CODEWITHSUNDEEP
regexawk
Sudeep
Sudeep

Reputation: 3185

awk print matching line and line before the matched

Following is what I am trying to do using awk. Get the line that matches the regex and the line immediately before the matched and print. I can get the line that matched the regex but not the line immediately before that:

awk '{if ($0!~/^CGCGGCTGCTGG/) print $0}'

Upvotes: 6

Views: 50551

Answers (6)

Zeus
Zeus

Reputation: 6566

I created the following awk script. Prints the matching line as well as the previous 2 lines. You can make it more flexible from this idea.

search.awk

{
    a[0]=$0;
    for(i=0;i<2;i++)
    {
       getline;
       if(i==0){
            a[1]=$0;
       }
       if(i==1){
            if($0 ~ /message received/){
                print a[0];     
                print a[1];
                print $0;
            }
       }
    }
}

Usage: 

awk '{print $0}' LogFile.log | awk -f search.awk

Upvotes: 1

IDDQD
IDDQD

Reputation: 3741

Maybe slightly off-topic, but I used the answer from belisarius to create my own variation of the above solution, that searches for the Nth entry, and returns that and the previous line.

awk -v count=1 '/abc/{{i++};if(i==count){print a;print;exit}};{a=$0}' file

Upvotes: 0

amateur
amateur

Reputation: 21

use more straightforward pattern search

gawk '{if (/^abc$/) {print x; print $0};x=$0}' file1 > file2

Upvotes: 2

Dr. belisarius
Dr. belisarius

Reputation: 61026

/abc/{if(a!="")print a;print;a="";next}
{a=$0}

Upvotes: 3

ocodo
ocodo

Reputation: 30248

Why not use grep -EB1 '^CGCGGCTGCTGG'

The awk to do the same thing is very long-winded, see Marco's answer.

Upvotes: 0

marco
marco

Reputation: 4675

In this case you could easily solve it with grep:

grep -B1 foo file

However, if you need to to use awk:

awk '/foo/{if (a && a !~ /foo/) print a; print} {a=$0}' file

Upvotes: 19

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