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How do I reuse previous arguments in reverse order in Bash?

Posted March 13, 2015 2.9k views
Linux BasicsGitLinux Commands

How do I do this in one line reusing and reversing the arguments? Also what’s the syntax to create a function that will do this?

mv newname.js oldname.js
git mv oldname.js newname.js

for example to make a folder then cd into it you

mkdir new-folder-name && cd $_

is it something like
mv newname.js oldname.js && git mv $2 $1

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gdi2290
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3 comments
  • sierracircle March 14, 2015
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    have you considered writing a script? It is fairly simple, and you can add variables when you run it.

    just create a text file, save it with whatever you want to name it, and then make it executable with:

     chmod +x   file-name

    a quick example of a script, which I can pass two variable to when I run it:

    #!/bin/bash

echo $1
echo $2

exit 0

    now if I run that script like this:

    file-name   "first variable"  "second variable"

    it should echo:

    first variable
    second variable

  • gdi2290 March 14, 2015
    Reply Report

    thanks, I made the file github.com/gdi2290/git-mv on github

  • sierracircle March 14, 2015
    Reply Report

    awesomer.

    One more fun step: you can make a line in your .bashrc file (if you dont have a .bashrc file, just create a file called .bashrc in your users home directory)

    add this:

    
#run script
alias funscriptname="/your/script-file"

    reload your bashrc:

    . ~/.bashrc

    and now you can run this:

    funscriptname  "variable one"  "variable two"

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1 answer
ryanpq June 9, 2016

This question was answered by @sierracircle:

have you considered writing a script? It is fairly simple, and you can add variables when you run it.

just create a text file, save it with whatever you want to name it, and then make it executable with:

chmod +x   file-name

a quick example of a script, which I can pass two variable to when I run it:

#!/bin/bash

echo $1
echo $2

exit 0

now if I run that script like this:

file-name   "first variable"  "second variable"

it should echo:

first variable
second variable

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