ios - 错误 : 'Cannot call value of non-function type' on Swift 3

Question

以下代码的第二行和第三行适用于 swift 2.3，自从我更新到 swift 3 后，我一直收到错误消息无法调用非函数类型的值 'Any?!' 对于他们两个:

let dic = try JSONSerialization.jsonObject(with: data!, options: JSONSerialization.ReadingOptions.mutableLeaves) as! NSDictionary                    

let lat = ((((dic["results"] as AnyObject).value(forKey: "geometry") as AnyObject).value(forKey: "location") as AnyObject).value(forKey: "lat") as AnyObject).object(0) as! Double
let lon = ((((dic["results"] as AnyObject).value(forKey: "geometry") as AnyObject).value(forKey: "location") as AnyObject).value(forKey: "lng") as AnyObject).object(0) as! Double

self.delegate.locateWithLongitude(lon, andLatitude: lat, andTitle: self.searchResults[(indexPath as NSIndexPath).row])

这是谷歌地图搜索的回调。

该方法正在读取的 JSON 是这样的:

{
   "results" : [
      {
         "address_components" : [
            {
               "long_name" : "São Paulo",
               "short_name" : "São Paulo",
               "types" : [ "locality", "political" ]
            },
            {
               "long_name" : "São Paulo",
               "short_name" : "São Paulo",
               "types" : [ "administrative_area_level_2", "political" ]
            },
            {
               "long_name" : "State of São Paulo",
               "short_name" : "SP",
               "types" : [ "administrative_area_level_1", "political" ]
            },
            {
               "long_name" : "Brazil",
               "short_name" : "BR",
               "types" : [ "country", "political" ]
            }
         ],
         "formatted_address" : "São Paulo, State of São Paulo, Brazil",
         "geometry" : {
            "bounds" : {
               "northeast" : {
                  "lat" : -23.3566039,
                  "lng" : -46.36508449999999
               },
               "southwest" : {
                  "lat" : -24.0082209,
                  "lng" : -46.825514
               }
            },
            "location" : {
               "lat" : -23.5505199,
               "lng" : -46.63330939999999
            },
            "location_type" : "APPROXIMATE",
            "viewport" : {
               "northeast" : {
                  "lat" : -23.3566039,
                  "lng" : -46.36508449999999
               },
               "southwest" : {
                  "lat" : -24.0082209,
                  "lng" : -46.825514
               }
            }
         },
         "partial_match" : true,
         "place_id" : "ChIJ0WGkg4FEzpQRrlsz_whLqZs",
         "types" : [ "locality", "political" ]
      }
   ],
   "status" : "OK"
}

有谁知道我做了什么改变才能让它发挥作用？

Answer 1

在解析 JSON 时，Swift 中有一些非常糟糕的事情:

• 转换为没有类型信息的 NSDictionary 或 NSArray。
• 转换为非常不确定的 AnyObject。
• valueForKey，这是一种键值编码方法，在这种情况下不是必需的。
• mutableleaves 选项在使用 Swift 类型时毫无意义，在只读取对象时从不需要。

起诉所有暗示不良编程习惯的教程。 ;-)

请考虑 AnyObject 在 Swift 3 中已被替换为 Any。

JSON 仅支持两种集合类型，array – 由 [] 表示 – 和 dictionary – 由 {} 表示.

要可靠地解析 JSON，您必须仔细阅读 JSON 以识别结构并告诉编译器节点的类型。

您的主要错误是关键 results 的值类型错误，它是一个数组。

此代码检查是否存在具有可选绑定(bind)的所有键，以及数组是否不为空以及 location 的字典是否包含 2 个项目:

if let jsonObject = try JSONSerialization.jsonObject(with: data!, options: []) as? [String:Any],
  let results = jsonObject["results"] as? [[String:Any]], !results.isEmpty,
  let geometry = results[0]["geometry"] as? [String:Any],
  let location = geometry["location"] as? [String:Double],
  let lat = location["lat"], let lng = location["lng"] {

  print("lat: \(lat) - lng: \(lng)")
} 

苹果发表综合文章Working with JSON in Swift