将表单导出数据加载到php数组中

Question

I am trying this way but it doesnt work:

  $sql = "INSERT INTO customers (userName,password,email,tel,companyName,idNumber,streetAddress,zipCode,city,coAdress) VALUES ($_POST['userName'],$_POST['password'],$_POST['email'] ,$_POST['tel'],$_POST['companyName'],$_POST['idNumber'],$_POST['streetAddress'],$_POST['zipCode'],$_POST['city'],$_POST['coAdress']);

 if ($conn->query($sql) === TRUE) {
     echo "New record created successfully";
 } else {
     echo "Error: " . $sql . "<br>" . $conn->error;
 }

I have now finally been able to see the output of:

Form.php

<form class="form" method="post" id="compReg" action="process.php">
<input type="submit">

connect.php

 <?php
 print_r($_POST);?>

The output when fields are exported is this:

Array ( [name] => Hedda Calhoun [email] => [e-mail] [tel] => [phone] [password] => Pa$$w0rd! [cpassword] => Pa$$w0rd! [companyName] => Underwood and Christian LLC [personalNumber] => 119 [streetAdress] => Enim officia et at numquam est voluptas placeat nulla et corporis dolorem in eum modi ratione incidunt nihil [zipCode] => 58366 [city] => Unde quis vel quam eiusmod rerum quaerat voluptatum labore provident [coAdress] => Nostrud culpa sint quis ab est in obcaecati illo consequat Ducimus nulla )

How can I load these into a php array so I can acces the and export them to the database?

Answer 1

You can access using Key

$_POST['name']
$_POST['email']
$_POST['tel']

For inserting records in database

<?php
$servername = "localhost";
$username = "username";
$password = "password";
$dbname = "myDB";

// Create connection
$conn = mysqli_connect($servername, $username, $password, $dbname);
// Check connection
if (!$conn) {
    die("Connection failed: " . mysqli_connect_error());
}

$sql = "INSERT INTO USERS (name, email, tel)
VALUES ('".$_POST['name']."', '".$_POST['email']."', '".$_POST['tel']."')";

if (mysqli_query($conn, $sql)) {
    echo "New record created successfully";
} else {
    echo "Error: " . $sql . "<br>" . mysqli_error($conn);
}

mysqli_close($conn);
?>