CODEWITHSUNDEEP
linuxbashgrep
newbie.my
newbie.my

Reputation: 341

passing grep into a variable in bash

I have a file named email.txt like these one :

Subject:My test
From:my email <[email protected]>

this is third test

I want to take out only the email address in this file by using bash script.So i put this script in my bash script named myscript:

#!/bin/bash

file=$(myscript)

var1=$(awk 'NR==2' $file)

var2=$("$var1" | (grep -Eio '\b[A-Z0-9._%+-]+@[A-Z0-9.-]+\.[A-Z]{2,4}\b'))

echo $var2

But I failed to run this script.When I run this command manually in bash i can obtain the email address:

echo $var1 | grep -Eio '\b[A-Z0-9._%+-]+@[A-Z0-9.-]+\.[A-Z]{2,4}\b'

I need to put the email address to store in a variable so i can use it in other function.Can someone show me how to solve this problem? Thanks.

Upvotes: 19

Views: 99818

Answers (5)

Peter Flynn
Peter Flynn

Reputation: 243

There's a cruder way:

cat $file | grep @ | tr '<>' '\012\012' | grep @

That is, extract the line(s) with @ signs, turn the angle brackets into newlines, then grep again for anything left with an @ sign.

Refine as needed...

Upvotes: 0

Prince John Wesley
Prince John Wesley

Reputation: 63708

Use Bash parameter expansion,

var2="${var1#*:}"

Upvotes: 0

flesk
flesk

Reputation: 7589

I think this is an overly complicated way to go about things, but if you just want to get your script to work, try this:

#!/bin/bash

file="email.txt"

var1=$(awk 'NR==2' $file)

var2=$(echo "$var1" | grep -Eio '\b[A-Z0-9._%+-]+@[A-Z0-9.-]+\.[A-Z]{2,4}\b')

echo $var2

I'm not sure what file=$(myscript) was supposed to do, but on the next line you want a file name as argument to awk, so you should just assign email.txt as a string value to file, not execute a command called myscript. $var1 isn't a command (it's just a line from your text file), so you have to echo it to give grep anything useful to work with. The additional parentheses around grep are redundant.

Upvotes: 28

gshilin
gshilin

Reputation: 676

There are very helpful flags for bash: -xv

The line with

var2=$("$var1" | (grep...

should be 

var2=$(echo "$var1" | (grep...

Also my version of grep doesn't have -o flag.

And, as far as grep patterns are "greedy" even as the following code runs, it's output is not exactly what you want.

#!/bin/bash -xv
file=test.txt
var1=$(awk 'NR==2' $file)

var2=$(echo "$var1" | (grep -Ei '\b[A-Z0-9._%+-]+@[A-Z0-9.-]+.[A-Z]{2,4}\b'))

echo $var2

Upvotes: 1

puckipedia
puckipedia

Reputation: 891

What is happening is this:

var2=$("$var1" | (grep -Eio '\b[A-Z0-9._%+-]+@[A-Z0-9.-]+\.[A-Z]{2,4}\b'))
       ^^^^^^^ Execute the program named (what is in variable var1).

You need to do something like this:

var2=$(echo "$var1" | grep -Eio '\b[A-Z0-9._%+-]+@[A-Z0-9.-]+\.[A-Z]{2,4}\b')

or even

var2=$(awk 'NR==2' $file | grep -Eio '\b[A-Z0-9._%+-]+@[A-Z0-9.-]+\.[A-Z]{2,4}\b')

Upvotes: 11

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