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我有以下注册码，它似乎可以工作，但实际上并没有将输入的信息插入我的表中。这一切都运行，没有出现任何错误，并且“echo 'end';” 正在显示。

编辑，更新代码： 现在得到这个错误

警告：mysqli_stmt::bind_param()：类型定义字符串中的元素数与第 19 行 C:\xampp\htdocs\ppa\test.php 中的绑定变量数不匹配

这是哪一行：

$insert_stmt->bind_param($email, $password, $random_salt, $user);

PHP：

   <?php
include "includes/db_connect.php";

if (isset($_POST['email'], $_POST['p'])) {
    $email = $_POST['email'];
    //Default user perms
    $perms = "user";

    $password = hash('sha512', $_POST['p']); //Need to add JavaScript to hash password before it gets here
    //Create random salt
    $random_salt = hash('sha512', uniqid(mt_rand(1, getrandmax()), true));

    //Create salted password
    $password = hash('sha512', $password.$random_salt);

    //Add insert to database script
    //Use prepared statements!
    if ($insert_stmt = $mysqli->prepare("INSERT INTO users (email, password, salt, perms) VALUES (?, ?, ?, ?)")) {
        $insert_stmt->bind_param($email, $password, $random_salt, $perms);
        $insert_stmt->execute();
    }
    echo "Email: ".$email."<br />";
    echo "Password: ".$password."<br />";
    echo "Random Salt: ".$random_salt."<br />";
    echo "Permissions: ".$perms."<br />";
}
?>

这是我的 db_connect.php 页面

<?php
define("HOST", 'localhost');
define("USER", 'ppa_user');
define("PASSWORD", 'password');
define("DATABASE", 'ppa');

$mysqli = new mysqli(HOST, USER, PASSWORD, DATABASE);

if ($mysqli->connect_errno) {
    //No database found, redirect to setup
    $url = "http://".$_SERVER['HTTP_HOST'].'/ppa/setup.php';
    header('Location: '.$url);
}
?>
4

3 回答 3

1

替换以下内容：

//Add insert to database script
    //Use prepared statements!
    if ($insert_stmt = $mysqli->prepare("INSERT INTO users (email, password, salt, perms) VALUES (?, ?, ?, ?)"));
    $insert_stmt->bind_param('ssss', $_POST['email'], $password, $random_salt, $user);

//Execute the prepared query
$insert_stmt->execute();
echo "end";

和：

//Add insert to database script
//Use prepared statements!
if ($insert_stmt = $mysqli->prepare("INSERT INTO users (email, password, salt, perms) VALUES (?, ?, ?, ?)")) {
    $insert_stmt->bind_param($_POST['email'], $password, $random_salt, $user);
    $insert_stmt->execute();
    echo "end";
}
于 2013-11-13T18:05:06.237 回答
1

检查这个 

if ($insert_stmt = $mysqli->prepare("INSERT INTO users (email, password, salt, perms) VALUES (?, ?, ?, ?)"));
$insert_stmt->execute(array($_POST['email'], $password, $random_salt, $user));
于 2013-11-13T18:15:20.093 回答
1

为了结束这个问题的回答，得出的结论是 OP 需要使用以下代码：

$insert_stmt->bind_param("ssss", $email, $password, $random_salt, $perms);
于 2013-11-13T18:41:39.217 回答