how to insert details in the table

Question

<html>
<head></head>

<body>

<form action="register.php" method="post">
    Username: <input type="text" name="uname" >

    Password: <input type="text" name="upass" >

    Email: <input type="text" name="uemail" >


    <input type="submit" value="register" name="register" >

</form>
</body>


</html>

<?php
$con= mysqli_connect("localhost","root","","users") or die('Error Connecting to Database');

    if(isset($_POST['register']))  
    {  
        $user_name=$_POST['uname'];  
        $user_pass=$_POST['upass'];  
        $user_email=$_POST['uemail'];

if($user_name=='')  
    {  

        echo"<script>alert('Please enter the name')</script>";  
exit();  
    }  

    if($user_pass=='')  
    {  
        echo"<script>alert('Please enter the password')</script>";  
exit();  
    }  

    if($user_email=='')  
    {       
    echo"<script>alert('Please enter the email')</script>";  
exit();  
    }


    $check_email="select * from accounts WHERE email='$user_email' and username='$user_name'";  
    $result=mysqli_query($con,$check_email);    

    $count=mysqli_num_rows($result) or die('Connection error');//here is the cause of the error

        if($count>0)  
        {  
            echo "<script>alert('User already exists , Please try another one!')</script>";  
            exit();  
        }  


        $insert_user="insert into accounts (username,password,email) VALUE ('$user_name','$user_pass','$user_email')";  
    if(mysqli_query($con,$insert_user))  
    {  
        echo"<script>window.open('welcome.php','_self')</script>";  
    }  



}  

?>

Your script is at risk of [SQL Injection Attack](https://stackoverflow.com/q/60174/5914775). Have a look at what happened to [Little Bobby Tables](http://bobby-tables.com/). Even [if you are escaping inputs, its not safe!](https://stackoverflow.com/q/5741187/5914775). Use [prepared parameterized statements](https://php.net/manual/en/mysqli.quickstart.prepared-statements.php) instead. — Tom Udding, Nov 23 '17 at 10:06
`or die(mysqli_connect_error());` try this so that you will get exact connection error details and able to rectify that.Make sure all parameters provided for connection is correct — Alive to die - Anant, Nov 23 '17 at 10:07

Raghav · Answer 1 · 2017-11-23T12:00:46.580

As per the Code Observed that

1) You are trying to fetch records for new users whose account doesn't exist in the DB. So If you enter new Username, Email , Password It will automatically goes to

die('Connection error');

Because count will be '0' here. So PHP script Execution stops and the record will not be inserted in DB.

Try this Below Code. !!

<?php
$con= mysqli_connect("localhost","root","root","testdb") or die('Error Connecting to Database');

    if(isset($_POST['register']))  
    {  
        $user_name=$_POST['uname'];  
        $user_pass=$_POST['upass'];  
        $user_email=$_POST['uemail'];

if(empty($user_name))
    {  
        echo"<script>alert('Please enter the name')</script>";  
        exit();  
    }  

    if(empty($user_pass))
    {  
        echo"<script>alert('Please enter the password')</script>";  
        exit();  
    }  

    if(empty($user_email))
    {       
        echo"<script>alert('Please enter the email')</script>";  
        exit();  
    }

       $check_email="select * from accounts WHERE email='$user_email' OR name='$user_name'";  
       $result = mysqli_query($con,$check_email);    

        while($row = $result->fetch_array()){
            if($user_email == $row['email'] || $user_name == $row['name']){
                echo "<script>alert('User already exists , Please try another one!')</script>";  
               exit();  
            }
        }

        $insert_user="insert into accounts (name,password,email) VALUE ('$user_name','$user_pass','$user_email')";  
        if(mysqli_query($con,$insert_user))  
            {  
                echo"<script>window.open('welcome.php','_self')</script>";  
            } 
}  

?>

Hope this helps..!!!

thanks @raghav !! i tried it but it still didnt change anything! — kayjordan, Nov 23 '17 at 10:48
when i enter details of existing user and submit the page shows no error and just reloads — kayjordan, Nov 23 '17 at 11:04
ur code helped @raghav so thanks a lot but the only error im getting is the one with existing user check — kayjordan, Nov 23 '17 at 11:07
Fatal error: Call to a member function fetch_array() on boolean in C:\wamp64\www\try\login.php on line 38 <==this is the error i got @raghav — kayjordan, Nov 23 '17 at 16:05

how to insert details in the table

1 Answers1