モバイル開発のための PHP Web サービス開発と API 設計

<?php

// 连接数据库
$conn = new mysqli('localhost', 'root', 'password', 'database');

// 处理 POST 请求来创建新用户
if ($_SERVER['REQUEST_METHOD'] === 'POST') {
    $data = json_decode(file_get_contents('php://input'), true);
    $stmt = $conn->prepare("INSERT INTO users (name, email) VALUES (?, ?)");
    $stmt->bind_param('ss', $data['name'], $data['email']);
    $stmt->execute();
    echo json_encode(['id' => $conn->insert_id]);
} else {
    // 处理 GET 请求以获取所有用户
    $stmt = $conn->prepare("SELECT * FROM users");
    $stmt->execute();
    $result = $stmt->get_result();
    $users = [];
    while ($row = $result->fetch_assoc()) {
        $users[] = $row;
    }
    echo json_encode($users);
}

?>

// Swift 代码示例
import Foundation

// 创建 URL 请求
let url = URL(string: "http://example.com/api/users")!
var urlRequest = URLRequest(url: url)

// 设置 HTTP 方法和标头
urlRequest.httpMethod = "POST"
urlRequest.setValue("application/json", forHTTPHeaderField: "Content-Type")

// 准备 JSON 数据
let jsonData = try! JSONEncoder().encode(userData)

// 设置请求主体
urlRequest.httpBody = jsonData

// 发送请求并处理响应
let task = URLSession.shared.dataTask(with: urlRequest) { data, response, error in
    if let data = data {
        let jsonDecoder = JSONDecoder()
        let response = try! jsonDecoder.decode([User].self, from: data)
        // 使用响应中的数据
    }
}

task.resume()