小编Fre*_*ric的帖子

我有一段时间了解使用Swift 3的简单JSON序列化原则.我可以帮助解决从网站解码JSON到阵列所以我可以访问它jsonResult["team1"]["a"]等吗？这是相关代码:

let httprequest = URLSession.shared.dataTask(with: myurl){ (data, response, error) in

self.label.text = "RESULT"

    if error != nil {

        print(error)

    } else {

        if let urlContent = data {

            do {

                let jsonResult = try JSONSerialization.jsonObject(with: urlContent, options: 
                    JSONSerialization.ReadingOptions.mutableContainers)

                print(jsonResult) //this part works fine

                print(jsonResult["team1"])

                } catch {

                    print("JSON Processing Failed")
                }
            }
        }
    }
    httprequest.resume()

传入的JSON是:

{
team1 = {
    a = 1;
    b = 2;
    c = red;
};

team2 = {
    a = 1;
    b …

我刚刚在我的新 MacBook 上安装了 Xcode。Swift iOS 的新起始模板需要 iOS SDK 13+（即它们需要 UISceneDelegate 或 SwiftUI）。有没有办法使用与 iOS 12 兼容的内置模板创建新应用程序？

我正在学习使用谓词进行过滤.我找到了一个教程,但是在Swift 3中有一个方面对我不起作用.这是一些特定的代码:

let ageIs33Predicate01 = NSPredicate(format: "age = 33") //THIS WORKS
let ageIs33Predicate02 = NSPredicate(format: "%K = 33", "age") //THIS WORKS
let ageIs33Predicate03 = NSPredicate(format: "%K = %@", "age","33") //THIS DOESN'T WORK
let ageIs33Predicate04 = NSPredicate(format: "age = %@","33") //THIS DOESN'T WORK

所有4编译,但最后2编译没有结果,即使我有一个年龄= 33的情况.这是教程中的测试完整测试代码:

import Foundation

class Person: NSObject {
    let firstName: String
    let lastName: String
    let age: Int

    init(firstName: String, lastName: String, age: Int) {
        self.firstName = firstName
        self.lastName = lastName
        self.age = age
    }

    override var description: String …