Grep: Show all the file up to the match

Question

grep --before-context 5 shows 5 lines before the match.

I want to show everything before the match.
Doing grep --before-context 99999999 would work but it is not very... professional.

How to show all the file up to the match?

Answer 1

sed can replace most of grep's functionality.

sed -n '1,/<pattern>/ p' <file>

This means print from the first line until pattern is matched.

A couple of range examples

sed -n '/<pattern>/,$ p' <file> # from pattern to end of file
sed -n '/<pattern1>/,/<pattern2>/ p' <file> # from pattern1 to pattern2

Answer 2

Sed is better for that.

Just do:

sed '/PATTERN/q' FILE

It works like this:

For each line, we look if it matches /PATTERN:

• if yes, we print it and quit
• otherwise, we print it

This is the most efficient solution, because as soon as it sees PATTERN, it quits. Without q, sed would continue to read the rest of the file, and do nothing with it. For big files it can make a difference.

This trick can also be used to emulate head:

sed 10q FILE

Answer 3

print up to and including the match:

awk '{print} /pattern/ {exit}' filename

print up to BUT NOT including the match:

awk '/pattern/ {exit} {print}' filename

Disclaimer: I know awk better than sed -- I bet there is a sed way to do both.

Answer 4

For people like me, who choose to remember only the basic usage of tools, and willing to accept less elegant and less efficient solutions:

head -n $(grep -n pattern filename | cut -d: -f1) filename