CODEWITHSUNDEEP
regexawkgrep
Red Ant
Red Ant

Reputation: 375

How to print till the pattern match including pattern string

I want to print the line till pattern is matched but including the pattern

for example:

http://google.com/search

I want to use ".com" as match string and just print http://google.com , ignoring rest of the line.

awk or grep would be helpful.

I tried codes from posts but didnt get exactly that i wanted.

request your help

Upvotes: 0

Views: 328

Answers (4)

shellter
shellter

Reputation: 37258

Another approach. Assuming that your goal is to harvesting top-level domains that all end in .com, then you can just delete everything at the end, including .com and replace it with only .com, i.e.

echo "http://google.com/search
1
2" | awk '/\.com/{sub(/\.com.*/, ".com", $0) ;print $0}'

output

http://google.com

I've included extra lines of input to show that only lines containing .com are processed.

If you want to print out all lines of input, remove the initial match to reg-ex, i.e. 

echo "http://google.com/search
1
2" | awk '{sub(/\.com.*/, ".com", $0) ;print $0}'

output 2

http://google.com
    1
    2

IHTH

Upvotes: 0

Jotne
Jotne

Reputation: 41446

Here is an awk

echo "http://google.com/search" | awk -F/ -vOFS=/ '/\.com/ {$NF="";sub(/\/$/,"");print}'
http://google.com

Upvotes: 0

anubhava
anubhava

Reputation: 784938

Use grep -o to output only matched string:

s='http://google.com/search'

grep -o '.*\.com' <<< "$s"
http://google.com

Upvotes: 2

stevejpurves
stevejpurves

Reputation: 933

use the following regex: /(.*.com)/

see this example

Upvotes: 0

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