there, In my bash script, I define a command line function:
rep() {
find ./ -type f -exec sed -i -e 's/$1/$2/g' {} \;
}
After source ~/.bashrc
when I type: rep get foo
It doesn't work. Does anyone know what's going on here?
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"It doesn't work" is very unspecific. Until I read the answer I thought your code failed to define the function somehow. My bad for being on the wrong track, but as the instructions in the help section amplify many times over, you want to be as specific as possible next time.tripleee– tripleee2015-06-26 03:41:56 +00:00Commented Jun 26, 2015 at 3:41
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1 Answer
Single quotes around the sed 's/$1/$2/g' prevent bash from evaluating the $1 and $2 function arguments into strings - your command right now will swap a file with $1 in it for $2.
Try using double quotes instead.
rep() {
find . -type f -exec sed -i -e "s/$1/$2/g" {} \;
}
1 Comment
tripleee
Actually, the OP's attempt would not swap literal
$1 because $ in a regex represents the end of line. There can obviously not be anything after end of line, so $1 would never match anything; and so the OP's script would do nothing at all (except needlessly copy the file over itself).