PHP MySQL Insert into table [duplicate] - php

Question

This question already has answers here:
Can I mix MySQL APIs in PHP?
(4 answers)
Closed 6 years ago.
I've followed a tutorial tot he exact code and i can't seem to get this to work
When I submit the form nothing gets put into my table
I can't seem to find the problem, ive changed a bunch of stuff but i cant seem to find the answer.
If anyone can have a look at my code and help me resolve the issue i would be very thankfull
Code
<form class="form" method="post" action="index.php">
<h2>
Post new account...
</h2>
<div class="form-group">
<label for="title">Title</label>
<input required type="text" class="form-control" name="title" placeholder="Enter a title for the account...">
</div>
<div class="form-group">
<div class="row">
<div class="col-md-6">
<label for="server">Server</label>
<input required type="text" class="form-control" name="server" placeholder="Enter Server...">
</div>
<div class="col-md-6">
<label for="price">Price</label>
<input required type="number" class="form-control" name="price" placeholder="Enter Price...">
</div>
</div>
</div>
<div class="form-group">
<div class="row">
<div class="col-md-6">
<label for="Energy">Energy</label>
<input required type="number" class="form-control" name="energy" placeholder="Enter Energy...">
</div>
<div class="col-md-6">
<label for="Stamina">Stamina</label>
<input required type="number" class="form-control" name="stamina" placeholder="Enter Stamina...">
</div>
</div>
</div>
<div class="form-group">
<div class="row">
<div class="col-md-6">
<label for="Jewels">Jewels</label>
<input required type="number" class="form-control" name="jewels" placeholder="Enter Jewels...">
</div>
<div class="col-md-6">
<label for="honor">Stamina</label>
<input required type="number" class="form-control" name="honor" placeholder="Enter Honor...">
</div>
</div>
</div>
<div class="form-group">
<label for="image">Image URL</label>
<input required type="text" class="form-control" name="image" placeholder="Enter the image link...">
</div>
<div class="form-group">
<label for="link">Link URL</label>
<input required type="text" class="form-control" name="link" placeholder="Enter the payment link...">
</div>
<div class="form-group">
<input class="btn btn-primary btn-md" type="submit" name="submit">
</div>
</form>
<?php
if (isset($_POST['submit'])) {
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("We're having problems at the moment, please come back later!");
}
$query = "INSERT INTO accounts (title,price,energy,stamina,jewels,honor,damage,server,image,link) VALUES ('$_POST[title]','$_POST[price]','$_POST[energy]','$_POST[stamina]','$_POST[jewels]','$_POST[honor]','$_POST[damage]','$_POST[server]','$_POST[image]','$_POST[link]')";
mysql_query($query,$conn);
mysql_close($conn);
};
?>

Answer 1

Check your connection:
if (!$conn) {
die("Connection failed: " . mysqli_connect_error());
}
Then check if it is inserting into the database or not. If there's an error, it will let you know as well.
if (mysqli_query($conn, $sql)) {
echo "New record created successfully";
} else {
echo "Error: " . $sql . "<br>" . mysqli_error($conn);
}
Also, don't mix mysql_ with mysqli_. Start using mysqli_ instead of mysql_. mysql_ is deprecated.