not sure what is going on here with your quotes, but a suitable replacement to the line meaing you get what you need for html compliance: from: <a href='<? echo "$website"; ?>' target="_blank"><img src="homepage.gif" border="0"></a> to: <a href="'<?=$website;?>'" target="_blank"><img src="homepage.gif" border="0"></a> removing echo and using just the equals apparently makes it easier to debug and looks nicer. ________________________________ From: Totto <zealion101@xxxxxxxxx> To: php-objects@xxxxxxxxxxxxxxx Sent: Wednesday, 19 May, 2010 7:54:52 Subject: Link to a image Hi there, I am making a Guest Book of my own. People wrote comments to my guest book is saved in mysql database. There is a field in the table named "website". I retrive all the data from the table and display it on the page. Everything is fine but the website address I want to link with a picture so if someone wants to see the web site of the guest can visit there. My code to achive this is as under: <a href='<? echo "$website"; ?>' target="_blank"><img src="homepage.gif" border="0"></a> What is happening that when mouse goes over the picture it reads the link along with my web site address for example if $website is holding data from the field www.websitelink.com and my web site address is http://www.example.com it read as http://www.example.com/SOME FOLDER NAME/guestbook/www.websitelink.com which gives a 404 error. How can I eleminate my web site address from the actual link that it will appear as www.websitelink.com only? I will realy appriciate if someone can help me. Best regards and thanks [Non-text portions of this message have been removed]
