Error On Line 116 In My Script!!

Anon

Hey Everyone, I am currently making a pretty neat bookmarks script using PHP/MySQL but When I Uploaded it, I am getting a

Parse Error: Parse error on line 116

Here Is My Code.. If You See whats wrong, PLEASE tell me..Or E-mail Me.

<?
if (!$submit)
{
?>

<head>
<title>Bookmarks2Go Sign Up</title>
</head>
<br>
Ever thought it would be cool to have a list of your favorite and most useful websites online? So you can visit it anytime you want, and even share it with your friends? Bookmarks 2 go is that! Each member gets a unique members name, and can let their friends login and view and add bookmarks at their choosing! Please Sign Up Below!<br><br>

<h3>Sign Up!</h3>
<form method ="post" action="push.php">
<table>
<tr>
<td>
Username
</td>
<td>
<input name="username" length="10" maxlength="30">
</td>
</tr>
<tr>
<td>
Website Name
</td>
<td>
<input name="title1" length="30" maxlength="30">
</td>
</tr>
<tr>
<td>
URL
</td>
<td>
<input name="url1" length="30">
</td>
</tr>
<tr>
<td>
Website Name
</td>
<td>
<input name="title2" length="30" maxlength="30">
</td>
</tr>
<tr>
<td>
URL
</td>
<td>
<input name="url2" length="30">
</td>
</tr>
<tr>
<td>
Website Name
</td>
<td>
<input name="title3" length="30" maxlength="30">
</td>
</tr>
<tr>
<td>
URL
</td>
<td>
<input name="url3" length="30">
</td>
</tr>
<tr>
<td>
Website Name
</td>
<td>
<input name="title4" length="30" maxlength="30">
</td>
</tr>
<tr>
<td>
URL
</td>
<td>
<input name="url4" length="30">
</td>
</tr>
<tr>
<td>
Website name
</td>
<td>
<input name="title5" length="30" maxlength="30">
</td>
</tr>
<tr>
<td>
URL
</td>
<td>
<input name="url5" length="30">
</td>
</tr>
</table>
</form>
</body>

<?
}
else
{
?>
<?

$connection = mysql_connect("localhost", "xpcorp_chris", "galloway") or die("Invalid server or user!");

mysql_select_db("xpcorp_test",$connection");

$query = "insert into
url_list(uid,title1,url1,title2,url2,title3,url3,title4,url4,title5,url5)
values('$username','$title1','$url1','$title2','$url2','$title3','$url3','$title4','$url4','$title5','$url5')";

$result = mysql_query($query,$connection) or die("Error in query!");

<head>
<basefont face="arial">
</head>
<body>
<center>
<h3>Success!</h3>
<table>
<tr>
<td>
<? echo $username; ?>'s bookmarks and account has been created!
</td>
</tr>
</center>
</body>

<?
}
?>

Anon

Probably the row that looks like this:
mysql_select_db("xpcorp_test",$connection");

Remove the last qoute ( " ), so the line looks like:
mysql_select_db("xpcorp_test",$connection);

and it should work.

/Staffan

Anon

Ok, Well Atleast its bringing up my form now.. But when I enter all my bookmarks, I Get "Error In Query!" Which is one of the errors I set.. the url is http://www.xpcorponline.com/push.php and here is the code..

<?
if (!$submit)
{
?>

<?
}
else
{
?>
<?

$connection = mysql_connect("localhost", "xpcorp_chris", "galloway") or die("Invalid server or user!");

mysql_select_db("xpcorp_test",$connection);

$result = mysql_query($query,$connection) or die("Error in query!");

<head>
<basefont face="arial">
</head>
<body>
<center>
<h3>Success!</h3>
<table>
<tr>
<td>
<? echo $username; ?>'s bookmarks and account has been created!
</td>
</tr>
</center>
</body>

<?
}
?>

khenriks

You should have this line:

$result = mysql_query($query,$connection) or die("Error in query!");

READ:

$result = mysql_query($query,$connection) or die("Error in query! ".mysql_error());

This will print out the error text from MySQL, it should help you figure out what's wrong.

-Kevin