Match patternA and print it only when patternB is matched including the following line

awksedtext processing

I am looking to get all the lines which have the word 'search_string' + the line after it + the line matching 'mod' before it.
I tried:

grep -n 'mod\|search_string' ip | grep --before 1 search_string> inter  
grep -n --after 1 search_string ip >> inter  
sort -t':' -k1,1n -u inter -o op

Is there a better way?

File:

mod start1  
some lines  
mod start2  
other lines  
mod start3  
 many other lines  
 search_string yada yada  
 hello  
 many other lines  
 search_string yada yada  
 bye  
mod start4  
 search_string baba baba  
 this too  
mod start5

Expected output :

mod start3  
 search_string yada yada   
 hello  
 search_string yada yada  
 bye  
mod start4  
 search_string baba baba  
 this too

Best Answer

awk '
   $0 ~ /mod/ { md=$0 }
   $0 ~ /search_string/ { if(md!="") { print md }; md="" ; print; getline; print }
   '

Explanation:

A line containing mod is saved as md.
A line containing search_string triggers printing the previously saved md, the line itself and the next line.
if(md!="") and md="" are there to make sure you don't get duplicated mod lines when there are many search_string-s under a single mod (mod start3 in your example).

Note:

A line containing both mod and search_string will break this logic.

Related Solutions

Text Processing with Sed and Grep – Return Only the Portion of a Line After a Matching Pattern

The canonical tool for that would be sed.

sed -n -e 's/^.*stalled: //p'

Detailed explanation:

-n means not to print anything by default.
-e is followed by a sed command.
s is the pattern replacement command.
The regular expression ^.*stalled: matches the pattern you're looking for, plus any preceding text (.* meaning any text, with an initial ^ to say that the match begins at the beginning of the line). Note that if stalled: occurs several times on the line, this will match the last occurrence.
The match, i.e. everything on the line up to stalled:, is replaced by the empty string (i.e. deleted).
The final p means to print the transformed line.

If you want to retain the matching portion, use a backreference: \1 in the replacement part designates what is inside a group $…$ in the pattern. Here, you could write stalled: again in the replacement part; this feature is useful when the pattern you're looking for is more general than a simple string.

sed -n -e 's/^.*\(stalled: \)/\1/p'

Sometimes you'll want to remove the portion of the line after the match. You can include it in the match by including .*$ at the end of the pattern (any text .* followed by the end of the line $). Unless you put that part in a group that you reference in the replacement text, the end of the line will not be in the output.

As a further illustration of groups and backreferences, this command swaps the part before the match and the part after the match.

sed -n -e 's/^\(.*\)\(stalled: \)\(.*\)$/\3\2\1/p'

Grep/Sed/Awk – Print Line Only if Next Line Does Not Match

Here is an awk alternative:

awk '
  /^Starting/ { I[$5] = $0                  }
  /^ID/       { delete I[$2]                }
  END         { for (key in I) print I[key] }
' infile

Output:

Starting activity for ID 33367

The I associative array keeps track of what ids have been seen.

Best Answer

Related Solutions

Text Processing with Sed and Grep – Return Only the Portion of a Line After a Matching Pattern

Grep/Sed/Awk – Print Line Only if Next Line Does Not Match

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