First SQL Database not working - php

Question

I am working on a personal project of mine and I am accepting a name and email so I created a database called:
emailtemp with columns of "name" and "email"
PHPMyAdmin Page
HTML Form:
<form method="post" action="connect.php" class="contact100-form validate-form">
<div class="wrap-input100 m-b-10 validate-input" data-validate = "Name is required">
<input class="s2-txt1 placeholder0 input100" type="text" name="username" placeholder="Your Name">
<span class="focus-input100"></span>
</div>
<div class="wrap-input100 m-b-20 validate-input" data-validate = "Email is required: ex#abc.xyz">
<input class="s2-txt1 placeholder0 input100" type="text" name="emailAddress" placeholder="Email Address">
<span class="focus-input100"></span>
</div>
<div class="w-full">
<button class="flex-c-m s2-txt2 size4 bg1 bor1 hov1 trans-04">
Subscribe
</button>
</div>
</form>
PHP File:
<?php
$connect = mysqli_connect("mysql.*****.net","*****","****>","emailtemp");
//Sending form data to sql db.
mysqli_query($connect,"INSERT INTO posts (customerName, customerEmail)
VALUES ('$_POST[username]', '$_POST[emailAddress]')";
?>
Based off everything I read and looked at it it should work but after I click "subscribe" it pulls up connect.php and says "is currently unable to handle this request.
HTTP ERROR 500"
Is there a way to have this process in the background and stay on the same page and just say successful?

Answer 1

http://php.net/manual/en/mysqli.prepare.php
$connect = mysqli_connect("mysql.*****.net","*****","****>","emailtemp");
if ($stmt = $mysqli_prepare($connect,"INSERT INTO posts (customerName, customerEmail) VALUES (?, ?)")) {
mysqli_stmt_bind_param($stmt, "ss", $_POST['username'],$_POST['emailAddress']);
mysqli_stmt_execute($stmt);
}

Answer 2

Looks like you're missing a ) at the end of your mysqli_query() call.
If your DB is really like this:
emailtemp with columns of "name" and "email"
.. then you should match those in your query:
mysqli_query($connect,"INSERT INTO emailtemp (name, email)
VALUES ('$_POST[username]', '$_POST[emailAddress]')");
Recommended to check for user/email first, if they're somehow empty that will throw an exception.
like this:
$username = $_POST[username];
$email = $_POST[emailAddress];
if ($username && $email) {
// stuff here
} else {
// other stuff here
}

Answer 3

You forgot to close the mysqli_query function with an ) and you forgot the brackets around username and emailAdress
mysqli_query($connect,"INSERT INTO posts (customerName, customerEmail)
VALUES ('$_POST[username]', '$_POST[emailAddress]')");
Also this code makes you vulnerable to an SQL injection
Users can inject other commands to the SQL server and they get executed with admin rights.
You can use PDO, a secure implementation of SQL in PHP (if used right of course)
In your example it would be:
$dbhost = "your-host";
$dbname = "your-db-name";
$dbusername = "your-username";
$dbpassword = "your-password";
$conn = new PDO("mysql:host=$dbhost;dbname=$dbname", $dbusername, $dbpassword);
$statement = $conn->prepare("INSERT INTO posts (customerName, customerEmail)
VALUES (:username, :emailAddress");
$statement->execute(array(
"username" => $_POST['username'],
"emailAdress" => $_POST['emailAdress']
));