Question

所以我试图将数据发布到服务器，然后我得到一个400代码。我知道这意味着它是一个糟糕的请求，而且由于我认为这是一个错误的json。所以这就是我被告知要形成json的方式

curl -i -u admin:admin -H "Content-Type: application/json" -X POST –d '{"customer_id":1, "device_id":"1234", "notification_id":"NOTIFICATIONID123456", "operating_system_type":"IOS"}' localhost:8090/api/1.0/notificationInformation

所以现在我不知道服务器是问题还是我的代码是问题。我以前遇到过android HttpUrlConnection的问题。

所以这是我的http处理程序类：

public class MyHttpPost extends AsyncTask<String, Void, String> {



private String username;
private String password;

private String json =  "{\"customer_id\": \"000234\" , \"device_id\":\"1234443\", \"notification_id\":\"fLqDFPgBHcs:APA91bHHIB7kyTRqR18pK78k81AbV211jdhIyNlWK-CmejFKIK6FZJgx4R-7uzyfYLTqi_jhqclks07nkkQFnORXOU28wJ5qC3GIIY_WPaNxKCxIUTltMaWihPGcvaeOgyW7669M3K1n\", \"operating_system_type\":\"Android\"}";
///api/1.0/notificationInformation

public MyHttpPost(){}
public MyHttpPost(String username, String password)
{
    this.username = username;
    this.password = password;
}
@Override
protected String doInBackground(String... params) {
    String url = params[0];
    postHttp(url);
    return null;
}

   public void postHttp(String myUrl)
   {


   Log.d("Button ", "Pushed");
   try {

       URL url = new URL(myUrl);
       if (username!=null)
       {
           Authenticator.setDefault(new Authenticator() {
               @Override
               protected PasswordAuthentication getPasswordAuthentication() {
                   return new PasswordAuthentication(username, password.toCharArray());
               }
           });
       }

       HttpURLConnection connection = (HttpURLConnection) url.openConnection();

       // Allow Outputs (sending)
       connection.setDoOutput(true);
       Log.d("Allows input: ", Boolean.toString(connection.getDoInput()));
       connection.setUseCaches(false);


       // Enable POST method
       connection.setRequestMethod("POST");
       connection.setRequestProperty("Content-Type", "application/json");
       connection.setRequestProperty("charset", "utf-8");
       DataOutputStream printout = new DataOutputStream(connection.getOutputStream());
       //printout.write(json.toString().getBytes("UTF8"));
       printout.write(json.getBytes("UTF8"));



       printout.flush();
       printout.close();
       int statuscode = connection.getResponseCode();
       Log.d("Response code: " , Integer.toString(statuscode));


   } catch (MalformedURLException e) {
       e.printStackTrace();
   } catch (UnsupportedEncodingException e) {
       e.printStackTrace();
   } catch (ProtocolException e) {
       e.printStackTrace();
   } catch (IOException e) {
       e.printStackTrace();
   }
   }

   }

对此有任何帮助将不胜感激。

Answer 1

试试这个

  URL url = new URL(url);
                    HttpURLConnection conn = (HttpURLConnection)url.openConnection();
                    conn.setRequestMethod("POST");
                    conn.setRequestProperty("Content-Type", "application/json");
                    conn.setDoOutput(true);
                JSONObject postDataParams = new JSONObject();
                    try {
                        postDataParams.put("username","myname");
                        postDataParams.put("password", "mypwd");
                    } catch (JSONException e) {
                        e.printStackTrace();
                    }

                    PrintWriter out = new PrintWriter(conn.getOutputStream());
                    out.print(postDataParams);
                    out.close();

                    int responseCode=conn.getResponseCode();

当我尝试发布时，Android URLConnection失败

1 个答案: