[SOLVED] Can't get data into MySQL database!!!

[GOLDfish_74]

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Anywho, a summary of the story goes, I am trying to run a really simple script that would place users in my database when they registered:

<?php 
// Connects to your Database 
mysql_connect("localhost", "goldfish_root", "password") or die(mysql_error()); 
mysql_select_db("goldfish_retaildefault") or die(mysql_error()); 

//This code runs if the form has been submitted
if (isset($_POST['submit'])) { 

//This makes sure they did not leave any fields blank
if (!$_POST['username'] | !$_POST['pass'] | !$_POST['pass2'] ) {
die('You did not complete all of the required fields');
}

// checks if the username is in use
if (!get_magic_quotes_gpc()) {
$_POST['username'] = addslashes($_POST['username']);
}
$usercheck = $_POST['username'];
$check = mysql_query("SELECT user_name FROM users WHERE user_name = '$usercheck'") 
or die(mysql_error());
$check2 = mysql_num_rows($check);

//if the name exists it gives an error
if ($check2 != 0) {
die('Sorry, '.$_POST['username'].', you are already registered.');
}

// this makes sure both passwords entered match
if ($_POST['pass'] != $_POST['pass2']) {
die('Your passwords did not match.');
}

// here we encrypt the password and add slashes if needed
$_POST['pass'] = md5($_POST['pass']);
if (!get_magic_quotes_gpc()) {
$_POST['pass'] = addslashes($_POST['pass']);
$_POST['username'] = addslashes($_POST['username']);
}

// now we insert it into the database
$insert = "INSERT INTO users (user_id, user_name, user_pass, user_sec)
VALUES ('".$_POST['userid']."', '".$_POST['username']."', '".$_POST['pass']."', '".$_POST['security']."')";
mysql_query($insert);
?>


<h1>Registered</h1>
<p>Thank you, you have registered - you may now login</a>.</p>
<p><? echo ($_POST['pass']); echo ($_POST['security']); echo ($_POST['username']); echo ($_POST['userid']) ?></p>

<?php 
} 
else 
{ 
?>


<form action="<?php echo $_SERVER['PHP_SELF']; ?>" method="post">
<table border="0">
<tr><td>Your User ID is:</td><td>
<? $userid = rand(1000, 9999); echo($userid); ?>
<input type="hidden" name="userid" value="<? echo($userid); ?>">
</td></tr>
<tr><td>Full Name:</td><td>
<input type="text" name="username" maxlength="60">
</td></tr>
<tr><td>Password:</td><td>
<input type="password" name="pass" maxlength="10">
</td></tr>
<tr><td>Confirm Password:</td><td>
<input type="password" name="pass2" maxlength="10">
</td></tr>
<tr><td>Security Level</td><td>
<input type="radio" name="security" value="2">Supervisor<br><input type="radio" name="security" value="1">Employee</td></tr>
<tr><th colspan=2><input type="submit" name="submit" value="Register"></th></tr> </table>
</form>

<?php
}
?> 

 

This did not report any errors but also failed to put any data in the database.

As this failed to work I figured I'd just try a simple alternative:

<?
error_reporting(E_ALL);  
mysql_connect("localhost", "goldfish_root", "password") or die(mysql_error()); 
mysql_select_db("goldfish_retaildefault") or die(mysql_error()); 
$insert = "INSERT INTO users (user_id, user_name, user_pass, user_sec)
VALUES ('1234, Josh, 3883, 1')";
$add_member = mysql_query($insert);
?>

This also failed to work with no errors, and now I have no idea what is going wrong.

 

These questions all came about when I tried to use EasyPHP to test my script while I was on public transport and away from home. When it seemed they did not like Vista, I got a free web host (qupis.com) and the problems are still the same. I am sure there is nothing wrong with my code, so what could be going wrong?

 

Regards,

 

THEfish.

[Richzilla]

try removing this part and see what happens - (user_id, user_name, user_pass, user_sec)

 

[GOLDfish_74]

Removing that line made absolutely no difference what so ever. It ran without reporting an error and still failed to put the data into the database.

 

Regards,

 

THEfish!

[TNGuy03]

If you have an 'id' field for the records in your db table which is auto incremented, be sure to pass 'NULL' for that field.

[radar]

Here is how I insert things to a database...

 

$insert = "INSERT INTO users SET

user_id = '',

user_name = 'Josh',

user_pass = '3883',

user_sec = '1'";

 

you can also use this to update. by changing the syntax from insert to update.. such as..

 

$update = "UPDATE users SET

user_id = '',

user_name = "Jump",

user_pass = '3434324',

user_sec = '2' WHERE user_id = '$user_id'";

 

hope it helps..

 

 

[penguin0]

If user_id is auto incrementing, there is no reason to insert it at all.

[GOLDfish_74]

Just to let you know, I have done a fair bit of PHP and MySQL before. And I know what auto_increment is... AND ITS NOT SET TO THAT. If you had cared to look, the id is set by a random number generator, not by the database.

 

If you have any more suggestions that could work, please post.

 

Regards,

 

THEfish!

 

PS: I tried using SET instead of VALUES with no difference in result.

[GOLDfish_74]

Turns out it was the most obvious thing in the world!!! user_pass, was meant to be user_pin. and I had error display off. That and I agree EasyPHP is not what I should be using... lol, THANKS for all your patience to find such a pointless flaw.

 

Ciao,

 

THEfish!