Question

以下代码的第二行和第三行正在使用swift 2.3，因为我更新到swift 3，我已经收到错误无法调用非函数类型的值＆＃39;任何？！＆＃39; 两个人：

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。

该方法正在阅读的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"
}

有谁知道我做了什么改变让它起作用？

Answer 1

在解析JSON时，Swift中有一些非常糟糕的东西：

投射到NSDictionary或NSArray，但没有类型信息。
投射到AnyObject这是非常未指定的。
valueForKey这是一种键值编码方法，在这种情况下不是必需的。
选项mutableleaves在使用Swift类型时没有意义，在仅读取对象时从不需要。

所有教程都表明编程习惯不佳。 ; - ）

请注意，AnyObject已在Swift 3中替换为Any。

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

为了可靠地解析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。

您的主要错误是密钥results的值类型错误，这是数组。

此代码检查是否存在具有可选绑定的所有键，如果数组不为空且location的字典包含2个项：

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Working with JSON in Swift

错误：＆＃39;无法调用非功能类型的值＆＃39;在Swift 3上

1 个答案: