is the simplest reproduction I could come up with for my question.



The .bashrc lines that I have to integrate the git-prompt script:



source ~/.git-prompt.sh
PS1="[33[01;32m]u@h[33[00m]:[33[01;34m]w[33[00m]$(__git_ps1)$ "









share|improve this question















I'm trying to have a part of my prompt set dynamically by a function, so in my .bashrc I have:



asdf ()
{
echo -n $(pwd)
}
PS1="u@h:w $(asdf)$ "


Opening a shell gives me what I expect at first:



$ bash
darthbith@server:~/test /home/darthbith/test$


However, when I change directory, the part defined by the function doesn't change:



darthbith@server:~/test /home/darthbith/test$ cd ~/test2
darthbith@server:~/test2 /home/darthbith/test$


My actual goal is to use the git-prompt.sh script to show the branch of my git repository when I'm in one with pretty colors and everything, but the problem is that it never updates the branch name when I change repositories. The trivial example above is the simplest reproduction I could come up with for my question.



The .bashrc lines that I have to integrate the git-prompt script:



source ~/.git-prompt.sh
PS1="[33[01;32m]u@h[33[00m]:[33[01;34m]w[33[00m]$(__git_ps1)$ "






command-line bash bashrc prompt ps1






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edited Nov 22 at 20:45









wjandrea

7,87342258




7,87342258










asked Jul 23 '15 at 18:28









darthbith

15019




15019












  • On a side note, if you want a git prompt I would recommend github.com/magicmonty/bash-git-prompt/blob/master/README.md
    – mgor
    Jul 23 '15 at 18:38


















  • On a side note, if you want a git prompt I would recommend github.com/magicmonty/bash-git-prompt/blob/master/README.md
    – mgor
    Jul 23 '15 at 18:38
















On a side note, if you want a git prompt I would recommend github.com/magicmonty/bash-git-prompt/blob/master/README.md
– mgor
Jul 23 '15 at 18:38




On a side note, if you want a git prompt I would recommend github.com/magicmonty/bash-git-prompt/blob/master/README.md
– mgor
Jul 23 '15 at 18:38










2 Answers
2






active

oldest

votes

















up vote
14
down vote



accepted










According to Bash prompt Howto:




[21:58:33][giles@nikola:~]$ PS1="[$(date +%H%M)][u@h:w]$ "
[2159][giles@nikola:~]$ ls
bin mail
[2200][giles@nikola:~]$


It's important to notice the backslash before the dollar sign of the command substitution. Without it, the external command is executed exactly once: when the PS1 string is read into the environment.







share|improve this answer























  • Thank you! Now if only I could get it to print the colors instead of the escape sequences returned from the function...
    – darthbith
    Jul 23 '15 at 20:23


















up vote
6
down vote













When you used $(..) in double-quotes, the shell evaluated the command substitution before assigning to PS1. Thus, PS1 contained only the output, not the command substitution itself. Instead, either use single-quotes, or escape the $, so that the string is passed as-is to PS1, and then evaluated when the prompt is set:



$ PS1='$(pwd) $ '
/tmp $ cd /var
/var $ echo "$PS1"
$(pwd) $


Compare:



/var $ PS1="$(pwd) $ "
/var $ echo "$PS1"
a /var $ a
/var $





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    2 Answers
    2






    active

    oldest

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    2 Answers
    2






    active

    oldest

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    active

    oldest

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    up vote
    14
    down vote



    accepted










    According to Bash prompt Howto:




    [21:58:33][giles@nikola:~]$ PS1="[$(date +%H%M)][u@h:w]$ "
    [2159][giles@nikola:~]$ ls
    bin mail
    [2200][giles@nikola:~]$


    It's important to notice the backslash before the dollar sign of the command substitution. Without it, the external command is executed exactly once: when the PS1 string is read into the environment.







    share|improve this answer























    • Thank you! Now if only I could get it to print the colors instead of the escape sequences returned from the function...
      – darthbith
      Jul 23 '15 at 20:23















    up vote
    14
    down vote



    accepted










    According to Bash prompt Howto:




    [21:58:33][giles@nikola:~]$ PS1="[$(date +%H%M)][u@h:w]$ "
    [2159][giles@nikola:~]$ ls
    bin mail
    [2200][giles@nikola:~]$


    It's important to notice the backslash before the dollar sign of the command substitution. Without it, the external command is executed exactly once: when the PS1 string is read into the environment.







    share|improve this answer























    • Thank you! Now if only I could get it to print the colors instead of the escape sequences returned from the function...
      – darthbith
      Jul 23 '15 at 20:23













    up vote
    14
    down vote



    accepted







    up vote
    14
    down vote



    accepted






    According to Bash prompt Howto:




    [21:58:33][giles@nikola:~]$ PS1="[$(date +%H%M)][u@h:w]$ "
    [2159][giles@nikola:~]$ ls
    bin mail
    [2200][giles@nikola:~]$


    It's important to notice the backslash before the dollar sign of the command substitution. Without it, the external command is executed exactly once: when the PS1 string is read into the environment.







    share|improve this answer














    According to Bash prompt Howto:




    [21:58:33][giles@nikola:~]$ PS1="[$(date +%H%M)][u@h:w]$ "
    [2159][giles@nikola:~]$ ls
    bin mail
    [2200][giles@nikola:~]$


    It's important to notice the backslash before the dollar sign of the command substitution. Without it, the external command is executed exactly once: when the PS1 string is read into the environment.








    share|improve this answer














    share|improve this answer



    share|improve this answer








    edited Nov 22 at 20:46









    wjandrea

    7,87342258




    7,87342258










    answered Jul 23 '15 at 18:36









    mgor

    881410




    881410












    • Thank you! Now if only I could get it to print the colors instead of the escape sequences returned from the function...
      – darthbith
      Jul 23 '15 at 20:23


















    • Thank you! Now if only I could get it to print the colors instead of the escape sequences returned from the function...
      – darthbith
      Jul 23 '15 at 20:23
















    Thank you! Now if only I could get it to print the colors instead of the escape sequences returned from the function...
    – darthbith
    Jul 23 '15 at 20:23




    Thank you! Now if only I could get it to print the colors instead of the escape sequences returned from the function...
    – darthbith
    Jul 23 '15 at 20:23












    up vote
    6
    down vote













    When you used $(..) in double-quotes, the shell evaluated the command substitution before assigning to PS1. Thus, PS1 contained only the output, not the command substitution itself. Instead, either use single-quotes, or escape the $, so that the string is passed as-is to PS1, and then evaluated when the prompt is set:



    $ PS1='$(pwd) $ '
    /tmp $ cd /var
    /var $ echo "$PS1"
    $(pwd) $


    Compare:



    /var $ PS1="$(pwd) $ "
    /var $ echo "$PS1"
    a /var $ a
    /var $





    share|improve this answer

























      up vote
      6
      down vote













      When you used $(..) in double-quotes, the shell evaluated the command substitution before assigning to PS1. Thus, PS1 contained only the output, not the command substitution itself. Instead, either use single-quotes, or escape the $, so that the string is passed as-is to PS1, and then evaluated when the prompt is set:



      $ PS1='$(pwd) $ '
      /tmp $ cd /var
      /var $ echo "$PS1"
      $(pwd) $


      Compare:



      /var $ PS1="$(pwd) $ "
      /var $ echo "$PS1"
      a /var $ a
      /var $





      share|improve this answer























        up vote
        6
        down vote










        up vote
        6
        down vote









        When you used $(..) in double-quotes, the shell evaluated the command substitution before assigning to PS1. Thus, PS1 contained only the output, not the command substitution itself. Instead, either use single-quotes, or escape the $, so that the string is passed as-is to PS1, and then evaluated when the prompt is set:



        $ PS1='$(pwd) $ '
        /tmp $ cd /var
        /var $ echo "$PS1"
        $(pwd) $


        Compare:



        /var $ PS1="$(pwd) $ "
        /var $ echo "$PS1"
        a /var $ a
        /var $





        share|improve this answer












        When you used $(..) in double-quotes, the shell evaluated the command substitution before assigning to PS1. Thus, PS1 contained only the output, not the command substitution itself. Instead, either use single-quotes, or escape the $, so that the string is passed as-is to PS1, and then evaluated when the prompt is set:



        $ PS1='$(pwd) $ '
        /tmp $ cd /var
        /var $ echo "$PS1"
        $(pwd) $


        Compare:



        /var $ PS1="$(pwd) $ "
        /var $ echo "$PS1"
        a /var $ a
        /var $






        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered Jul 23 '15 at 18:34









        muru

        134k19282483




        134k19282483






























             

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