I know how to use the utility sed in bash, and grep is also good. But for the output, they always output a line containing the pattern.
Is there a way in bash to cut out a particular string containing the pattern I want (using regexp)?
3 Answers
That's why you have --help flag, grep --help:
-o, --only-matching show only the part of a line matching PATTERN
Now you can
$ echo 'hello world' | grep -o hello
hello
Using sed:
$ echo "hello world" | sed 's/.*\(hello\).*/\1/'
5 Comments
Chromium
Is there any way to use
sed to do that? I am just curious, not for a particular goal.Chromium
Is the
\1 affecting the result? I don't see any about this in the sed manual page.
Maroun
@Chromium
\1 is the first group, which is the hello. It's a regex thing.Chromium
so what does the command
sed 's/.*\(hello\).*/\1/' really means?Chromium
It doesn't mean to translate the
regexp to group first, right?With GNU grep you can use -o:
-o Print only the matched (non-empty) parts of a matching line, with each such part on a separate output line.
Example:
$ echo ab cd ef gh | grep a
ab cd ef gh
$ echo ab cd ef gh | grep -o a
a
Comments
Through sed,
$ echo 'hello world' | sed 's/\(hello\)\|./\1/g'
hello
$ echo 'hello world' | sed -r 's/(hello)|./\1/g'
hello
grep -ois what you are looking for.