如何将我的网站连接到MySQL数据库？

Question

下面是我在Sublime中的代码，但数据库没有被调用。

<?php$username="root"; 
$password="changedpassword";$database="User"; 
$field1-name=$_POST['name']; 
$field2-name=$_POST['password']; 
$field3-name=$_POST['email']; 
$field4-name=$_POST['sex']; 
$field5-name=$_POST['school']; 
$field6-name=$_POST['birth']; 
mysql_connect(localhost,$username,$password); 
@mysql_select_db($database) or die("Unable to select database"); 
$query = "INSERT INTO create_user (name, password, email, sex, school, birth) VALUES('','$field1-name','$field2-name', 
'$field3-name','$field4-name','$field5-name','$field6-name')";mysql_query($query);mysql_close();?> 

Answer 1

让我们一步一步来看看。首先，这里是你当前的代码，整理了能够阅读：

<?php 
$username = "root"; 
$password = "changedpassword"; 
$database = "User"; 
$field1_name = $_POST['name']; 
$field2_name = $_POST['password']; 
$field3_name = $_POST['email']; 
$field4_name = $_POST['sex']; 
$field5_name = $_POST['school']; 
$field6_name = $_POST['birth']; 
mysql_connect(localhost, $username, $password); 
@mysql_select_db($database) or die("Unable to select database"); 
$query = " 
    INSERT INTO 
      create_user 
       (
        name, 
        password, 
        email, 
        sex, 
        school, 
        birth 
       ) 
      VALUES 
       (
        '', 
        '$field1_name', 
        '$field2_name', 
        '$field3_name', 
        '$field4_name', 
        '$field5_name', 
        '$field6_name' 
      ) 
"; 
mysql_query($query); 
mysql_close(); 
?> 

我所做的只有两个变化（整理的空白，并用_name代替-name，如PHP变量不能包含连字符），但它已经一个很大的改进。代码不再是一个眼睛。它没有语法错误，并且可读。尽管如此，还是有很多问题。

首先，您会看到我们将七个值插入六列。这将是一个问题。修正了通过去除第一空白值：

$query = " 
    INSERT INTO 
      create_user 
       (
        name, 
        password, 
        email, 
        sex, 
        school, 
        birth 
       ) 
      VALUES 
       (
        '$field1_name', 
        '$field2_name', 
        '$field3_name', 
        '$field4_name', 
        '$field5_name', 
        '$field6_name' 
      ) 
"; 

现在我们有一些可能实际工作。这是一种痛苦的不安全感，有很大的SQL注入攻击的可能性，并且它不适用于最新的PHP，因为mysql_函数已被删除，但它实际上可能适合某处工作。你不想把它投入生产，但出于测试目的，我们正在某处。

Answer 2

MySQL自PHP 5.6起弃用，且不安全，请改为使用PDO或MySQLi。

与库MySQLi

 <?php 


     //MySQLi information 

     $db_host  = "localhost"; 
     $db_username = "username"; 
     $db_password = "password"; 

     //connect to mysqli database (Host/Username/Password) 
     $connection = mysqli_connect($db_host, $db_username, $db_password) or die("Error " . mysqli_error()); 

     //select MySQLi dabatase table 
     $db = mysqli_select_db($connection, "table") or die("Error " . mysqli_error()); 

    $field1_name = $_POST['name']; 
    $field2_name = $_POST['password']; 
    $field3_name = $_POST['email']; 
    $field4_name = $_POST['sex']; 
    $field5_name = $_POST['school']; 
    $field6_name = $_POST['birth']; 

    $query = mysqli_query($connection, "INSERT INTO create_user 
      (name, password, email, sex, school, birth) VALUES 
        (
         '$field1_name', 
         '$field2_name', 
         '$field3_name', 
         '$field4_name', 
         '$field5_name', 
         '$field6_name' 
       ) 
    "); 

使用连接这一点，你就会好的。我希望这对你有帮助！