Error checking Empty() for PHP

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Below is my piece of code that i am trying to work with.

PHP

<?php if (empty(userphoto($curauth))){echo "<img src='http://localhost/wordpress/wp-content/uploads/2013/01/NoImage.png' /?>"} else { userphoto($curauth)}   ?>

I am checking if userphoto($curauth) is null, i am replacing it with a default image.

Everytime i do this, i get the below error.

"Fatal error: Can't use function return value in write context in..."

Can someome guide me what is the problem exactly?

Posted 30-Jan-13 8:36am

Vamshi Krishna Naidu

Updated 30-Jan-13 8:47am

v2

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DinoRondelly 30-Jan-13 15:06pm

Why not drop empty and just check to see if its null?

If(userphote($curauth) !== null)
{

}

Not sure what version of php you are using but i found this

Prior to PHP 5.5, empty() only supports variables; anything else will result in a parse error. In other words, the following will not work: empty(trim($name)). Instead, use trim($name) == false.

http://www.php.net/manual/en/function.empty.php

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Matej Hlatky · Accepted Answer · 2013-01-30T09:09:00

I tried your code and the main problem was in the if part:

PHP

if (empty(userphoto($curauth)))...

First, your userphoto function SHOULD not print/echo value, but only return path to user image.
Second, you need to assign return value from userphoto($curauth) function call into variable!
Third, you are missing semicolons after echo and userphoto function calls.

Check my updated code:

PHP

<?php
function userphoto($_) {
  return null; // Your code goes here
}

$photo = userphoto($curauth);

if (empty($photo))
{
  $photo = 'http://localhost/wordpress/wp-content/uploads/2013/01/NoImage.png';
}

echo "<img src='{$photo}' />";
?>