RxAlamofire make post call with json body

Question

I want to make post call using RxAlamofire not able to find any method to do so

tried using requestJSON method but there is no paramter to pass post json in

 RxAlamofire.requestJSON(.post, url)

how to make post call and pass json data to post call in RxAlamofire

Answer 1

Use following code

  var request = URLRequest(url: URL(string: "https://some_url")!)
    //Following code to pass post json 
    request.httpBody = json
    request.httpMethod = "POST"
    request.setValue("application/json", forHTTPHeaderField: "Content-Type")
    RxAlamofire.request(request as URLRequestConvertible).responseJSON().asObservable()

Answer 2

use this function with proper parameters ancoding

public func urlRequest(_ method: Alamofire.HTTPMethod,
                   _ url: URLConvertible,
                   parameters: [String: Any]? = nil,
                   encoding: ParameterEncoding = URLEncoding.default,
                   headers: [String: String]? = nil)

Answer 3

Using jsonEncoding

RxAlamofire.requestJSON(.post, url, encode: JsonEncoding.default)

It works for me~

Answer 4

amodkanthe's solution works. But its return value is not friendly. So I changed it a litte bit

    var request = URLRequest(url: URL(string: "https://some_url")!)
    request.httpBody = jsonData  // jsonData is a Data type
    request.httpMethod = "POST"
    request.setValue("application/json", forHTTPHeaderField: "Content-Type") // this line is important
    RxAlamofire.requestJSON() //=> returns a Observable<(HttpURLResponse, Any)>, the `Any` type is actually a dictionary

p.s. the version I'm using is RxAlamofire(6.1.1)