Detect empty variable in PHP not working - php

Question

I have been fighting with the following code and variants, but have not been able to come up with anything that works. The idea is that I detect if the variable $largeimg6 (the URL of an image) is empty or not (whether an image has been loaded) and if not, load the value of a default variable ($defaultimage) which holds the URL of a default image..
<?php if(!empty($largeimg6)) : ?>
<img class="rpPicture" id="rpPicture<?php echo $key; ?>" onload="showPicture(this, <?php echo SHOW_PICTURES; ?>)" width="60" height="60" src="<?php echo $largeimg6; ?>" alt="Buy CD!" title="Buy CD!" />
<?php endif; ?>
The problem is that I do not know where to add the arguement for $largeimg6 being empty and therefore adding:
$largeimg6 = $defaultimage
Any help would be appreciated.
Thanks.

Answer 1

You can try this before that if statement:
<?php $largeimg6 = $largeimg6 != '' ? $largeimg6 : $defaultimage; ?>
The != '' comparision may be change according to your need to isset(), strlen(), is_null()(as Ville Rouhiainen suggested) or whatever.

Answer 2

You can use the alternative (and trim as already suggested):
$largeimg6 = trim($largeimg6);
if (isset($largeimg6)&&strlen($largeimg6)>0)
but you'll probably be doing better by filtering the url:
$largeimg6 = trim($largeimg6);
if(filter_var($largeimg6, FILTER_VALIDATE_URL))
More info: http://php.net/manual/es/book.filter.php

Answer 3

Solution
This answer assumes that you have a variable $largeimg6 which is either going to be set to a url or be empty.
That being the case it's a fairly simple fix. You need to remove the if statement entirely and replace your:
<?php echo $largeimg6; ?>
with:
<?php echo (empty($largeimg6)) ? '/default/image.jpg' : $largeimg6; ?>
Explanation
The above is equivalent to:
if(empty($largeimg6)){
echo '/default/image.jpg';
}
else{
echo $largeimg6;
}
But in the form:
(IF) ? THEN : ELSE;