[SOLVED] How to print a range of lines between two patterns only if another pattern is included in this range?

Issue

I'm working on text files on Linux. I want to be able to print all the lines between two patterns (including the lines where the patterns are found) only if another pattern is found in those lines.

For instance:

PatternStart
line1
line2
PatternInside
line3
line4
PatternEnd

PatternStart
line1
line2
line3
PatternEnd

I want to only get the first block of lines because it contains the PatternInside. Right now, all I can do is extract the lines between my boundary patterns with

awk '/PatternStart/,/PatternEnd/' file

But this will extract the two blocks of lines.

Solution

You can use

awk 'flag{
    buf = buf $0 ORS;
    if (/PatternEnd/ && buf ~ /PatternInside/)
       {printf "%s", buf; flag=0; buf=""}
}
/PatternStart/{buf = $0 ORS; flag=1}' file

Here, the /PatternStart/{buf = $0; flag=1}' finds the line that matches the PatternStart pattern, starts writing the output value to buf, and sets the flag. If the flag is true, subsequent lines are appended to buf, and once there is a line where PatternEnd matches and the PatternInside finds a match in the buf, the match is printed, buf gets cleared and the flag is reset.

See the online demo that yields

PatternStart
line1
line2
PatternInside
line3
line4
PatternEnd

Answered By - Wiktor Stribiżew

Answer Checked By - Mary Flores (WPSolving Volunteer)