我的目标是创建一个简单的函数,我传入一个url并返回JSON。我环顾四周,发现了Alamofire实现完成处理程序的小例子。
我也在使用Swifty Json帮助解析它。
如何将此处的内容转换为返回Json的函数。
func request() {
Alamofire.request(.GET, API_END_POINT)
.responseJSON {
response in
// swiftyJsonVar is what I would like this function to return.
let swiftyJsonVar = JSON(response.result.value!)
}
}
答案 0 :(得分:9)
Swift 3 + 和 Alamofire 4 +
// Call function
myFunction("bodrum") { response in
print(response["yourParameter"].stringValue)
}
// POST
func myFunction(_ cityName:String, completion: @escaping (JSON) -> ()) {
let token = "token"
let parameters = ["city" : cityName]
let headers = ["Authorization": "token"]
let url = URL(string: "url")!
let reqUrl = URLRequest(url: url)
Alamofire.request(reqUrl, method: .post, parameters: parameters, encoding: URLEncoding.default, headers: headers)
.validate()
.responseJSON { response in
switch response.result {
case .Success:
let jsonData = JSON(data: response.data!)
completion(jsonData)
case .Failure(let error):
MExceptionManager.handleNetworkErrors(error)
completion(JSON(data: NSData()))
}
}
}
Swift 2 和 Alamofire 3 +
// POST
func myFunction(cityName:String, completion : (JSON) -> ()) {
Alamofire.request(.POST, "url", parameters: ["city" : cityName], encoding: ParameterEncoding.JSON, headers: ["Authorization": "token"])
.validate()
.responseJSON { response in
switch response.result {
case .Success:
let jsonData = JSON(data: response.data!)
completion(jsonData)
case .Failure(let error):
MExceptionManager.handleNetworkErrors(error)
completion(JSON(data: NSData()))
}
}
}
// GET
func myFunction(cityName:String, completion : (JSON) -> ()) {
Alamofire.request(.GET, "url", parameters: ["param1" : cityName], headers: ["Authorization": "token"])
.validate()
.responseJSON { response in
switch response.result {
case .Success:
let jsonData = JSON(data: response.data!)
completion(jsonData)
case .Failure(let error):
MExceptionManager.handleNetworkErrors(error)
completion(JSON(data: NSData()))
}
}
}
// Call function
myFunction("bodrum") { response in
print(response["yourParameter"].stringValue)
}