把各種變量納入使用PHP

Question

我有兩個PHP文件，Database_conn和findOutMore

在database_conn PHP的一個數據庫表：

<?php 

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



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

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

而且在findOutMore

<?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); 




?> 

我的PHP m實際上沒有得到任何錯誤，回聲正在按照他們應該的方式工作（返回用戶輸入的值）。但是數據從未真正進入數據庫。

我認爲這可能是在值的開始時使用NULL，但我不確定如何更改它，因爲它應該是主鍵，條目的ID。

Answer 1

您需要實際運行查詢;）

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

mysqli_query($conn, $query);

我想，雖然指出，你的腳本是非常不安全的。你應該看看使用預處理語句，並有可能切換到PDO：

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

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