how to use xargs with sed in search pattern

Question

I need to use the output of a command as a search pattern in sed. I will make an example using echo, but assume that can be a more complicated command:

echo "some pattern" | xargs sed -i 's/{}/replacement/g' file.txt

That command doesn't work because "some pattern" has a whitespace, but I think that clearly illustrate my problem.

How can I make that command work?

Thanks in advance,

Answer 1

Use command substitution instead, so your example would look like:

sed -i "s/$(echo "some pattern")/replacement/g" file.txt

The double quotes allow for the command substitution to work while preventing spaces from being split.

Answer 2

This might work for you (GNU sed):

echo "some pattern" | sed 's|.*|s/&/replacement/g|' | sed -f - -i file.txt

Essentially turn the some pattern into a sed substitution command and feed it via a pipe to another sed invocation. The last sed invocation uses the -f switch which accepts the sed commands via a file, the file in this case being the standard input -.

If you are using bash, the here-string can be employed:

<<<"some pattern" sed 's|.*|s/&/replacement/g|' | sed -f - -i file.txt

N.B. the sed separators | and / should not be a part of some pattern otherwise the regexp will not be formed properly.

Answer 3

You need to tell xargs what to replace with the -I switch - it doesn't seem to know about the {} automatically, at least in some versions.

echo "pattern" | xargs -I '{}' sed -i 's/{}/replacement/g' file.txt

Answer 4

this works on Linux(tested):

find . -type f -print0 | xargs -0 sed -i 's/str1/str2/g'