How to swap parameters in Shell command

Question

I have a command with parameters in shell. like

Command -a -b

assume a b are very long directory name, and they are not always the first and last parameters.

However, I need to use this command again. I use ctrl+p to go back the command, but this time I need to reverse a b order.

Command -b -a

Instead of retyping them again, is there any way to fast swap them?

Thanks in advance.

Answer 1

Yes you can use history substitution:

$ echo foo bar
foo bar
$ !:0 !:2 !:1          # previous command with second arg then first arg
echo bar foo
bar foo
$ 

Answer 2

You can use Alt + t to swap the current and the previous word.

Here are some more shortcuts.

Mind that you swap words, not parameters. If for instance you have mixed up the order grep takes its parameters as in grep ~/Documents/myFile searchString and want to correct that to grep searchString ~/Documents/myFile, Alt + t won't help you.

Go for Paul's answer instead and do !:0 !:2 !:1.

If that's too inconvenient for you to type (it is for me) you can create an alias for that in ~/.bash_alias:

# Swap the first and the second argument of the last command
alias swapArgs='$(history -p !:0 !:2 !:1)'

Answer 3

Instead of swapping the args, you can easily cut and paste the last arg into a different place. Ctrl-W will cut the last argument and Ctrl-Y will paste it.

$ # Cursor position is shown as |
$ ls foo bar|

$ # Press <Ctrl-W> to cut the argument just before the cursor
$ ls foo |

$ # Now press <Alt-B> to move the cursor one argument back
$ ls |foo

$ # Now press <Ctrl-Y> to paste the cut argument
$ ls bar|foo

$ # Now press <space> to insert a space between the arguments
$ ls bar |foo

$ # You can safely press <enter> while the cursor is in the middle of the line
$ ls bar |foo
bar foo