PHP RegEx Capture with Single Quotes

Question

Hi all,

I'm trying to capture from the following string:

var MyCode = "jdgsrtjd";
var ProductId = 'PX49EZ482H';
var TempPath = 'Media/Pos/';

What I'd like to get is the variable length value between the single quoted ProductId value

PX49EX482H

I had this, and I think it is close, but the single quotes are tripping me up. I'm not sure how to escape them properly.

preg_match('/var ProductID ='(.*?)';/', $str, $matches);

Thanks in advance!

Answer 1

Characters are escaped within strings in PHP (and virtually all C-syntax languages) with backslashes:

'This is a string which contains \'single\' quotes';
"This is a \"double\" quoted string";

In your example:

preg_match('/var ProductID =\'(.*?)\';/', $str, $matches);

Note that you don't have to escape single quotes in a double quoted string:

preg_match("/var ProductID ='(.*?)';/", $str, $matches);

Answer 2

Try this:

preg_match('/var ProductID = \'(.*?)\';/im', $str, $matches);

Answer 3

Alternatively you can use " in place of ' this way you need not escape the ' found in the pattern:

preg_match("/var ProductID ='(.*?)';/", $str, $matches);

Also the pattern you are looking for var ProductID ='(.*?)'; does not match your input string because:

• there is no space after =
• ProductID does not match ProductId

To fix 1, you can give a space after =. If you don't know the number of spaces you can use \s* for arbitrary space.

To fix 2, you can make the match case insensitive by using the i modifier.

preg_match("/var ProductID\s*=\s*'(.*?)';/i", $str, $matches);
                          ^^  ^^          ^