I recently saw a video where someone executed ^foo^bar in Bash. What is that combination for?
-
6Possible duplicate of ^x^y unix trick for all instances in last command? and relatedJulien Lopez– Julien Lopez2017-12-19 09:25:21 +00:00Commented Dec 19, 2017 at 9:25
Add a comment
|
2 Answers
Bash calls this a quick substitution. It's in the "History Expansion" section of the Bash man page, under the "Event Designators" section (online manual):
^string1^string2^Quick substitution. Repeat the previous command, replacing string1 with string2. Equivalent to
!!:s/string1/string2/
So ^foo^bar would run the previously executed command, but replace the first occurence of foo with bar.
Note that for s/old/new/, the bash man page says "The final
delimiter is optional if it is the last character of the
event line." This is why you can use ^foo^bar and aren't required to use ^foo^bar^.
(See this answer for a bunch of other designators, although I didn't mention this one there).
-
1Really nice. Didn't know this exists.PesaThe– PesaThe2017-12-18 21:20:42 +00:00Commented Dec 18, 2017 at 21:20
^foo^bar executes that last command, replacing the first instance of foo with bar. For example:
$ ech "hello"
-bash: ech: command not found
$ ^ech^echo
echo "hello"
hello
