PHP 函数如何与 Swift 交互：构建 iOS 友好型解决方案

let url = URL(string: "http://example.com/api/get-user-info.php")!
var request = URLRequest(url: url)
request.httpMethod = "GET"

request.setValue("application/json", forHTTPHeaderField: "Content-Type")

URLSession.shared.dataTask(with: request) { (data, response, error) in
    if let error = error {
        // Handle the error
    } else if let data = data {
        do {
            // Decode the JSON response
            let user = try JSONDecoder().decode(User.self, from: data)
            // Update the UI with the user's details
        } catch {
            // Handle the decoding error
        }
    }
}.resume()

<?php
// Get the user's ID from the request body
$userId = $_POST['user_id'];

// Query the database for the user's details
$user = getUserById($userId);

// Encode the user's details as JSON
$json = json_encode($user);

// Send the JSON response to the client
echo $json;