如何将数据插入PHP中的列到MariaDB？

Question

我熟悉MySql数据库,但有一个名为MariaDB的新数据库.我尝试从PHP代码插入数据而我不能,所以你可以帮我插入数据吗？我在服务器上的PHP是5.4.32,我的MySQL版本是10.0.20-MariaDB-cll-live.

这是我用来尝试将数据插入MariaDB的代码.

$connection = mysql_connect("localhost", "root", ""); // Establishing Connection with Server
$db = mysql_select_db("database", $connection); // Selecting Database from Server

if(isset($_POST['submit'])){ // Fetching variables of the form which travels in URL
    $name = $_POST['name'];
    $email = $_POST['email'];
    if($name !=''||$email !=''){
       //Insert Query of SQL
       $query = mysql_query("insert into VIP Membership(Name, Email) values ('$name', '$email')");
       echo "<br/><br/><span>Data Inserted successfully...!!</span>";
    }
    else{
       echo "<p>Insertion Failed <br/> Some Fields are Blank....!!</p>";
    }
}

mysql_close($connection); // Closing Connection with Server

编辑:我现在使用mysqli代码,但它给我一个错误500.

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

// Create connection
$conn = mysqli_connect($servername, $username, $password, $dbname);
// Check connection
if (!$conn) {
    die("Connection failed: " . mysqli_connect_error());
}
if(isset($_POST['submit'])){ // Fetching variables of the form which travels in URL
$name = $_POST['name'];
$email = $_POST['email'];
if($name !=''||$email !=''){

$sql = "INSERT INTO 'VIP Membership' (Name, Email)
VALUES ('$name', '$email')";

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

mysqli_close($conn);

Answer 1

你需要使用反引号(而不是单引号)来包围你的表名,因为它有一个空格.

$sql = "INSERT INTO `VIP Membership` (Name, Email) VALUES ('$name', '$email')";

更新:你应该真的是使用预处理语句.就目前而言,这段代码非常不安全.

$sql = mysqli_prepare($conn, "INSERT INTO `VIP Membership` (Name, Email) VALUES (?, ?)");
if($sql !== FALSE){
    mysqli_stmt_bind_param($sql, "ss", $name, $email);
    if(mysqli_stmt_execute($sql)){
        echo "New record created successfully";
    } else {
        echo mysqli_stmt_error($sql);
    }
} else{
    echo mysqli_error($conn);
}