tehwolf Posted April 10, 2012 Share Posted April 10, 2012 As title says, im trying to get the LightboxV2 Script to work. I've got a database with stored info (title, author, link and screenshot) about some projects, and a php script displaying those info in a table via mysql_fetch_object() which does work. I haven't saved the screenshots in a BLOB, just the http-link to them. The actual problem is that when clicking the image it opens in a new window instead of the Lightbox appearing, which is not really good. My Code: <?php include "include/config.php"; ?> <html> <head> <title><?php echo $pagetitle; ?></title> <link rel="stylesheet" href="css/lightbox.css" type="text/css" media="screen" /> <script type="text/javascript" src="http://tehwolf.te.funpic.de/js/prototype.js"></script> <script type="text/javascript" src="http://tehwolf.te.funpic.de/js/scriptaculous.js?load=effects,builder"></script> <script type="text/javascript" src="http://tehwolf.te.funpic.de/js/lightbox.js"></script> </head> <body> <?php $verbindung = mysql_connect($mysql_host, $mysql_user, $mysql_password) or die("Verbindung zur Datenbank konnte nicht hergestellt werden"); mysql_select_db($mysql_database) or die ("Datenbank konnte nicht ausgewählt werden"); ?> <table border="1" cellpadding="0" cellspacing="0"> <tr> <th>Titel:</th> <?php $anf1 = "SELECT title FROM style WHERE id='1'"; $erg1 = mysql_query($anf1); while($row1 = mysql_fetch_object($erg1)) { echo "<th>".$row1->title."</th>"; } ?> </tr> <tr><td>Autor:</td> <?php $anf2 = "SELECT autor FROM style WHERE id='1'"; $erg2 = mysql_query($anf2); while($row2 = mysql_fetch_object($erg2)) { echo "<td>".$row2->autor."</td>"; } ?> </tr> <tr><td>Screenshot:</td> <?php $anf3 = "SELECT screen FROM style WHERE id='1'"; $erg3 = mysql_query($anf3); while($row3 = mysql_fetch_object($erg3)) { echo "<td><a href=".$row3->screen." rel='lightbox'>Screenshot</a></td>"; } ?> </tr> <tr><td>Link:</td> <?php $anf4 = "SELECT link FROM style WHERE id='1'"; $erg4 = mysql_query($anf4); while($row4 = mysql_fetch_object($erg4)) { echo "<td>".$row4->link."</td>"; } ?> </tr> </table> <br /> <br /> <br /> </body> </html> Quote Link to comment Share on other sites More sharing options...
bspace Posted April 10, 2012 Share Posted April 10, 2012 if the link is opening the image in a new window that usually means the javascript isn't working are you sure the js is loading? you might have better luck asking in the javascript subforum Quote Link to comment Share on other sites More sharing options...
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