2017-03-10 60 views
0

我向一个有时会返回一个简单的JSON对象的数组的API发出请求,我用一个简单的“for i in count”循环解析,其中I使用SwiftyJSON来附加json [i] [“city”]。例如:SwiftyJSON-遍历ID不是数组的对象

[{"city":"Lakefront","code":"NEW","country":"United States","municipality":"New Orleans","isChild":false,"hasChild":false}, 
{"city":"Auckland - Auckland International Airport","code":"AKL","country":"New Zealand","municipality":"Auckland","isChild":false,"hasChild":false}, 
{"city":"Blenheim","code":"BHE","country":"New Zealand","municipality":"Blenheim","isChild":false,"hasChild":false}] 

然而,在某些情况下,该API将返回与多个双键和对象值,其中我无法与SwiftyJSON解析的对象。例如:

{"2":{"city":"New York","code":"NYC","country":"United States","municipality":"New York","isChild":false,"hasChild":true}, 
"32":{"city":"John F. Kennedy - NY","code":"JFK","country":"United States","municipality":"New York","isChild":true,"hasChild":false}, 
"414":{"city":"LaGuardia - NY","code":"LGA","country":"United States","municipality":"New York","isChild":true,"hasChild":false}} 

在第二种情况下,有通过与SwiftyJSON第一对象的方法来循环,并且获取对象值指出的是,我将不知道ID(2,32,414)中预先?

谢谢!

回答

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在你的第一个案例中,你有一个字典数组。在第二种情况下,你有一本字典词典。我没有长时间使用SwiftyJSON,所以我不记得它是如何工作的,但这应该足以让你走了。

+0

这是答案的基础。但我找不到将JSON对象转换为[String:Any]字典的方法。 –

+0

忽略SwiftyJSON,你只需说'guard let dict = aJSONObject as [String:Any] else {return}' –