[eluser]krislec[/eluser]
Hi everybody,
Is it possible to get this code line to work ?
Code:
<img src="<?php echo base_url();?>application/views/test_img.php" alt="graphique dynamique" title="graphique" />
firebug tells me :
"Failed to load given URL"
but the files is in the right places !
How can I do if I want to use dynamic image made by mysql data ?
Thanx.
