android - 在 Web 服务中将 JSON 对象作为参数传递 - 返回 "Bad Request"响应

Question

Android 新手，我需要在我的网络服务调用中使用以下参数。我理解参数实际上是 JSON 对象。

当它应该返回登录用户信息时，下面的代码返回带有“title: Bad request”的 XML。 logcat 的值显示为 --> json: {"Query":"com.androidatc.customviewindrawer.Query@f1eb09f","includeUserMiscInfo":true} 表示我的参数不正确。如何正确传递？

protected void sendJson(final String email, final String pwd) {
        Thread t = new Thread() {

            public void run() {
                Looper.prepare(); //For Preparing Message Pool for the child Thread
                HttpClient client = new DefaultHttpClient();
                HttpConnectionParams.setConnectionTimeout(client.getParams(), 10000); //Timeout Limit
                HttpResponse response;
                JSONObject json = new JSONObject();

                try {
                    HttpPost post = new HttpPost("http://192.168.0.102/SDService_SAFTI/ServiceSD.svc/LoginUser");
                    Query queryObj = new Query();
                    queryObj.setLogin("WT");
                    queryObj.setPassword("3");


                    json.put("Query", queryObj);
//                  json.put("email", email);
//                  json.put("password", pwd);
                    json.put("includeUserMiscInfo", true);


                    StringEntity se = new StringEntity( json.toString());
                    se.setContentType(new BasicHeader(HTTP.CONTENT_TYPE, "application/json"));
                    post.setEntity(se);
                    response = client.execute(post);

                    /*Checking response */
                    if(response!=null){


                        InputStream in = response.getEntity().getContent(); //Get the data in the entity
                         Toast.makeText(getActivity().getApplicationContext(), "Response:" + convertStreamToString(in),Toast.LENGTH_LONG).show();

                    }

                } catch(Exception e) {
                    e.printStackTrace();
//                  getActivity().createDialog("Error", "Cannot Estabilish Connection");
                }

                Looper.loop(); //Loop in the message queue
            }
        };

        t.start();
    }

    private static String convertStreamToString(InputStream is) {

        BufferedReader reader = new BufferedReader(new InputStreamReader(is));
        StringBuilder sb = new StringBuilder();

        String line = null;
        try {
            while ((line = reader.readLine()) != null) {
                sb.append((line + "\n"));
            }
        } catch (IOException e) {
            e.printStackTrace();
        } finally {
            try {
                is.close();
            } catch (IOException e) {
                e.printStackTrace();
            }
        }
        return sb.toString();
    }

查询.java

public class Query {
    String login;

    public void setPassword(String password) {
        this.password = password;
    }

    public void setLogin(String login) {
        this.login = login;
    }

    String password;

    public String getPassword() {
        return password;
    }

    public String getLogin() {
        return login;
    }




}

如有任何建议，我们将不胜感激。非常感谢。

Answer 1

作为您的 Json 类型，您可能必须这样做，

JSONArray array = new JSONArray();
array.put("login=WT&password=3");

json.put("Query", array);
json.put("includeUserMiscInfo", "true");

我想建议你也尝试替换这条线，

json.put("includeUserMiscInfo", true);

与

json.put("includeUserMiscInfo", "true");