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error parameter

dean012

By dean012
in PHP Coding Help

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dean012 Member

dean012

Posted
Posted

error on line 66

 

 

 

<html>
    <head>
        <title>Registration Form</title>
    </head>

<body>
<form method='post' action='registration.php'>
    <table width='400' border='5' align='center'>
    
        <tr>
        <td><h1>Registration Form</h1></td>
        
    </tr>
    
    <tr>
        <td>User Name:</td>
        <td><input type='text' name='name'/></td>
    </tr>
    
    <tr>
        <td>Password:</td>
        <td><input type='password' name='pass'/></td>
    </tr>
    <tr>
        <td>Email:</td>
        <td><input type='text' name='email'/></td>
    </tr>
    <tr>
        <td><input type='submit' name='submit' value='Sign Up'/></td>
        
    </tr>
    
    </table>
</form>    
</body>
</html>
<?php
mysql_connect("localhost","root","");
mysql_select_db("user_db");
    if(isset($_POST['submit'])){
    
    $user_name = $_POST['name'];
    $user_pass = $_POST['pass'];
    $user_email = $_POST['email'];
    
    if($user_name==''){
    echo "<script>alert('Please enter your Username')</script>";
    exit();
    }
    
    if($user_pass==''){
    echo "<script>alert('Please enter your password')</script>";
    exit();
    }
    
    if($user_email==''){
    echo "<script>alert('Please enter your email')</script>";
    exit();
    }
    
    $check_email="select *select* from users where
    user_email='$user_email'";
    
    $run = mysql_query($check_email);
    
    if(mysql_num_rows($run)>0){
    
    echo"<script>alert('Email $user_email
    is already exist in our databse, please try another
    one')</script>";
    exit();
    }
    
    $query = "insert into users
    (user_name,user_pass,user_email) values('
    $user_name','$user_pass','$user_email')";
    if(mysql_query($query)){
    
    echo "<script>alert('Registration
    Successfull!')</script>";
    }
    
}
    


?>

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gristoi Newbie

gristoi

Posted
Posted

i mean, correct - you do have an error there. If you would like someone to comment on how to fix it then please make sure you put a bit more information in your post as to what the problem is and what you have done to solve the issue so far. To start you off, ensure 

 $user_email

is not returning something that is causing the string to escape

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jcbones Newbie

jcbones

Posted
Posted

If the parameter is NOT a resource, and you are passing it the result of a query, then the query must be returning false, which is boolean.  A query returning false means it failed.  The first recourse of action for this would be to return the query error using mysql_error. This would tell you that you have a malformed query string, which would lead you to looking up the proper http://dev.mysql.com/doc/refman/5.0/en/select.html'>SELECT syntax for a MySQL query string.

 

You should be using mysqli_error because you should be using mysqli or PDO at this point.

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