In bash, how do I bind a function key to a command?

BashShellBinding

Bash Problem Overview


Example: I want to bind the F12 key to the command echo "foobar" such that every time I hit F12 the message "foobar" will be printed to screen. Ideally it could be any arbitrary shell command, not just builtins. How does one go about this?

Bash Solutions


Solution 1 - Bash

You can determine the character sequence emitted by a key by pressing Ctrl-v at the command line, then pressing the key you're interested in. On my system for F12, I get ^[[24~. The ^[ represents Esc. Different types of terminals or terminal emulators can emit different codes for the same key.

At a Bash prompt you can enter a command like this to enable the key macro so you can try it out.

bind '"\e[24~":"foobar"'

Now, when you press F12, you'll get "foobar" on the command line ready for further editing. If you wanted a keystroke to enter a command immediately, you can add a newline:

bind '"\e[24~":"pwd\n"'

Now when you press F12, you'll get the current directory displayed without having to press Enter. What if you've already typed something on the line and you use this which automatically executes? It could get messy. However, you could clear the line as part of your macro:

bind '"\e[24~":"\C-k \C-upwd\n"'

The space makes sure that the Ctrl-u has something to delete to keep the bell from ringing.

Once you've gotten the macro working the way you want, you can make it persistent by adding it to your ~/.inputrc file. There's no need for the bind command or the outer set of single quotes:

"\e[24~":"\C-k \C-upwd\n"

Edit:

You can also create a key binding that will execute something without disturbing the current command line.

bind -x '"\eW":"who"'

Then while you're typing a command that requires a username, for example, and you need to know the names of user who are logged in, you can press Alt-Shift-W and the output of who will be displayed and the prompt will be re-issued with your partial command intact and the cursor in the same position in the line.

Unfortunately, this doesn't work properly for keys such as F12 which output more than two characters. In some cases this can be worked around.

The command (who in this case) could be any executable - a program, script or function.

Solution 2 - Bash

You can define bash key bindings in ~/.inputrc (configuration file for the GNU Readline library). The syntax is

<keysym or key name>: macro

for example:

Control-o: "> output"

will create a macro which inserts "> output" when you press ControlO

 "\e[11~": "echo foobar"

will create a macro which inserts "echo foobar" when you press F1... I don't know what the keysym for F11 is off hand.

Edit:

.inputrc understands the \n escape sequence for linefeed, so you can use

 "\e[11~": "echo foobar\n"

Which will effectively 'press enter' after the command is issued.

Solution 3 - Bash

This solution is specific to X11 environments and has nothing to do with bash, but adding the following to your .Xmodmaps

 % loadkeys
 keycode 88 = F12
 string F12 = "foobar"
 %

will send the string "foobar" to the terminal upon hitting F12.

Solution 4 - Bash

I wanted to bind Ctrl+B to a command. Inspired by an answer above, I tried to use bind but could not figure out what series of cryptic squiggles (\e[24~ ?) translate to Ctrl+B.

On a Mac, go to Settings of the Terminal app, Profiles -> Keyboard -> + then press the keyboard shortcut you're after and it comes out. For me Ctrl+B resulted in \002 which i successfully bound to command

bind '"\002":"echo command"'

Also, if you want the command to be executed right-away (not just inserted in to the prompt), you can add the Enter to the end of your command, like so:

bind '"\002":"echo command\015"'

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Content TypeOriginal AuthorOriginal Content on Stackoverflow
QuestionSiegeXView Question on Stackoverflow
Solution 1 - BashDennis WilliamsonView Answer on Stackoverflow
Solution 2 - BashBarton ChittendenView Answer on Stackoverflow
Solution 3 - BashWesley RiceView Answer on Stackoverflow
Solution 4 - BashPeter PerháčView Answer on Stackoverflow