如何简单地简单地获取请求,而标头和主体位于原始的空块内

问题描述 投票:0回答:1

在iOS Swift中,我使用一个简单的请求,但无法获得响应:

var headers: HTTPHeaders = ["Content-Type": "application/json"]
        let url = "http://18.139.224.233/api/promoted"
        Alamofire.request(url, method: .get, parameters: [:], encoding: JSONEncoding.default, headers: headers).responseJSON { (response) in
              switch response.result {
              case .success:
                  print("SUKCES with \(response)")
              case .failure(let error):
                  print("ERROR with '\(error)")
              }
          }

我在邮递员上尝试了我的API,在那里得到响应,因为在此设置了该值

{

}

在未加工的体内,

但是iOS我不知道如何添加它

这就是我收到此回复的原因:

[Response Body]: 
<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 3.2 Final//EN">
<title>400 Bad Request</title>
<h1>Bad Request</h1>
<p>Invalid JSON data: ''</p>

[Result]: FAILURE

以下是邮递员请求成功执行API屏幕截图

enter image description here enter image description here

ios swift alamofire
1个回答
0
投票

尝试一下。

import UIKit
import Alamofire

class ViewController: UIViewController {

 override func viewDidLoad() {
   super.viewDidLoad()
   var headers: HTTPHeaders = ["Content-Type": "application/json"]
        let url = "http://18.139.224.233/api/promoted"
        Alamofire.request(url, method: .get, parameters: [:], encoding: JSONEncoding.default, headers: headers).responseJSON { (response) in
              switch response.result {
              case .success:
                  print("SUKCES with \(response)")
              case .failure(let error):
                  print("ERROR with '\(error)")
              }
          }
   }
}
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