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Darksevear
Forum Newbie
Posts: 11
Joined: Sat May 13, 2006 10:38 pm

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Post by Darksevear »

hawleyjr | Please use

Code: Select all

,

Code: Select all

and [syntax="..."] tags where appropriate when posting code. Your post has been edited to reflect how we'd like it posted. Please read:  [url=http://forums.devnetwork.net/viewtopic.php?t=21171]Posting Code in the Forums[/url] to learn how to do it too.[/color]


Hello, I get an error when ever i try this script. It says "$end expected on line 82" I dont understand what im missing. 

Can someone please help me : )

Cheers, daniel

Script:

Code: Select all

<html>
<head>
<basefont face="Arial">
</head>
<body>

<?php

if (!isset($_POST['submit'])) {
// form not submitted
?>

    <form action="<?=$_SERVER['PHP_SELF']?>" method="post">
    Username: <input type="text" name="user"> 
    Password: <input type="text" name="pass"> <br>
	<input type="text" name="email1"> 
	@<input type="text" name="email2"> <br>
	First Name: <input type="text" name="name1">
	Last Name: <input type="text" name="name2"><br>
	Country: <input type="text" name="country">
	
    <input type="submit" name="submit">
    </form>

<?php
}
else {


// form submitted
// set server access variables
    $host = 'localhost';
    $username = "user";
	$password = "password";
    $db = "wintersword";
    
// get form input
    // check to make sure it's all there
    // escape input values for greater safety
    $user = empty($_POST['user']) ? die ("Please Enter A Username") : mysql_escape_string($_POST['user']);
    $pass = empty($_POST['pass']) ? die ("Please Enter A Password") : mysql_escape_string($_POST['pass']);
	$email1 = empty($_POST['email1']) ? die ("Please Enter Your Email Address") : mysql_escape_string($_POST['email1']);
    $email2 = empty($_POST['email2']) ? die ("Please Enter Your Email Address") : mysql_escape_string($_POST['email2']);
	$name1 = empty($_POST['name1']) ? die ("Please Enter Your First Name") : mysql_escape_string($_POST['name1']);
    $name2 = empty($_POST['name2']) ? die ("Please Enter Your Last Name") : mysql_escape_string($_POST['name2']);
	$country = empty($_POST['country']) ? die ("Please Enter What Country You Are In") : mysql_escape_string($_POST['country']);
	
    // open connection
	$connection = mysql_connect("$host", "$username", "$password") or die ("Unable to connect!");

if(!empty($_POST)){

    $user = mysql_real_escape_string($_POST['user']);

    $result = mysql_query("SELECT count(*) FROM `users` WHERE `user` = '$user'") or die(mysql_error());
    $count = mysql_result($result,0);

    if($count > 0){
        die('This username is already in use.  Please select a different username.');
    }
    else {
	
	    // create query
    $query = "INSERT INTO users (user, pass, email1, email2, name1, name2, country) VALUES ('$user', '$pass', '$email1', '$email2', '$name1', '$name2', '$country')";
    
    // execute query
    $result = mysql_query($query) or die ("Error in query: $query. ".mysql_error());
    
    // print message with ID of inserted record
    echo "New record inserted with ID ".mysql_insert_id();
    
    // close connection
    mysql_close($connection);
	
	}

} 
?>
</body>
</html>
hawleyjr | Please use

Code: Select all

,

Code: Select all

and [syntax="..."] tags where appropriate when posting code. Your post has been edited to reflect how we'd like it posted. Please read:  [url=http://forums.devnetwork.net/viewtopic.php?t=21171]Posting Code in the Forums[/url] to learn how to do it too.[/color]

Woops, I'm sorry.
Last edited by Darksevear on Sun May 14, 2006 4:23 pm, edited 1 time in total.
User avatar
andym01480
Forum Contributor
Posts: 390
Joined: Wed Apr 19, 2006 5:01 pm

Post by andym01480 »

You need another curly bracket at the end for the else at line 27.
BTW you had better put your code in the php tags before a moderator sees it!!!!
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