grab multiple lines after a matching target line

Question

This question is similar to How to show lines after each grep match until other specific match? thus answers may be similar to those there.

I am trying to grab and extract from a file the lines that are after the lines that are matching a TARGET value ("forms=2", in my case) and the next empty line. The segment of my file is like below:

forms=1
Code=00416T0
Code=00416T0

forms=2       #Target**
Code=06538T0  #grab this line
Code=06538T0  #grab this line
Code=11288T0  #grab this line
Code=11288T0  #grab this line

forms=1
Code=00549T0
Code=00549T0
Code=00549T0

forms=2      #Target**
Code=00553T0 #grab this line
Code=02576T0 #grab this line
Code=02576T0 #grab this line

forms=1
Code=11099T0

So I would like to find a way according to Target "forms=2" to have the following lines even if same of those grabbing lines are identical

Code=06538T0 #grab this line
Code=06538T0 #grab this line
Code=11288T0 #grab this line
Code=11288T0 #grab this line
Code=00553T0 #grab this line
Code=02576T0 #grab this line
Code=02576T0 #grab this line

Any help please?

possible duplicate of How to show lines after each grep match until other specific match? — Mat, Mar 01 '14 at 11:10
@Mat I guess that you need a better duplicate; the referenced one includes the lines matching the pattern whereas in this case you need to exclude those. — devnull, Mar 01 '14 at 11:58

score 2 · Answer 1 · edited Apr 13 '17 at 12:36

2

The following might work for you:

sed -n '/forms=2/,/^[^C]/{/^[^C]/b;p}' filename

or, as suggested by Graeme:

sed -n '/^forms=2/,/^[^C]/ {/^Code=/p}' filename

For your input, it'd produce:

Code=06538T0  #grab this line
Code=06538T0  #grab this line
Code=11288T0  #grab this line
Code=11288T0  #grab this line
Code=00553T0 #grab this line
Code=02576T0 #grab this line
Code=02576T0 #grab this line

For handling cases where you might lines like form=20, you could say:

sed -n '/^forms=2\b/,/^[^C]/ {/^Code=/p}' filename

edited Apr 13 '17 at 12:36

Community

1

answered Mar 01 '14 at 11:09

devnull

10,691

This is good, I would do it like this though sed -n '/^forms=2/,/^[^C]/ {/^Code=/p}'. Only thing I can't figure out though is how to exclude sections with forms=20 without using extended regex. – Graeme Mar 01 '14 at 11:31
@Graeme Yes, your suggestion makes sense. See the edit above for handling forms=20 case. – devnull Mar 01 '14 at 11:54
That works, I have no idea why though. \b is backspace, right? – Graeme Mar 01 '14 at 12:02
1

@Graeme Word boundary. – devnull Mar 01 '14 at 12:03

cuonglm · Answer 2 · 2014-03-01T18:13:20.837

Using perl:

%perl -lne 'if(/forms=2/.../^$/ and $_!~/forms=2|^$/){print}' file
Code=06538T0  #grab this line
Code=06538T0  #grab this line
Code=11288T0  #grab this line
Code=11288T0  #grab this line
Code=00553T0 #grab this line
Code=02576T0 #grab this line
Code=02576T0 #grab this line

Using awk:

awk '/forms=2/,/^$/{if(!/forms=2/&&!/^$/)print}' file

or:

awk '/^$/{flag=0};flag;/forms=2/{flag=1}' file

Using sed (GNU sed):

sed '/forms=2/,/^$/{//!b};d'

score 0 · Answer 3 · 2014-03-01T12:26:36.573

0

Another sed:

sed -n '/forms=2 /,/^$/ {/^.*forms.*$/d; /^$/d; s/  *#grab/ #grab/}' input

.

Code=06538T0 #grab this line
Code=06538T0 #grab this line
Code=11288T0 #grab this line
Code=11288T0 #grab this line
Code=00553T0 #grab this line
Code=02576T0 #grab this line
Code=02576T0 #grab this line

edited Mar 01 '14 at 12:26

answered Mar 01 '14 at 11:13

What if there was a line like this forms=20? – Graeme Mar 01 '14 at 11:33
Also, I think this can be condensed to sed -n '/forms=2/,/^$/ {s/[ ][ ]*#grab/ #grab/p}'. You don't need to d if you already have -n, just add the p to the end of the substitution. – Graeme Mar 01 '14 at 11:42
Thats is also what i was looking!! – user61677 Mar 03 '14 at 10:36

grab multiple lines after a matching target line

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