php preg_match fails when used with a variable

Question

Excuse me, I am making a terrible mistake somewhere, but this is the situaton:

In php i have:

$ln = "A/RADIUS ADMITS VALUE 20";
$trg = "A/RADIUS";
$matches = array();
$zz = preg_match('#$trg\sADMITS\s(VALUE)\s([^\s]+)#',$ln,$matches);

I want to capture the word "VALUE" without quotes and the last word, here the string 20, given by ([^\s]+) . That is not a whitespace repeated more than once right?

But $zz is 0, indicating no match and $matches is empty. I also tried with

 $zz = preg_match('#'.$trg.'\sADMITS\s(VALUE)\s([^\s]+)#',$ln,$matches);

same problem.

Where is the mistake I am stupidly making?

preg_quote($trg, '#').

h2ooooooo
– h2ooooooo

2014-04-25 12:35:36 +00:00
Commented Apr 25, 2014 at 12:35 — h2ooooooo
– h2ooooooo, Commented Apr 25, 2014 at 12:35

Erik Schierboom · Accepted Answer · 2014-04-25 12:36:20Z

1

You are using single quotes to embed the $trg variable in the string, which only works when using double quotes to enclose the string. This should work:

$zz = preg_match("#$trg\sADMITS\s(VALUE)\s([^\s]+)#",$ln,$matches);

answered Apr 25, 2014 at 12:36

Erik Schierboom

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3 Comments

Erik Schierboom Over a year ago

@h2ooooooo Odd, as I can confirm it working as expected. See: 3v4l.org/ETYd5

Sean Over a year ago

while using single quote then using the '.' to concat the variable is NOT working, embedding the variable in a double quoted string IS working. Thank you a lot.

Sean Over a year ago

Yet another quetion, about $zz = preg_match('#'.$trg.'\sADMITS\s(VALUE)\s([^\s]+)#', $ln , $matches); failing : is it a bug in PHP ? should I file a report?

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