How can extract lines that match the regexp string ^li from the text (not a file) below using sed or something?
linux
loan
litmus
launch
I tried grep but I couldn't find a way to search within a quoted text, not a text file.
grep -n -i "^li" "linux
loan
litmus
launch"
Returns No such file or directory. And I don't want to save this text as a file before searching, if possible.
You need to use a herestring (<<<) here to pass the string as input to grep, herestring returns a file descriptor, grep can then operate on that:
$ grep -ni "^li" <<<"linux
loan
litmus
launch"
Output:
1:linux
3:litmus
If your shell doesn't support herestrings, many shells don't, you can print your string and pipe it to grep:
$ echo "linux
loan
litmus
launch" | grep -n -i "^li"
1:linux
3:litmus
Or use heredoc (<<):
$ grep -ni "^li" <<EOF
> linux
> loan
> litmus
> launch
> EOF
1:linux
3:litmus
You can just do:
grep \^li
And then start typing your stuff. grep will read the terminal. sed will do the same. Use CTRL+D to stop them, and ... >saved_output_file output redirections to save the effects.
With sed you can use the sed -n '/^li/p'
Using @heemayl example
$ echo "linux
> loan
> litmus
> launch" | sed -n '/^li/p'
linux
litmus
