CODEWITHSUNDEEP
bashshell
Manish
Manish

Reputation: 35

Extract Specific word from Text file between Specifc words in BASH

VAR="10.101.10.5"

I have a text file abc.txt in the below format:

Name : www.ayz.com  { Destination : 10.101.10.2  tcpmask        members {     10.101.10.5 :http { address 10.101.10.5 }  10.101.10.3 :http { address 10.101.10.3 } } 

Name : www.abc.com   { Destination : 10.101.10.5  tcpmask        members {     10.101.10.5 :http { address 10.101.10.5 }  10.101.10.3 :http { address 10.101.10.3 } }

The requirement is when I grep for a specific IP for example 10.101.10.5. It should the search only the line which has the IP 10.101.10.5 after the word Destination and before the word members.

So basically I need a command which just lists the second line only when I grep for 10.101.10.5, it should not return first line because it has IP after members.

I tried using sed but it's not helping:

sed -n '/destination : $VAR /,/<members>/p' /abc.txt

Upvotes: 1

Views: 239

Answers (2)

Saboteur
Saboteur

Reputation: 1428

Use double quotas and variable name. You can use also some regexp, for example grep "Destination\s*:\s*$VAR" to be more flexible for extra spaces after Destination

$ VAR="10.101.10.5"
$ grep "Destination : $VAR" file.txt
Name : www.abc.com   { Destination : 10.101.10.5  tcpmask        members {     10.101.10.5 :http { address 10.101.10.5 }  10.101.10.3 :http { address 10.101.10.3 } }

Upvotes: 1

Cyrus
Cyrus

Reputation: 88563

I suggest:

grep 'Destination .* 10\.101\.10\.5 .* members' abc.txt

Update:

grep "Destination .* ${VAR//./\\.} .* members" abc.txt

See: Difference between single and double quotes in bash

Upvotes: 1

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