使用PHP

Question

我有两个PHP文件，Database_conn和findOutMore

database_conn中的PHP：

<?php

$webserver = 'localhost';
$password = 'xxxx';
$username = 'xxxx';
$database = 'xxxx';



$conn = mysqli_connect( $webserver, $username, $password, $database);

if (!$conn) {
die("Connection failed: ".mysqli_connect_error());
}
?>

和findOutMore中的PHP

<?php
include 'database_conn.php';

$forename = $_REQUEST['forename'];
$surname = $_REQUEST['surname'];
$email = $_REQUEST['email'];
$landline = $_REQUEST['landLineTelNo'];
$mobile = $_REQUEST['MobileTelNo'];
$address = $_REQUEST['postalAddress'];
$method = $_REQUEST ['sendMethod'];

echo"<h3> These are your details </h3>";
echo"Name: ";
echo $forename;
echo " ";
echo $surname;
echo "<br>";
echo "Email: ";
echo $email;
echo "<br>";
echo "Landline: ";
echo $landline;
echo"<br>";
echo "Mobile: ";
echo $mobile;
echo "<br>";
echo "Address: ";
echo $address;
echo "<br>";
echo"Contact Method: ";
echo $method;


$query="INSERT INTO Database_Table
    (forename, surname, postalAddress, landLineTelNo, MobileTelNo, email, sendMethod)
     values (NULL, '$forename','$surname','$address','$landline','$mobile','$email','$method')";


    if (mysqli_query($conn, $query)) {
     echo "New record created successfully";
}

mysqli_close($conn);




?>

我实际上并没有收到任何错误，并且回声正常工作（返回用户输入的值。）但数据实际上从未输入数据库。

我认为可能是在值的开头处使用NULL，但我不确定如何更改它，因为它应该是主键，即ID条目。

Answer 1

您需要实际运行查询;）

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

mysqli_query($conn, $query);

我想指出，你的剧本非常不安全。您应该查看使用预准备语句，并可能切换到PDO：

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

http://php.net/manual/en/book.pdo.php