Reputation: 539
I have a problem here that deals with the displaying of a photo inside a variable within a div class.
Here's my code below and help me find solutions. Thanks
if($newimage){
$url = 'http://www.client.jaobuilders.com/uploads/profile_picture/upload_photos/$newimage';
} else {
$url = 'http://www.client.jaobuilders.com/images/blank_photo.jpg';
}
return '
<div class="comment">
<div class="avatar">
'.$link_open.'
<img src="'.$url.'" width="50px" height="50px"/>
'.$link_close.'
</div>
$newimage
is a variable and the value will depend on the user who logged in.
I really don't know what to do. Help me.
Upvotes: 0
Views: 383
Reputation: 137320
You have a problem with quotes - some of them are lacking. Also you did not close <div>
tag:
return '
<div class="comment">
<div class="avatar">
'.$link_open.'
<img src="'.$url.'" width="50px" height="50px"/>
'.$link_close.'
</div>
</div>';
I can only hope this return
statement is enclosed in some kind of function or method. Or at least it is the only return
statement in a file that has been properly included somewhere (like $my_divs = include('some_file.php');
).
Upvotes: 2