PHP fails to post to MySQL Database - php

Question

I have a formText.php file that contains a form with the following code form code:
<form action="insert.php" method="post">
<p>
<label for="theNames">Name:</label>
<input type="text" name="theName" id="theName">
</p>
<p>
<label for="theCitys">City:</label>
<input type="text" name="theCity" id="theCity">
</p>
<p>
<label for="theAges">Are you over eighteen?(Y/N)</label>
<input type="text" name="theAge" id="theAge">
</p>
<p>
<label for="theDates">Date:</label>
<input type="text" name="theDate" id="theDate">
</p>
<input type="submit" value="Submit">
</form>
Then I have an insert.php file with the following script:
<?php
/* Attempt MySQL server connection. Assuming you are running MySQL
server with default setting (user 'root' with no password) */
$link = mysqli_connect("localhost", "root", "root","phpteste");
// Check connection
if($link === false){
die("ERROR: Could not connect. " . mysqli_connect_error());
}
// Escape user inputs for security (EDITED)
$theName = mysqli_real_escape_string($link, $_POST['theName']);
$theCity = mysqli_real_escape_string($link, $_POST['theCity']);
$theAge = mysqli_real_escape_string($link, $_POST['theAge']);
$theDate = mysqli_real_escape_string($link, date("Y-m-d h:i:s",$_POST['theDate']));
// attempt insert query execution
$sql = "INSERT INTO tabelateste (id, name, city, overeighteen, date) VALUES (NULL, '$theName', '$theCity', '$theAge', '$theDate')";
if(mysqli_query($link, $sql)){
echo "Records added successfully.";
} else{
echo "ERROR: Could not able to execute $sql. " . mysqli_error($link);
}
// close connection
mysqli_close($link);
?>
My database is called phpteste and my table name is tabelateste.
What am I doing wrong here?
Whenever I click Submit nothing comes up and nothing gets added to the database.

Answer 1

Your post data name fields are wrong. SO you need to change below line:
// Escape user inputs for security
$theName = mysqli_real_escape_string($link, $_POST['theName']);
$theCity = mysqli_real_escape_string($link, $_POST['theCity']);
$theAge = mysqli_real_escape_string($link, $_POST['theAge']);
$theDate = mysqli_real_escape_string($link, date("Y-m-d h:i:s",$_POST['theDate']));
You need to change date to signup_date as per your database table structure.
$sql = "INSERT INTO tabelateste (name, city, overeighteen, signup_date) VALUES ('$theName', '$theCity', '$theAge', '$theDate')";

Answer 2

$sql = "INSERT INTO tabelateste (`name`, `city`, `overeighteen`, `date`) VALUES ('$theName', '$theCity', '$theAge', '$theDate')";
Use this code

Answer 3

I just tested your code (copied and pasted) and it works perfectly under my server configuration (Windows 10 - PHP 5.6) . My best guess is that you have a typo in either the table name or the MySQL configuration.
If you copied this code from another site. Please check that you created the database and the table , and that the MySQL configuration is correct.
A good to check for this kind of mistakes so is to read the PHP error logs

Answer 4

Try it like this maybe
if(isset($_POST['submit']) && !empty($_POST) ){
$theName = $_POST['theName'];
$theCity = $_POST['theCity'];
$theAge = $_POST['theAge'];
$theDate = $_POST['theDate'];
$servername = "localhost";
$username = "root";
$password = "root";
$dbname = "phpteste";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "INSERT INTO tabelateste (name, city, overeighteen, date)
VALUES ('$theName ', '$theCity ', '$theAge ', '$theDate ')";
if ($conn->query($sql) === TRUE) {
$last_id = $conn->insert_id;
} else {
echo "Error: " . $sql . "<br>" . $conn->error;
}
}