2017-03-10 22 views
0

我向一個有時會返回一個簡單的JSON對象的數組的API發出請求,我用一個簡單的「for i in count」循環解析,其中I使用SwiftyJSON來附加json [i] [「city」]。例如:SwiftyJSON-遍歷ID不是數組的對象

[{"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,在某些情況下,該API將返回與多個雙鍵和對象值,其中我無法與SwiftyJSON解析的對象。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}} 

在第二種情況下,有通過與SwiftyJSON第一對象的方法來循環,並且獲取對象值指出的是,我將不知道ID(2,32,414)中預先?

謝謝!

回答

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在你的第一個案例中,你有一個字典數組。在第二種情況下,你有一本字典詞典。我沒有長時間使用SwiftyJSON,所以我不記得它是如何工作的,但這應該足以讓你走了。

+0

這是答案的基礎。但我找不到將JSON對象轉換爲[String:Any]字典的方法。 –

+0

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