Code outputting html instead of an image?

Question

Here is the code that I am referring to:

<?php if ( is_archive() ) { echo '<img src="'.bloginfo('template_url').'/images/test.png" />'; }?>

This is what the code outputs: http://site.com/wp-content/themes/themename

I'd like it to output the actual image in the code. What part of this did I overlook?

Can you be more specific. The current code will output the <img src="whateverurl/image/test.png" /> — NT_
– NT_, Commented May 19, 2012 at 3:56
The code you have written isn't going to output what you say it does. It's either going to output an image tag or nothing — jprofitt
– jprofitt, Commented May 19, 2012 at 3:56
That's very strange. On my screen, it's showing exactly what I put up there. I was thinking there may possibly be an error in the code but I can't seem to find one. — John
– John, Commented May 19, 2012 at 3:58
Oh yes. It shows that because bloginfo actually echo's the output . it doesnt return string. — NT_
– NT_, Commented May 19, 2012 at 3:59

NT_ · Accepted Answer · 2012-05-19 04:01:34Z

1

bloginfo() doesn't output the string. It echo's it directly to output stream. So, the code should be:

<?php if ( is_archive() ) { ?>
   <img src="<?php bloginfo('template_url'); ?>/images/test.png" />'; 
<?php  }  ?>

Or else, you can use get_bloginfo() :

<?php if ( is_archive() ) { echo '<img src="'.get_bloginfo('template_url').'/images/test.png" />'; }?>

answered May 19, 2012 at 4:01

NT_

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Comments

Adriaan · Accepted Answer · 2012-05-19 04:13:58Z

0

Have you tried this:

$template_url = get_bloginfo('template_url');

<?php if ( is_archive() ) { echo '<img src="'.$template_url.'/images/test.png" />'; } ?>

answered May 19, 2012 at 3:58

Adriaan

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You can't use this. If you want to store values from bloginfo into variables , you are supposed to use get_bloginfo() not bloginfo()

I've already given the answer. No need to edit it to the correct code.