Data not displaying in mysql database

Parkie02 · November 27, 2013

Can somebody tell me why my data is not stored in the database whodidntpay in the table complainant. When entering it in the webpage and sign up button is selected it is suppose to add it to the db.

<html>
	<head>
		<title> 
			Registration Form
		</title>
	</head>
	
<body>
<form method='post' action='registration.php'>
	<table width='400' border='5' align='center'>
	<tr>
		<td align='center' colspan='5'><h1>Registration Form</h1></td>
	</tr>

	<tr>
		<td>Username:</td>
		<td><input type='text' name='username' /></td>
	</tr>
	
	<tr>
		<td>Password:</td>
		<td><input type='password' name='password' /></td>
	</tr>
	
	<tr>
		<td>Email:</td>
		<td><input type='text' name='email' /></td>
	</tr>
	
	<tr>
		<td>Company Name:</td>
		<td><input type='text' name='company' /></td>
	</tr>
	
	<tr>
		<td colspan='5' align='center'><input type='submit' name='submit' value='Sign Up' /></td>
	</tr>
	
	
	</table>
</form>
</body>
</html>
<?php
mysql_connect("localhost","root","mj2015");
mysql_select_db("whodidntpay");

	if(isset($_POST['submit']))
	{
		$com_username = $_POST['username'];
		$com_password = $_POST['password'];
		$com_email = $_POST['email'];
		$com_companyname = $_POST['company'];
		
		if($com_username=='')
		{
			echo "<script>alert('Please Enter Username')</script>";
			exit();
		}
		
		if($com_password=='')
		{
			echo "<script>alert('Please Enter Password')</script>";
			exit();
		}
		
		if($com_email=='')
		{
			echo "<script>alert('Please Enter Email')</script>";
			exit();
		}
		
		if($com_companyname=='')
		{
			echo "<script>alert('Please Enter Company Name')</script>";
			exit();
		}
		
		$check_email = "select * from complainant where email='$com_email'";
		$run = mysql_query($check_email);
		if(mysql_num_rows($run)>0)
		{
			echo"<script>alert('Email $com_email already exists, please try another email')</script>";
			exit();
		}
		
		$query = "insert into complainant(username,password,c_name,email) values ('$com_username','$com_password','$com_email','$com_companyname')";
		if (mysql_query($query))
		{
			echo "<script>alert('Registration Successful')</script>";
		}
	}	

?>

BrodaNoel · November 27, 2013

I cant see an error.

Please, do that:

En the first line of the file, write this:

if(isset($_POST['username'])){
var_dump($_POST);
}

And, append in each "mysql_query' function that: 'or die(mysql_error())'.

Example:

$check_email = "select * from complainant where email='$com_email'";
$run = mysql_query($check_email) or die(mysql_error());

Then, go to the site, write your data, push "submit" and paste here everything.

On the other side, check that: you have problems (in a future) with a vulnerability (security). read about: MySQL Injection.

Parkie02 · November 27, 2013

This is how i changed it, if it is not done correctly can you maybe help me, im new to php.

<html>
	<head>
		<title> 
			Registration Form
		</title>
	</head>
	
<body>
<form method='post' action='registration.php'>
	<table width='400' border='5' align='center'>
	<tr>
		<td align='center' colspan='5'><h1>Registration Form</h1></td>
	</tr>

	<tr>
		<td>Username:</td>
		<td><input type='text' name='username' /></td>
	</tr>
	
	<tr>
		<td>Password:</td>
		<td><input type='password' name='password' /></td>
	</tr>
	
	<tr>
		<td>Email:</td>
		<td><input type='text' name='email' /></td>
	</tr>
	
	<tr>
		<td>Company Name:</td>
		<td><input type='text' name='company' /></td>
	</tr>
	
	<tr>
		<td colspan='5' align='center'><input type='submit' name='submit' value='Sign Up' /></td>
	</tr>
	
	
	</table>
</form>
</body>
</html>
<?php
mysql_connect("localhost","root","mj2015");
mysql_select_db("whodidntpay");

if(isset($_POST['username'])){
var_dump($_POST);
}
	if(isset($_POST['submit']))
	{
		$com_username = $_POST['username'];
		$com_password = $_POST['password'];
		$com_email = $_POST['email'];
		$com_companyname = $_POST['company'];
		
		if($com_username=='')
		{
			echo "<script>alert('Please Enter Username')</script>";
			exit();
		}
		
		if($com_password=='')
		{
			echo "<script>alert('Please Enter Password')</script>";
			exit();
		}
		
		if($com_email=='')
		{
			echo "<script>alert('Please Enter Email')</script>";
			exit();
		}
		
		if($com_companyname=='')
		{
			echo "<script>alert('Please Enter Company Name')</script>";
			exit();
		}
		
		$check_email = "select * from complainant where email='$com_email'";
		$run = mysql_query($check_email) or die(mysql_error());
		if(mysql_num_rows($run)>0)
		{
			echo"<script>alert('Email $com_email already exists, please try another email')</script>";
			exit();
		}
		
		$query = "insert into complainant(username,password,c_name,email) values ('$com_username','$com_password','$com_email','$com_companyname')";
		if (mysql_query($query))
		{
			echo "<script>alert('Registration Successful')</script>";
		}
	}	

?>

This is what I get on the screen after entering information: and pressing signup:

array (size=5)
'username' => string 'Martin' (length=6)
'password' => string 'mj2015' (length=6)
'email' => string '[email protected]' (length=22)
'company' => string 'Baas' (length=4)
'submit' => string 'Sign Up' (length=7)

Barand · November 27, 2013

One thing I did notice, you are putting email in company field and vice versa

$query = "insert into complainant(username,password,c_name,email) values ('$com_username','$com_password','$com_email','$com_companyname')";

cyberRobot · November 27, 2013

Since the problem seems to be related to the insert query, display the result from mysql_error() some place after the following query:

$query = "insert into complainant(username,password,c_name,email) values ('$com_username','$com_password','$com_email','$com_companyname')";
if (mysql_query($query))

Parkie02 · November 27, 2013

Where do I append it? Can you maybe show me

cyberRobot · November 27, 2013

Try the following:

$query = "insert into complainant(username,password,c_name,email) values ('$com_username','$com_password','$com_email','$com_companyname')";
if (mysql_query($query))
{
     echo "<script>alert('Registration Successful')</script>";
}
echo mysql_error();

Parkie02 · November 27, 2013

Thank you got the problem, had error with my foreign key

cyberRobot · November 27, 2013

Glad you figured it out! Now just remember to remove the debugging code before going live.

BrodaNoel · November 27, 2013

I'm recommend everytimes push the "die(mysql_error())" in debug time.

That remove very much problems.

Just... for the next :

Change

$query = "insert into complainant(username,password,c_name,email) values ('$com_username','$com_password','$com_email','$com_companyname')";
        if (mysql_query($query))
        {
            echo "<script>alert('Registration Successful')</script>";
        }

for that:

$query = "insert into complainant(username,password,c_name,email) values ('$com_username','$com_password','$com_email','$com_companyname')";
$resTest = mysql_query($query) or die(mysql_error());
        if ($resTest)
        {
            echo "<script>alert('Registration Successful')</script>";
        }

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Data not displaying in mysql database

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