grab text by pattern with start and end tag in multiple lines [duplicate]

Question

I want to grab multiple lines in a file with a pattern that has a starting and an ending tag in an ungreedy way.

For example I have the following input:

file.txt

START
test1
test2
foo
END
some
more text

START
test3
bar
test4
test5
END
even more

START
baz
test6
END

Now I want to look for bar and print out everything between START and END, so that I will get:

START
test3
bar
test4
test5
END

What I have so far is the following grep command:

grep -Pzo '(?s)START.*?bar.*?END' file.txt

The problem is, that this expression is greedy and prints out:

START      # starts at first "START"-tag, not the next one
test1      #
test2      #
foo        #
END        #
some       #
more text  #

START
test3
bar
test4
test5
END

It's not done with the grep flags --before-context / --after-context, because the count of lines before and after can differ.

The tool used by text-processing doesn't matter. It should work on a general RedHat system. Also the faster the tool grabs the lines, the better it will be. Because I have big logfiles of each about 150MB.

Can someone tell me, how to achieve my goal the best way?

---

Update:

Okay, I got it. I just had to think about how to construct my command from don_crisstis link. Here's the solution:

ed -s file.txt <<< $'g/bar/?START?,/END/p\nq\n'

Thank you very much for all your very fast help!

And yes, finally it's a duplicate...

Answer 1

I think your problem is that your non-greedy matches can still swallow up more than you want them to, ie. ENDs and STARTs. This seems to work:

grep -Pzo '(?s)START(?:(?!END).)*?bar(?:(?!START).)*?END' file.txt

It covers all cases in your example and to be complete if you >> file.txt

bar
START
test7
END

Still works.

Answer 2

I would use awk, where you can specify the record separator. If the record separator is "END" (on its own line), then look for the record containing "bar":

awk 'BEGIN {RS = ORS = "\nEND\n"} /bar/' file.txt

---

Handling text that appears between the START and END markers. This change feels hacky, but it works for this scenario: using END as the record separator, remove any text before the START keyword

awk '
    BEGIN {RS = ORS = "\nEND\n"} 
    {sub(/^.*\nSTART\n/, "START\n")} 
    /bar/
' file.txt

This may not give desired results if "START" can appear more than once before END

foo
START
hello
START
bar
world
END
baz

will output as

START
bar
world
END

Answer 3

perl -nE 'BEGIN {$/="\nEND\n"} say /(START.*test.*)/s'

As @bobbel point out, replace say by print to avoid empty line separators.