CODEWITHSUNDEEP
phpawk
lxgeek
lxgeek

Reputation: 1814

PHP pass variable to awk

I'm a new one using PHP. I want to use awk in PHP. I have a file which is named a. The content is:

www.b.com * record=600,IN,A,1.2.3.4record=600,IN,A,5.6.7.8
www.1.com u record=600,IN,A,1.2.3.4
www.1.com w record=600,IN,A,1.2.3.4

Now I use PHP to handle it.

<?php


        $dn = "www.1.com";

        $hello = shell_exec("awk '{ if ( $1==".$dn.")print $1}' a") ;
        echo $hello;
?>

It doesn't work — can you tell me why? If I use this other way, it does work. 

$hello = shell_exec("awk '{ if ( $1==\"www.b.com\")print $1}' a") ;

Upvotes: 2

Views: 2243

Answers (1)

Michael Berkowski
Michael Berkowski

Reputation: 270677

awk sees $dn as a string literal, so you need to quote the string inside awk, which is what you did in your test example:

$dn = "www.1.com";

$hello = shell_exec("awk '{ if ( $1==\"".$dn."\")print $1}' a") ;
// ---------------------------------^^^-------^^^
// Surround in escaped double quotes...
echo $hello;

If the string $dn is to be sourced from any kind of user or browser input, be sure to properly escape it with escapeshellarg() or escapeshellcmd() first. Since escapeshellarg() adds single quotes, that probably isn't useful here, but escapeshellcmd() which is normally used to escape the entire command, could at least escape any internal quotes or other bad characters that might come in $dn.

Upvotes: 6

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