Error 500 when performing Mysqli Queries

Question

I've been working on a tiny form fetch and insert to MySQL database system with PHP. I've made this type of code work before on a local WAMP server and now I'm trying to do it for real this time on my hosting, but I keep getting the 500 Error Internal Server error with the following code.

HTML:

<!DOCTYPE html>
<html lang="en">
<head>
    <meta charset="utf-8">
    <meta http-equiv="X-UA-Compatible" content="IE=edge">
    <meta name="viewport" content="width=device-width, initial-scale=1">
    <link href="css/bootstrap.min.css" rel="stylesheet">
    <link rel="stylesheet" type="text/css" href="Main_Stylesheet.css">
    <script   src="https://ajax.googleapis.com/ajax/libs/jquery/1.11.3/jquery.min.js"></script>
    <script src="js/bootstrap.min.js"></script>
    <title>Contact</title>
</head>
<body>
    <nav class="navbar navbar-default" id="Body_Nav" style="margin-bottom: none;">
        <div class="container-fluid">
            <div class="navbar-header">
                <a class="navbar-brand" href="index.php">Filosofie in Amsterdam</a>
                <button type="button" class="navbar-toggle collapsed" data-toggle="collapse" data-target="#myNavbar">
                <span class="icon-bar"></span>
                <span class="icon-bar"></span>
                <span class="icon-bar"></span> 
                </button>
            </div>
            <div class="collapse navbar-collapse" id="myNavbar" >
            <ul class="nav navbar-nav">
                <li><a href="index.php">Home</a></li>
                <li><a href="bestel.php">Bestel</a></li>
                <li><a href="preview.php">Preview</a></li>
                <li class="active"><a href="contact.php">Contact</a></li>
                <li><a href="us.php">Over Ons</a></li>
                <li><a href="donate.php">Doneren</a></li>
            </ul>
            </div>
        </div>
    </nav>
    <div class="container-fluid" id="Contact_Main_Header">
        <div class="row">
            <div class="col-lg-12">
                <h1 class="text-center" id="Contact_Main_Header_H1">Contact</h1>
            </div>
        </div>    
    </div>
    <div class="container">
        <div class="row">
            <div class="col-lg-6">
                <h2 class="text-center">Informatie</h2>
            </div>
            <div class="col-lg-6">
                <h2 class="text-center">Nieuwsbrief</h2>
                <p class="text-justify">Op de hoogte zijn van voortgang of het laatste nieuws? Geef je dan op via de nieuwsbrief hieronder:</p>
                <form method="post" action="formvalid.php">
                    <div class="form-group">
                        <label for="Contact_Form_Email">Email Adres:</label>
                        <input type="email" name="Contact_Form_Email" class="form-control" id="Contact_Form_Email" placeholder="Plaats hier je email adres" required>
                    </div>    
                    <div class="form-group">
                        <label for="Contact_Form_First_Name">Voornaam:</label>
                        <input type="text" name="Contact_Form_First_Name" class="form-control" id="Contact_Form_First_Name" placeholder="Plaats hier je voornaam" required>   
                    </div>
                    <div class="form-group">
                        <label for="Contact_Form_Last_Name">Achternaam:</label>
                        <input type="text" name="Contact_Form_Last_Name" class="form-control" id="Contact_Form_Last_Name" placeholder="Plaats hier je voornaam" required>
                    </div>    
                    <button type="submit" class="btn btn-default">Verzenden</button>
                </form>
            </div>
        </div>
    </div>
</body>
</html>

And the PHP:

 <?php
 $Server = "localhost";
 $Username = "filoso1q_admin";
 $PW = "Foxtrot15";
 $DB = "filoso1q_nieuwsbrief";
 $connection = mysqli_connect($Server, $Username, $PW, $DB);

 if($connection = false) {
     die("Error: " . mysqli_error_connect());
 }  

 $Email = $_POST("Contact_Form_Email");
 $First_Name = $_POST("Contact_Form_First_Name");
 $Last_Name = $_POST("Contact_Form_Last_Name");

 $sql = "INSERT INTO `Users` ('User_ID', 'Email', 'First_Name', 'Last_Name')  VALUES (NULL, '$Email', '$First_Name',  '$Last_Name');";

 mysqli_query($sql);

 mysqli_close($connection);
 ?>

I'm pretty certain that some of this should work, but for some reason I can only run queries in PHPmyadmin itself. The host I have uses the Cpanel webadmin for the webserver.

Besides the dupe, you need to connect `mysqli_query($connection, $sql);` you have all your answers now. — Funk Forty Niner, Feb 19 '16 at 22:40
`$Email = $_POST("Contact_Form_Email");` should be `$Email = $_POST["Contact_Form_Email"];` and for all the other `$_POST` — Matt, Feb 19 '16 at 22:40
^ and that too. It's questions like these that make my head spin. — Funk Forty Niner, Feb 19 '16 at 22:40
Okay, did that, but still won't perform the actual query. No error, nothing just blank. — Bo Kamphues, Feb 19 '16 at 22:45
`if($connection = false)` that is an assignment. You need to `if($connection == false)` as a comparison. I'll just reopen the question and let someone else fix all of those errors. Nothing that hasn't been covered in comments and thttp://stackoverflow.com/questions/11321491/when-to-use-single-quotes-double-quotes-and-backticks — Funk Forty Niner, Feb 19 '16 at 22:50

score 2 · Accepted Answer · edited May 23 '17 at 12:23

I changed my mind and decided to post an answer after all.

You're assigning if($connection = false) with one equal sign, rather than comparing with two if($connection == false).

Or, consult the manual on mysqli_connect() for more options:

http://php.net/manual/en/function.mysqli-connect.php

You're using incorrect identifier qualifiers for the columns in your query. They should be ticks and not regular quotes.

http://dev.mysql.com/doc/en/identifier-qualifiers.html

Sidenote about the ticks: You don't even need those, since you are not using any of MySQL's reserved words, nor do they contain spaces or hyphens, or any other character that MySQL would complain about, so you can safely remove those for the column names.

Then you used the incorrect type of brackets for your POST arrays. You used (...) rather then square ones [...].

You also did not pass the connection to your query, which needs to be the first parameter.

http://php.net/manual/en/mysqli.query.php

Rewrite with error checking on the query:

<?php
 $Server = "localhost";
 $Username = "filoso1q_admin";
 $PW = "Foxtrot15";
 $DB = "filoso1q_nieuwsbrief";
 $connection = mysqli_connect($Server, $Username, $PW, $DB);

 if($connection == false) {
     die("Error: " . mysqli_error_connect());
 }  

 $Email = $_POST["Contact_Form_Email"];
 $First_Name = $_POST["Contact_Form_First_Name"];
 $Last_Name = $_POST["Contact_Form_Last_Name"];

 $sql = "INSERT INTO `Users` (`User_ID`, `Email`, `First_Name`, `Last_Name`)  VALUES (NULL, '$Email', '$First_Name',  '$Last_Name');";

 mysqli_query($connection, $sql) or die(mysqli_error($connection));

 mysqli_close($connection);
 ?>

Add error reporting to the top of your file(s) which will help find errors.

<?php 
error_reporting(E_ALL);
ini_set('display_errors', 1);

// Then the rest of your code

Sidenote: Displaying errors should only be done in staging, and never production.

and or die(mysqli_error($con)) to mysqli_query() for debugging.

Also make sure that you have the correct types chosen for your columns and that their lengths are long enough to accommodate the data.

Plus, make sure that your User_ID column does accept NULL values and isn't an AUTO_INCREMENT. Otherwise, using NULL will fail.

You would need to change your query to VALUES ('', '$Email', ...

Your present code is open to SQL injection. Use mysqli_* with prepared statements, or PDO with prepared statements.

An 500 error is a server error and can be viewed in your logs. Either check those, or use error reporting as outlined above in order to be viewed on screen.

Thanks, the error report actually saved me. Turned out that the username was simple being denied access because of the way the webhosting had originally set it up! — Bo Kamphues, Feb 19 '16 at 23:26
@BoKamphues You're welcome Bo, glad to know it's been resolved. However, once you've gone through all that, your code did contain many syntax errors and missing parameters and is the reason for my posting this answer. — Funk Forty Niner, Feb 19 '16 at 23:28
Yes, it was indeed very messy code. I'll get to it with the marking now! — Bo Kamphues, Feb 19 '16 at 23:33
@BoKamphues Cheers Bo. Have a read about using a prepared statement. There are links in my answer to have a look at. If and when (or already live with your project), you should be using those. Otherwise, your server/database could get compromised. *Stay safe*. ;-) — Funk Forty Niner, Feb 19 '16 at 23:36

Error 500 when performing Mysqli Queries

1 Answers1