Question

因此，我试图在Swift中制作一个应用程序，我需要登录该应用程序以登录OAuth以检索访问令牌。该API带有参数“ grant_type”和“ code”。

我尝试过：

let params = ["grant_type":"authorization_code", "code":"123456789"] as Dictionary<String, String>

var request = URLRequest(url: URL(string: "https://myschool.zportal.nl/api/v3/oauth/token")!)
request.httpMethod = "POST"
request.httpBody = try? JSONSerialization.data(withJSONObject: params, options: [])
request.addValue("application/json", forHTTPHeaderField: "Content-Type")

let session = URLSession.shared
let task = session.dataTask(with: request, completionHandler: { data, response, error -> Void in
    do {
        let json = try JSONSerialization.jsonObject(with: data!) as! Dictionary<String, AnyObject>
        print(json)
    } catch {
        print("error")
    }
})

task.resume()

这将返回：

["response": {
    data =     (
    );
    details = "class org.springframework.web.bind.MissingServletRequestParameterException: Required String parameter 'grant_type' is not present";
    endRow = 0;
    eventId = 216795;
    message = "Internal problems.";
    startRow = 0;
    status = 500;
    totalRows = 0;
}]

这很奇怪，因为我确实将'grant_type'参数解析为正确的值。

它应该返回如下内容：

{
  "response": {
    "status":200,
    "message":"",
    "startRow":0,
    "endRow":27,
    "totalRows":27,
    "data":[
      {
      },
      ...
    ]
  }
}

Answer 1

标准消息格式需要表单URL编码的主体，而不是JSON。也许像this Swift article中那样修改您的代码。

Answer 2

使其正常工作：

@IBAction func sendButtonPressed(_ sender: UIButton) {

        let urlString = "https://myschool.zportal.nl/api/v3/oauth/token"
        performRequest(urlString: urlString)

    }
    
    func performRequest(urlString: String) {
        
        if let url = URL(string: urlString) {
            
            var request = URLRequest(url: url)
            request.httpMethod = "POST"
            request.httpBody = "grant_type=authorization_code&code=1234567890".data(using: .utf8)
            
            let session = URLSession(configuration: .default)
            let task = session.dataTask(with: request, completionHandler: { data, response, error -> Void in
                //print(response!)
                do {
                    let json = try JSONSerialization.jsonObject(with: data!) as! Dictionary<String, AnyObject>
                    print(json)
                } catch {
                    print("error")
                }
            })
            
            task.resume()
            
        }
        
    }

不是最好的代码，但我稍后会对其进行完善。起初，我解析了JSON，但事实证明我需要解析一个String并将其转换为使用以下类型的数据类型：.data(using: .utf8)

尝试获取访问令牌时出现Swift'Required String parameter'grant_type'not present'错误

2 个答案: