swift - 如何处理 swift 3 中的错误？

Question

我正在使用以下代码:

   let jsonData1 = (try? Data(contentsOf: URL(string: "\(label2)")!)) as Data!
    var error : NSError?
    let ReadableJSON1 = JSON ( data:jsonData1!, options: JSONSerialization.ReadingOptions.mutableContainers, error: nil )


    if (error != nil){

        let myAlert = UIAlertController(title: "Alert", message: error?.localizedDescription, preferredStyle: .alert) //you can change message to whatever you want.
        let okAction = UIAlertAction(title: "Ok", style: .default, handler: nil)
        myAlert.addAction(okAction)
        self.present(myAlert, animated: true, completion: nil)

        return
    }

如果应用程序在从链接打开 json 文件时发现错误，则应向用户显示警报。目前它正在崩溃，似乎这些操作没有执行，这就是为什么我没有看到警报。您能帮我解决这个问题吗？

Answer 1

试试这个代码:

var json : [String: AnyObject]? // Property
do {
    let jsonData1 = (try? Data(contentsOf: URL(string: "\(label2)")!)) as Data!
    json = try JSONSerialization.jsonObject(with: jsonData1, options: JSONSerialization.ReadingOptions.mutableContainers) as! [String: AnyObject]
} 
catch let error as NSError {
    let myAlert = UIAlertController(title: "Alert", message: error?.localizedDescription, preferredStyle: .alert) //you can change message to whatever you want.
    let okAction = UIAlertAction(title: "Ok", style: .default, handler: nil)
    myAlert.addAction(okAction)
    self.present(myAlert, animated: true, completion: nil)

    return
}

希望对你有帮助!快乐编码!