alena1347 Posted January 29, 2013 Share Posted January 29, 2013 I have inserted a jquery in my html page but it is not fading out <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <link href="programmers.css" rel="stylesheet" type="text/css" /> <?php include('conn.php'); include('header.php'); ?> <head> <meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1" /> <title>User Allotment</title> <script src="//ajax.aspnetcdn.com/ajax/jQuery/jquery-1.8.3.min.js" type="text/javascript"></script> <script type="text/javascript"> $(document).ready(function(){ $("#detail").hide(); $("#view").click(function() { $("#form").fadeout(); $("#detail").fadeIn(); }); }); </script> </head> <body> <form id="form1" name="form1" method="post" action="" id="form"> <table align="center" width="100%" id="form"> <tr> <td height="25" width="100%" bgcolor="#CCCCCC" align="center"><a href="user_allotment.php">User Allotment</a> <a href="view_users.php">View Users</a> <a href="">Pending Aprovals</a> <a href="freelancers_pending.php">Freelancer Pending</a> <a href="search.php">Search</a> <a href="admin_login.htm">Logout</a></td> </tr> <tr> <td><br/> <table width="1019" border="1" align="center" bgcolor="#FFFFFF"> <tr> <td colspan="13" align="center"><strong>PENDING APPROVALS</strong></td> </tr> <tr align="center"> <td width="60" align="center" >Id</td> <td width="60" align="center">First Name</td> <td width="60" align="center">Last Name</td> <td width="65" align="center">Company</td> <td width="50" align="center">Phone</td> <td width="40" align="center">Ext</td> <td width="50" align="center">E-mail</td> <td width="70" align="center">Skill required</td> <td width="75" align="center">Experience</td> <td width="70" align="center">Duration of hire</td> <td width="70" align="center">Location of hire</td> <td width="110" align="center">Brief project description</td> <td width="60" align="center">view</td> </tr> <?php mysql_select_db("programmers") or die(mysql_error()); $data = mysql_query("SELECT * FROM hire") or die(mysql_error()); while($info = mysql_fetch_array( $data )) { ?> <tr> <td><?php echo $info['id'];?></td> <td><?php echo $info['fname'];?></td> <td><?php echo $info['lname'];?></td> <td><?php echo $info['cname'];?></td> <td><?php echo $info['cnum'];?></td> <td><?php echo $info['enum'];?></td> <td><?php echo $info['email'];?></td> <?php $id=$info['id']; $fetch=mysql_query("SELECT sname FROM skill WHERE id=$id"); while($row=mysql_fetch_array($fetch)) { $sname=$row['sname']; $sk[]=$sname; } $arr=implode(',',$sk); ?> <td><?php echo $arr;?></td> <td><?php echo $info['exnum'];?></td> <td><?php echo $info['dnum'];?></td> <td><?php echo $info['lhname'];?></td> <td><?php echo $info['task'];?></td> <td><label><input type="button" class="view" id="view" value="view"/></label></td> </tr> <?php } ?> </table> </td> </tr> </table> </form> </body> </html> <?php include('footer.php'); ?> Link to comment https://forums.phpfreaks.com/topic/273783-why-isnt-the-jquery-working/ Share on other sites More sharing options...
trq Posted January 29, 2013 Share Posted January 29, 2013 And you have done what to debug your code? Link to comment https://forums.phpfreaks.com/topic/273783-why-isnt-the-jquery-working/#findComment-1408928 Share on other sites More sharing options...
kicken Posted January 29, 2013 Share Posted January 29, 2013 <form id="form1" name="form1" method="post" action="" id="form"> You can't specify the id attribute twice. Link to comment https://forums.phpfreaks.com/topic/273783-why-isnt-the-jquery-working/#findComment-1408956 Share on other sites More sharing options...
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