Re: Using sample image where datafeeds don't provide one
What errors are you getting mate?
Wouldn't it be easier to stay in php and echo all the html statements?
Mark.
EDIT:
?php if (url_exists($row_Recordset1['imageURL'])){
echo '<a href="<?php echo $row_Recordset1['deepLink']; ?>&clickref=<?php echo $row_Recordset1['brand']; ?>" target="_blank">
<img src="<? echo $row_Recordset1['imageURL']; ?>" height="90" width="90" alt="<?php echo $row_Recordset1['brand']; ?>" title="<?php echo $row_Recordset1['name']; ?>" border="0" align="top" /></a>';<br />
}else{
echo '<a href="<?php echo $row_Recordset1['deepLink']; ?>&clickref=<?php echo $row_Recordset1['brand']; ?>" target="_blank">
<img src="images/noimage.jpg" height="90" width="90" alt="<?php echo $row_Recordset1['brand']; ?> " title="<?php echo $row_Recordset1['name']; ?>" border="0" align="top" /></a>'; }
?
Think I might have highlighted the problem by bolding one instance above.
Mark
|