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Possible Duplicate:
Annoying PHP error: “Strict Standards: Only variables should be passed by reference in”

I have this line of code, 

$extension=end(explode(".", $srcName));

when I fun my function I get

PHP Strict Standards: Only variables should be passed by reference in

I am not sure how to solve this

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Justin Erswell
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    Related: [Parentheses altering semantics of function call result](http://stackoverflow.com/q/6726589/367456) – hakre Nov 02 '12 at 14:43

1 Answers1

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The function end() requires a variable to be passed-by-reference and passing the return-value of a function doesn't acheive this. You'll need to use two lines to accomplish this:

$exploded = explode(".", $srcName);
$extension = end($exploded);

If you're simply trying to get a file-extension, you could also use substr() and strrpos() to do it in one line:

$extension = substr($srcName, strrpos($srcName, '.'));

Or, if you know the number of .'s that appear in the string, say it's only 1, you can use list() (but this won't work if there is a dynamic number of .'s:

list(,$extension) = explode('.', $srcName);
newfurniturey
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    *Pssst:* If it's about the file-extension, better link a question that is specifically about it. Like: [How to extract a file extension in PHP?](http://stackoverflow.com/q/173868/367456) – hakre Nov 02 '12 at 14:45
  • Billiant simple solution, makes sense and really helped me out there, thanks nefurniturey – Barry Connolly May 10 '13 at 10:07