Print all matching lines and the previous unindented line

awkgrepsedtext processing

For example, lets say I want to find all lines with 'matching' and the previous unindented lines in this text.

Container 1
   some text
   some text
   matching text
   some text
Container 2
   some text
   some text
Container 3
   some text
   matching text

The result I want would look like this

Container 1
   matching text
Container 3
   matching text

Is this possible?

Best Answer

awk '
    !/^[[:blank:]]/ {unindented = $0} 
    /matching/ && /^[[:blank:]]/ {print unindented; print}
' file

That remembers the last line that did not begin with whitespace. When you get to a matching line, use that value.

Related Solutions

Print lines between (and excluding) two patterns

Here is one way you could do it:

sed '1,/firstmatch/d;/secondmatch/,$d'

Explained: From the first line to the line matching firstmatch, delete. From the line matching secondmatch to the last line, delete.

Grep Pattern Matching – Show Lines Matching a Pattern and 4 Lines Before Each

Supposing you're on an elderly system, like HP-UX, that doesn't have GNU utilities, just the old, original BSD or AT&T "grep". You could do something like this:

#!/bin/sh

awk '/ORA-/ { print line1; print line2; print line3; print line4; print $0 }\
// {line1 = line2; line2 = line3; line3 = line4; line4 = $0}' $1

Yes, there's tons of edge conditions this doesn't get right, but whatta ya want for nothing? Also, given that you're working on some decroded, antiquated OS and hardware, you probably don't have the CPU horsepower for fancy error handling.

Best Answer

Related Solutions

Print lines between (and excluding) two patterns

Grep Pattern Matching – Show Lines Matching a Pattern and 4 Lines Before Each

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