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,
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.
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
this works on Linux(tested):
find . -type f -print0 | xargs -0 sed -i 's/str1/str2/g'
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.
来源：https://stackoverflow.com/questions/14402949/how-to-use-xargs-with-sed-in-search-pattern