Detect empty variable in PHP not working

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.

Where is $largeimg6 initially set? Will isset (php.net/manual/en/function.isset.php) work for you? — Ronnie
– Ronnie, Commented Oct 14, 2013 at 16:55

DontVoteMeDown · Accepted Answer · 2013-10-14 16:56:26Z

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.

answered Oct 14, 2013 at 16:56

DontVoteMeDown

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1 Comment

DontVoteMeDown Over a year ago

@omega1 nice, glad to help!

NotGaeL · Accepted Answer · 2013-10-14 17:08:04Z

1

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

edited Oct 14, 2013 at 17:08

answered Oct 14, 2013 at 16:58

NotGaeL

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Steven · Accepted Answer · 2013-10-14 17:14:47Z

1

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;

answered Oct 14, 2013 at 17:14

Steven

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