Reputation: 57
imagedisplay.php(view)
<html>
<body>
<h3>Your file was successfully uploaded!</h3>
<?php print_r($upload_data); ?> </br>
<?php $str=base_url()."images/".$upload_data['file_name'] ?> </br>
<?php $str=str_replace('http://','',$str) ?>
<?php echo $str; ?>
<img src= '$str'/> </br>
For echo $str; I got the string i need to display th image but when i pass it to img src.... i am not able to display it on the browswer
Is there any syntactical error or am i missing anything ...pls help?
Upvotes: 3
Views: 2331
Reputation: 41737
Just a small syntax problem here. Embed PHP echo command in the HTML code, like so:
<img src="<?php echo $str; ?>"/> </br>
or embed PHP echo short tags:
<img src="<?=$str?>"/> </br>
In other words: insert the PHP output at the positions, where you need it as HTML content.
Upvotes: 3