使用基本身份验证的POST在Android上失败但在C#中有效

时间:2011-08-04 18:55:28

标签: android http-post basic-authentication

我有一个正在开发的应用,需要我将数据发布到第三方API。我从一开始就一直在努力进行身份验证,并且不断进一步推迟,现在我被卡住了。

我已尝试使用Authenticator,但已阅读所有有关某些Android版本中似乎存在错误的信息:Authentication Example

我尝试了几种不同的选项,包括Apache Commons HTTP Library没有成功。毕竟,我决定确保API不是痛点。所以我编写了一个快速WinForms程序来测试API,它在第一次尝试时完美运行。因此,我正在使用的想法以及我使用的API看起来都很好,但我迫切需要一些指导,以解释为什么Java代码不起作用。

示例如下:

每天都有效的

C#代码:

        System.Net.ServicePointManager.Expect100Continue = false;
        // Create a request using a URL that can receive a post. 
        WebRequest request = WebRequest.Create(addWorkoutUrl);
        // Set the Method property of the request to POST.
        request.Method = "POST";
        // Create POST data and convert it to a byte array.
        string postData = "distance=4000&hours=0&minutes=20&seconds=0&tenths=0&month=08&day=01&year=2011&typeOfWorkout=standard&weightClass=H&age=28";
        byte[] byteArray = Encoding.UTF8.GetBytes(postData);
        // Set the ContentType property of the WebRequest.
        request.Headers["X-API-KEY"] = apiKey;
        request.Headers["Authorization"] = "Basic " + Convert.ToBase64String(Encoding.Default.GetBytes("username:password"));
        request.ContentType = "application/x-www-form-urlencoded";
        // Set the ContentLength property of the WebRequest.
        request.ContentLength = byteArray.Length;
        // Get the request stream.
        Stream dataStream = request.GetRequestStream();
        // Write the data to the request stream.
        dataStream.Write(byteArray, 0, byteArray.Length);
        // Close the Stream object.
        dataStream.Close();
        // Get the response.
        WebResponse response = request.GetResponse();
        // Display the status.
        MessageBox.Show(((HttpWebResponse)response).StatusDescription);
        // Get the stream containing content returned by the server.
        dataStream = response.GetResponseStream();
        // Open the stream using a StreamReader for easy access.
        StreamReader reader = new StreamReader(dataStream);
        // Read the content.
        string responseFromServer = reader.ReadToEnd();
        // Display the content.
        MessageBox.Show(responseFromServer);
        // Clean up the streams.
        reader.Close();
        dataStream.Close();
        response.Close();

Android的Java代码,当前返回500:内部服务器错误,但我相信这是我的错。

    URL url;
    String data = "distance=4000&hours=0&minutes=20&seconds=0&tenths=0&month=08&day=01&year=2011&typeOfWorkout=standard&weightClass=H&age=28";
HttpURLConnection connection = null;
    //Create connection
    url = new URL(urlBasePath);
    connection = (HttpURLConnection)url.openConnection();
    connection.setConnectTimeout(10000);
    connection.setUseCaches(false);
connection.setRequestProperty("User-Agent","Mozilla/5.0 ( compatible ) ");
    connection.setRequestProperty("Accept","*/*");
    connection.setRequestProperty("X-API-KEY", apiKey);
    connection.setRequestProperty("Authorization", "Basic " +
    Base64.encode((username + ":" + password).getBytes("UTF-8"), Base64.DEFAULT));
    connection.setRequestProperty("Content-Type", "application/x-www-form-urlencoded");
connection.setRequestProperty("Content-Length", "" + Integer.toString(data.getBytes("UTF-8").length));

    DataOutputStream wr = new DataOutputStream(connection.getOutputStream());
    wr.write(data.getBytes("UTF-8"));
    wr.flush();
    wr.close();
    statusCode = connection.getResponseCode();
    statusReason = connection.getResponseMessage();
    //At this point, I have the 500 error...

3 个答案:

答案 0 :(得分:3)

我找到了问题,最终解决了上述评论中提到的根本原因之后的解决方案。

我在我的示例中使用Base64.encode(),但我需要使用Base64.encodeToString()

不同之处在于encode()会返回byte[]encodeToString()会返回我期待的string

希望这会帮助被其抓住的其他人。

答案 1 :(得分:0)

这是一个更好的POST方法。

install-package HttpClient

然后:

public void DoPost()
{
    var httpClient = new HttpClient();
    var creds = string.Format("{0}:{1}", _username, _password);
    var basicAuth = string.Format("Basic {0}", Convert.ToBase64String(Encoding.UTF8.GetBytes(creds)));
    httpClient.DefaultRequestHeaders.Add("Authorization", basicAuth);
    var post = httpClient.PostAsync(_url, 
        new FormUrlEncodedContent(new Dictionary<string, string>
            {
                { "name", "Henrik" },
                { "age", "99" }
            }));

    post.Wait();
}

答案 2 :(得分:0)

我在java中试过这个

import java.io.*;
import java.net.*;

class download{   
    public static void main(String args[]){
        try{
            String details = "API-Key=e6d871be90a689&orderInfo={\"booking\":{\"restaurantinfo\":{\"id\":\"5722\"},\"referrer\":{\"id\": \"9448476530\" },   \"bookingdetails\":{\"instructions\":\"Make the stuff spicy\",\"bookingtime\": \"2011-11-09 12:12 pm\", \"num_guests\": \"5\"}, \"customerinfo\":{\"name\":\"Ramjee Ganti\",    \"mobile\":\"9345245530\",  \"email\": \"sajid@pappilon.in\",   \"landline\":{ \"number\":\"0908998393\",\"ext\":\"456\"}}}}";

            Authenticator.setDefault(new Authenticator() {
                protected PasswordAuthentication getPasswordAuthentication() {
                  return new PasswordAuthentication("admin", "1234".toCharArray());
                }
            });

            HttpURLConnection conn = null;
            //URL url = new URL("http://api-justeat.in/ma/orders/index");
                        URL url = new URL("http://api.geanly.in/ma/order_ma/index");
            conn = (HttpURLConnection) url.openConnection();

            conn.setDoOutput(true);
            conn.setDoInput (true);

            conn.setRequestMethod("POST");
            //conn.setRequestMethod(HttpConnection.POST);
            DataOutputStream outStream = new DataOutputStream(conn.getOutputStream());
            outStream.writeBytes(details);
            outStream.flush();
            outStream.close();          

            //Get Response  
            InputStream is = conn.getInputStream();
            BufferedReader rd = new BufferedReader(new InputStreamReader(is));
            String line;
            StringBuffer response = new StringBuffer(); 

            while((line = rd.readLine()) != null) {
                 System.out.println(line);
            }

            rd.close();
            System.out.println(conn.getResponseCode() + "\n\n");
        }catch(Exception e){
                    System.out.println(e);         
        }
    }
}

这可能会有所帮助。