无法在 Toast (Android) 中解析和显示结果？答案

【问题标题】：Unable to parse and display result in Toast (Android)?无法在 Toast (Android) 中解析和显示结果？
【发布时间】：2011-05-06 05:15:59
【问题描述】：

我是 Android 新手，所以我在 HTTP 请求的 JSON 解析中遇到问题。让我简要介绍一下。实际上我正在尝试登录，用户将输入“电子邮件”作为用户名和“密码”作为密码，只要他点击登录按钮，就会在 url http://excoflare.com/dev2010/bb_snet/baranzan/index.php?json=json&UM_email="+sUserName+"&UM_password="+sPassword 上发出 http 请求，如果成功Toast 消息将在同一布局上到达“欢迎用户名”。我有两个类 (i) HelloAndroid.java 和 (ii) RestJsonClient.java

HelloAndroid.java

 public class HelloAndroid extends Activity implements OnClickListener {
                   /** Called when the activity is first created. */
        @Override
        public void onCreate(Bundle icicle) {
                         super.onCreate(icicle);
                         setContentView(R.layout.main);
                         Button login = (Button)findViewById(R.id.login_button);
                        login.setOnClickListener(this);
                     }

                     public void onClick(View v) {
                         EditText usernameEditText = (EditText) findViewById(R.id.txt_username);
                         EditText passwordEditText = (EditText) findViewById(R.id.txt_password);
                         String sUserName = usernameEditText.getText().toString();
                         String sPassword = passwordEditText.getText().toString();
                         String address = "http://excoflare.com/dev2010/bb_snet/baranzan/index.php?json=json&UM_email="+sUserName+"&UM_password="+sPassword+"";
                         JSONObject json = RestJsonClient.connect(address); 
                          /* is something missing here if yes Please HELP*/
                         next();
                    }

                     private void next(){
                         Toast.makeText(HelloAndroid.this, "Welcome username", Toast.LENGTH_SHORT).show();
                    }

                   }

RestJsonClient

public class RestJsonClient {

    public static JSONObject connect(String url)
    {

        HttpClient httpclient = new DefaultHttpClient();

        // Prepare a request object
        HttpGet httpget = new HttpGet(url); 

        // Execute the request
        HttpResponse response;

        JSONObject json = new JSONObject();

        try {
            response = httpclient.execute(httpget);

            HttpEntity entity = response.getEntity();

            if (entity != null) {

                // A Simple JSON Response Read
                InputStream instream = entity.getContent();
                String result= convertStreamToString(instream);

                json=new JSONObject(result);

                instream.close();
            }

        } catch (ClientProtocolException e) {
            // TODO Auto-generated catch block
            e.printStackTrace();
        } catch (IOException e) {
            // TODO Auto-generated catch block
            e.printStackTrace();
        } catch (JSONException e) {
            // TODO Auto-generated catch block
            e.printStackTrace();
        }

        return json;
    }
    /**
     *
     * @param is
     * @return String
     */
    public 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();
    }

}

现在我只尝试使用正确/有效的电子邮件和密码，现在只有我想要的。请帮忙。非常感谢

【问题讨论】：

检查您是否在 logcat 中遇到任何异常......
方法convertStreamToString返回什么？

标签： android json parsing

【解决方案1】：

JSONObject json = RestJsonClient.connect(address); 
                      /* is something missing here if yes Please HELP*/
                     next();

这里你应该有一个 if-else 条件来检查你从 json 响应中得到的数据。在您的情况下，您应该检查用户身份验证。如果用户通过了身份验证，那么您应该不显示 Toast。

【讨论】：