php to extract a string from double quote-ThrowExceptions

Exception or error:

I have a string:

This is a text, “Your Balance left $0.10”, End 0

How can I extract the string in between the double quotes and have only the text (without the double quotes):

Your Balance left $0.10

I have tried preg_match_all() but with no luck.

How to solve:

As long as the format stays the same you can do this using a regular expression. "([^"]+)" will match the pattern

  • Double-quote
  • At least one non-double-quote
  • Double-quote

The brackets around the [^"]+ means that that portion will be returned as a separate group.


$str  = 'This is a text, "Your Balance left $0.10", End 0';

//forward slashes are the start and end delimeters
//third parameter is the array we want to fill with matches
if (preg_match('/"([^"]+)"/', $str, $m)) {
    print $m[1];   
} else {
   //preg_match returns the number of matches found, 
   //so if here didn't match pattern

//output: Your Balance left $0.10


For everyone hunting for a full featured string parser, try this:


Use in preg_match:

$haystack = "something else before 'Lars\' Teststring in quotes' something else after";


    [0] => 'Lars\' Teststring in quotes'

This works with single and double quoted string fragments.


Try this :

preg_match_all('`"([^"]*)"`', $string, $results);

You should get all your extracted strings in $results[1].


Unlike other answers, this supports escapes, e.g. "string with \" quote in it".

$content = stripslashes(preg_match('/"((?:[^"]|\\\\.)*)"/'));


The regular expression '"([^\\"]+)"' will match anything between two double quotes.

$string = '"Your Balance left $0.10", End 0';
preg_match('"([^\\"]+)"', $string, $result);
echo $result[0];


Just use str_replace and escape the quote:



Sorry, didnt see that there was text after the 2nd quote. In that case, I’d simply to 2 searches, one for the first quote and one for the 2nd quote, and then do a substr to extra all stuff between the two.

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