How handle async URLRequest finished?

Question

I try to create request in func and after call the func and return responсe json. But because request is async response arrived after func returns any data.

How handle request finishing?

score 0 · Answer 1 · answered Mar 19 '18 at 17:54

0

Pass a closure to your function, and call that closure with the result of the URL request.

answered Mar 19 '18 at 17:54

Gereon

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Please, can you give to me some code? I can add my request to ask if need – Golovanov Dmytrii Mar 19 '18 at 18:03

score 0 · Answer 2 · answered Mar 19 '18 at 18:42

You have asked for code so I am giving you some code snippets. Please modify it as per your requirement. These codes are using Alamofire but mainly you need to know how to handle response in callback.

   func requestYourApi(_ requestParams:[String:AnyObject], callback:@escaping (Dictionary<String, Any>, Error?)-> Void)
{
    let parameters:Parameters = requestParams
    Alamofire.request("https:yourUrl", method: .post, parameters: parameters, encoding: JSONEncoding.prettyPrinted).responseJSON    { response in
        debugPrint("response: \(response)")

        switch response.result
        {
        case .success(let value):
            callback(value as! Dictionary, nil)
        case .failure(let error):
            callback([:], error)
        }
    }
}

And call the above method like this, wherever required. You'll get the response in responseObj and error in error.

     requestYourApi(requestParams as [String : AnyObject]) { (responseObj, error) in
        DispatchQueue.main.async
            {
                if responseObj.count != 0
                {
                  // Do something with your response

                }
        }
    }

That's cool, but my code working without result of request. So code after calling this func don't have result of this func in time. — Golovanov Dmytrii, Mar 20 '18 at 11:52

How handle async URLRequest finished?

2 Answers2