在 Swift 語言中更好的處理 JSON 資料：SwiftyJSON

標籤：

SwiftyJSON能夠讓在Swift語言中更加簡便處理JSON資料。

With SwiftyJSON all you have to do is:

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1234	let json = JSONValue(dataFromNetworking)if let userName = json[0]["user"]["name"].string{  //Now you got your value}

And don‘t worry about the Optional Wrapping thing, it‘s done for you automatically

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1234	let json = JSONValue(dataFromNetworking)if let userName = json[999999]["wrong_key"]["wrong_name"].string{  //Calm down, take it easy, the ".string" property still produces the correct Optional String type with safety}

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let json = JSONValue(jsonObject)switch json["user_id"]{case .JString(let stringValue):    let id = stringValue.toInt()case .JNumber(let numberValue):    let id = numberValue.integerValuedefault:    println("ooops!!! JSON Data is Unexpected or Broken")

Error Handling?

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let json = JSONValue(dataFromNetworking)["some_key"]["some_wrong_key"]["wrong_name"]if json{  //JSONValue it self confirm to Protocol "LogicValue", with JSONValue.JInvalid produce false and others produce true}else{  println(json)  //> JSON Keypath Error: Incorrect Keypath "some_wrong_key/wrong_name"  //It always tells you where your key starts went wrong  switch json{  case .JInvalid(let error):    //An NSError containing detailed error information  }}

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在 Swift 語言中更好的處理 JSON 資料：SwiftyJSON