在 Swift (2.0) 中正确处理 NSJSONSerialization (try catch)？

Question

arowmy init 在 Swift Call can throw, but it is not marked with 'try' and the error is not handled let anyObj = NSJSONSerialization.JSONObjectWithData(data, options: NSJSONReadingOptions.MutableContainers) as! [String:AnyObject] 收到一条错误消息。我认为就我而言，我不能使用 try catch 块，因为此时 super 尚未初始化。 “尝试”需要一个抛出的函数。

这是我的功能：

required init(coder aDecoder : NSCoder)
{
    self.name  = String(stringInterpolationSegment: aDecoder.decodeObjectForKey("name") as! String!)
    self.number = Int(aDecoder.decodeIntegerForKey("number"))
    self.img = String(stringInterpolationSegment: aDecoder.decodeObjectForKey("image") as! String!)
    self.fieldproperties = []

    var tmpArray = [String]()
    tmpArray = aDecoder.decodeObjectForKey("properties") as! [String]


    let c : Int = tmpArray.count
    for var i = 0; i < c; i++
    {
        let data : NSData = tmpArray[i].dataUsingEncoding(NSUTF8StringEncoding)!

         // Xcode(7) give me error: 'CAll can thorw, but it is not marked with 'try' and the error is not handled'
        let anyObj =  NSJSONSerialization.JSONObjectWithData(data, options: NSJSONReadingOptions.MutableContainers) as! [String:AnyObject]

        let label = anyObj["label"] as AnyObject! as! String
        let value = anyObj["value"] as AnyObject! as! Int
        let uprate = anyObj["uprate"] as AnyObject! as! Int
        let sufix = anyObj["sufix"] as AnyObject! as! String

        let props = Fieldpropertie(label: label, value: value, uprate: uprate, sufix: sufix)
        self.fieldproperties.append(props)
    }
}

Xcode 的意思是： let anyObj = try NSJSONSerialization.JSONObjectWithData(data, options: NSJSONReadingOptions.MutableContainers) as! [String:AnyObject]

但我不知道在这里根据这个文档做正确的想法https://developer.apple.com/library/content/documentation/Swift/Conceptual/Swift_Programming_Language/ErrorHandling.html

Answer 1

jsonObject 可以出现throw 错误，因此将其放在do 块中，使用try 和catch 引发的任何错误。在 Swift 3 中：

do {
    let anyObj = try JSONSerialization.jsonObject(with: data) as! [String: Any]

    let label = anyObj["label"] as! String
    let value = anyObj["value"] as! Int
    let uprate = anyObj["uprate"] as! Int
    let sufix = anyObj["sufix"] as! String

    let props = Fieldpropertie(label: label, value: value, uprate: uprate, sufix: sufix)

    // etc.
} catch {
    print("json error: \(error.localizedDescription)")
}

或者，在 Swift 4 中，您可以通过使 struct 符合 Codable 来简化代码：

struct Fieldpropertie: Codable {
    let label: String
    let value: Int
    let uprate: Int
    let suffix: String
}

然后

do {
    let props = try JSONDecoder().decode(Fieldpropertie.self, from: data)
    // use props here; no manual parsing the properties is needed
} catch {
    print("json error: \(error.localizedDescription)")
}

对于 Swift 2，请参阅 previous revision of this answer。

Answer 2

JSONSerialization.JSONObject 抛出 ErrorType 而不是 NSError。

所以正确的捕获是

do {
    let anyObj = try JSONSerialization.JSONObject(with: data, options: []) as! [String:AnyObject]
    // use anyObj here
} catch let error {
    print("json error: \(error)")
}

catch let error 中error 的类型是ErrorType

Answer 3

不知道能不能解决你的问题，方法不就是JSONObjectWithData:options:error:吗？我认为您缺少 error 参数。