我在我的应用程序中使用Almofire。如何在使用.get
方法时使用字典。我的代码如下:
func validMemberId(completion: @escaping CompletionHandler){
guard let cookie = self.cookies else {
return
}
let isRegisterMemberUrl = createUrl(with: checkRegisterMemberIdUrl)
var isRegisterMemberRequest = URLRequest(url: isRegisterMemberUrl)
isRegisterMemberRequest.httpMethod = HTTPMethod.get.rawValue
isRegisterMemberRequest.allHTTPHeaderFields = HTTPCookie.requestHeaderFields(with: cookie)
Alamofire.request(isRegisterMemberRequest).responseString { (response) in
{ (response) in
if response.result.isSuccess && response.response?.statusCode == 200 {
}
}
}
我想使用Get方法和URLRequest
,我的词典[String: Any] = ["ID": "001110"]
答案 0 :(得分:0)
参数应为Dictionary
(键,值)
对于获取请求,您无需设置方法
let parameters: Parameters = ["foo": "bar"]
Alamofire.request("https://httpbin.org/get", parameters: parameters) // encoding defaults to `URLEncoding.default`
如果您想提出post
请求,只需添加method
let parameters: Parameters = ["foo": "bar"]
Alamofire.request("https://httpbin.org/post", method: .post, parameters: parameters)
答案 1 :(得分:0)
<强> Manual Parameter Encoding of a URLRequest 强>
可以在发出网络请求之外使用ParameterEncoding
API。
let url = URL(string: "https://httpbin.org/get")!
var urlRequest = URLRequest(url: url)
let parameters: Parameters = ["foo": "bar"]
let encodedURLRequest = try URLEncoding.queryString.encode(urlRequest, with: parameters)
Alamofire.request(encodedURLRequest)