PHP：传递变量的正确方法

Question

我将这段 PHP 代码安装在我还安装了数据库的服务器上。当我尝试通过 URL http://ptyxiaki2016.eu.pn/login.php?email=tade&password=1234 运行它时，所有值都为 NULL。

tade 1234 {"error":false,"uid":null,"user":{"name":null,"surname":null,"country":null,"email":null,"password" :null,"电话":null}}

我的 PHP 代码如下：

<?php
require_once 'DB_Functions.php';
require_once 'DB_Connect.php';
$db = new DB_Functions();

// json response array
$response = array("error" => FALSE);

if (isset($_GET['email']) && isset($_GET['password'])) {

// receiving the post params
$email = $_GET['email'];
$password = $_GET['password'];

// get the user by email and password
$user = $db->getUserByEmailAndPassword($email, $password);


if ($user != false) {
    // user is found
    $response["uid"] = $user["oid"];
    $response["error"] = FALSE;
    $response["uid"] = $user["oid"];
    $response["user"]["name"] = $user["name"];
    $response["user"]["surname"] = $user["surname"];
    $response["user"]["country"] = $user["country"];
    $response["user"]["email"] = $user["email"];
    $response["user"]["password"] = $user["password"];
    $response["user"]["telephone"] = $user["telephone"];
    echo json_encode($response);
} else {
    // user is not found with the credentials
    $response["error"] = TRUE;
    $response["error_msg"] = "Login credentials are wrong. Please try again!";
    echo json_encode($response);
}
} else {
// required post params is missing
$response["error"] = TRUE;
$response["error_msg"] = "Required parameters email or password is missing!";
echo json_encode($response);
}
?>

这就是我在登录时调用的函数getUserByEmailAndPassword()。它在我包含的 DB_Functions.php 中。

public function getUserByEmailAndPassword($email, $password) {

    $stmt = $this->conn->prepare("SELECT * FROM owner WHERE email = ?");
    $stmt->bind_param("s", $email);

    if ($stmt->execute()) {
        $user = $stmt->bind_result($oid, $name, $surname, $country, $email, $password, $telephone);
        while ($stmt->fetch()) {
                printf("%s %s\n", $email, $password);
    }
        $stmt->close();
        return $user;
        }
    }

我在这里缺少什么？我的数据库没问题，但是，我该如何检索数据？

Answer 1

错误在这里：

$user = $stmt->bind_result($oid, $name, $surname, $country, $email, $password, $telephone);

您将结果绑定到变量（$oid、$name、...），但返回 $user。

bind_result返回运算结果（真/假），但不返回查询结果。

所以调用 getUserByEmailAndPassword() 的结果是布尔值。如果你运行而不是 echo json_encode($response) echo json_encode($user) 你会看到它。

UPD：试试这个代码而不是你的代码

if ($stmt->execute()) {
    $stmt->bind_result($user["oid"], $user["name"], ...)
    while ($stmt->fetch()) {
            printf("%s %s\n", $email, $password);
}