ios - 当没有键而只有整数/字符串值时如何解析JSON？

Question

如何解析 this JSON文件？当键和值都可用时，我的代码可以正常工作。

到目前为止我的代码:

let url = URL(string: "http://uhunt.felix-halim.net/api/uname2uid/felix_halim")
let task = URLSession.shared.dataTask(with: url!, completionHandler: {
    (data, response, error) in
    print("Task Started")

    if error != nil {
        print("In Error!")
    } else {
        if let content = data {
            do {
                let myJSON =
                    try JSONSerialization.jsonObject(with: content, options: .mutableContainers) as AnyObject
                print(myJSON)
            } catch {
                print("In Catch!")
            }
        }
    }
})
task.resume()
print("Finished")

Answer 1

如果 JSON 的根对象不是字典或数组，你必须将 .allowFragments 作为选项传递(顺便说一句，永远不要传递 .mutableContainers，它在 Swift 中是没有意义的)

let url = URL(string: "http://uhunt.felix-halim.net/api/uname2uid/felix_halim")!
let task = URLSession.shared.dataTask(with: url) { data, response, error in
    print("Task Started")

    guard error == nil else {
        print("In Error!", error!)
        return
    }

    do {
        if let myJSON = try JSONSerialization.jsonObject(with: data!, options: .allowFragments) as? Int {
            print(myJSON)
        }
    } catch {
        print("In Catch!", error)
    }

}
task.resume()
print("Finished")