PHP echo function changing order of html output?

Question

The function getUserAvatar outputs the following :

 <img src="http://pathtoimage/image.jpg" height="16" width="16" alt="avatar" />

I'm calling it within a particular div.

When I call it like this is works perfectly :

 echo '<div class="avatar">';
 echo getUserAvatar($avatar_id);
 echo '</div>';

Resulting in the following html output :

 <div class="avatar"><img src="http://pathtoimage/image.jpg" height="16" width="16" alt="avatar" /></div>

But when I try to put it all on one line :

 echo '<div class="avatar">' .getUserAvatar($avatar_id). '</div>';

It results in the following html, executing the function before rendering the div tags? :

 <img src="http://pathtoimage/image.jpg" height="16" width="16" alt="avatar" /><div class="avatar"></div>

Could somebody please explain why this is and offer the correct solution?

are you returning or printing content in your getUserAvatar function? — mishu
– mishu, Commented Mar 22, 2015 at 11:22

jvitasek · Accepted Answer · 2015-03-22 11:24:21Z

1

<?php
function getUserAvatar($avatar_id) {
    // your function body
    return $your_return_id;
}

echo '<div class="avatar">' .getUserAvatar($avatar_id). '</div>';

answered Mar 22, 2015 at 11:24

jvitasek

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