使用Swift将JSON answer放入变量中

import Foundation

// 假设有以下JSON数据
let jsonString = """
{
    "name": "John Doe",
    "age": 30,
    "email": "johndoe@example.com"
}
"""

// 将JSON字符串转换为Data类型
guard let jsonData = jsonString.data(using: .utf8) else {
    print("Invalid JSON string")
    return
}

do {
    // 解析JSON数据
    let jsonObject = try JSONSerialization.jsonObject(with: jsonData, options: [])
    
    // 将解析后的JSON数据转换为字典类型
    guard let jsonDict = jsonObject as? [String: Any] else {
        print("Invalid JSON format")
        return
    }
    
    // 从字典中获取需要的值
    let name = jsonDict["name"] as? String
    let age = jsonDict["age"] as? Int
    let email = jsonDict["email"] as? String
    
    // 打印结果
    print("Name: \(name ?? "")")
    print("Age: \(age ?? 0)")
    print("Email: \(email ?? "")")
} catch {
    print("JSON serialization error: \(error)")
}