如何在Echo中添加onError？

Question

i just have a php script that show my slideshow image, i make it dynamic so the slideshow must match the image in folder to database before display it, when the database is empty or the image is not found, it just display an error message, i just want to add onError to anticipate that but i cant do it, can anyone help me??? here's the code

echo "<img src='".'slideshows/'.$rec2->image."'  title='".$rec2->name."' alt='".$rec2->description."' />";

i want to make it like

echo "<img src='".'slideshows/'.$rec->image."'  title='".$rec->name."' alt='".$rec->description."' onError="this.onerror=null;this.src='error.png'; this.onclick=null;" />";

but the onError didn't work, help me please... sorry for my bad language...

Answer 1

echo sprintf(
  '<img src="slideshows/%s" title="%s" alt="%s" onError="%s" />', 
  $rec->image,
  $rec->name,
  $rec->description,
  "this.onerror=null;this.src='error.png'; this.onclick=null;"
);