PHP code for displaying alternate image if custom field is not populated

Question

I am trying to display what might be called a placeholder image if an image custom field has not been provided

I've got this far but am a bit stuck. The output is showing the code. Here is the PHP I am using.

$my_image = get_post_meta($post->ID, 'ecpt_images', true);

if( isset($my_image) && $my_image != '') {
    echo '<img src="'.$my_image.'" width="230px" height="175px" />';
} else { 
    echo '<img src="/wp-content/uploads/2013/02/Poor.png">';
}

Thank you very much in advance

Are you sure that you're surrounding your PHP code with <?php /* code */ ?>? — Chris Forrence
– Chris Forrence, Commented Feb 12, 2013 at 20:12
please clarify what you mean by The output is showing code — d-_-b
– d-_-b, Commented Feb 12, 2013 at 20:13
Try rewriting if statement like if strlen ($myimg) > 0 , assuming image path is correct and it exists — Shridhar
– Shridhar, Commented Feb 12, 2013 at 20:24
The better way IMO is to have a container element with a css background image (which will be hidden if there is an actual image). This might even help with your problem (like iight I do not fully understand what the problem actially is). — Eike Pierstorff
– Eike Pierstorff, Commented Feb 12, 2013 at 20:24

rbncha · Accepted Answer · 2013-02-12 20:26:51Z

1

Check if you used short open tag <? /* your code */ ?>. This setting is by default disabled and not recommended. Make sure you wrapped your code properly within <?php /* your code */ ?>.

answered Feb 12, 2013 at 20:26

rbncha

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