PHP Error - Strict Standards: Only variables should be passed by reference

Question

I'm getting the following error:

Strict Standards: Only variables should be passed by reference in /home/bridgesh/public_html/includes/functions/html_output.php on line 45

From the following lines of code:

44. while ($val = current($new_get)){
45.     if($val==end(array_reverse ($new_get)) )$new_getstr.='?'.key($new_get).'='.$val;
46.     else $new_getstr.='&amp;'.key($new_get).'='.$val;
47.     next($new_get);     
48. }

I've read through similar questions & answers on the subject here, but cannot figure out how to break the line down correctly.

_“but cannot figure out how to break the line down correctly”_ – to actually break it down into multiple lines, so that you can find out which of the statements used there actually causes the error, would be a start. — CBroe, Oct 08 '13 at 14:16

score 0 · Answer 1 · answered Oct 08 '13 at 14:16

0

end works on a reference to an array, so it expects a variable, that contains an array array_reverse ($new_get) however is not a variable, but a function call You have to do:

   $reversed = array_reverse ($new_get)
   if($val==end($reversed) )$new_getstr.='?'.key($new_get).'='.$val;

answered Oct 08 '13 at 14:16

Maxim Krizhanovsky

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Brilliant Darkhazer, that worked perfectly! Thanks for your help! – pedro42 Oct 16 '13 at 10:38

score -1 · Answer 2 · answered Oct 08 '13 at 14:15

-1

Try this:

$new_getstr = http_build_query($new_get);

Documentation

answered Oct 08 '13 at 14:15

Niet the Dark Absol

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I'd love to see a reason for downvoting this answer. – Niet the Dark Absol Oct 08 '13 at 14:21
I didn't do this, but it's obviously because you missed an explanation. – Yang Oct 08 '13 at 14:31

PHP Error - Strict Standards: Only variables should be passed by reference

2 Answers2