错误：在 Swift 3 上'Cannot call value of non-function type'

Question

下面代码的第二和第三行是在swift 2.3上工作的，自从我更新到swift 3以来，我一直得到错误不能调用非函数类型的值'Any?!:

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])

这是来自google地图搜索的回调。

方法读取的JSON是:

{
   "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"
}

有人知道我做了什么改变使它工作吗?

Answer 1

在解析JSON时，Swift中有几件事非常糟糕:

• 转换为没有类型信息的NSDictionary或NSArray。
• cast to AnyObject，这是非常未指定的。
• valueForKey，这是一个键值编码方法，在这种情况下是不必要的。
• 选项mutableleaves，当使用Swift类型时没有意义，当对象只被读取时不需要。

休所有教程，建议不良的编程习惯。: -)

考虑AnyObject在Swift 3中已经被Any取代。

JSON只支持两种集合类型，array(由[]表示)和dictionary(由{}表示)。

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

您的主要错误是键results的值类型错误，这是一个数组。

这段代码检查是否存在所有带有可选绑定的键，以及数组是否为空，location的字典是否包含2项:

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)")
} 

Apple发布了一篇全面的文章