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bashgrep

Print everything on line after match

I have a large text file that contains a unique string in the middle. I want to print everything AFTER the string by using grep.

cat textfile | grep "target_string"
This highlights target_string but prints the whole file

cat textfile | grep -o "target_string"
This prints only target_string

cat textfile | grep -o "target_string*"
This prints only target_string

How can I print everything after "target_string" and nothing before?

Solution

  • You forgot the '.' in the regular expression:

    cat textfile | grep -o "target_string.*"