class 切换后 startActivity 不工作

Question

我想启动 activity 并在解析 JSON 响应后发出 Toast 消息。

public class Sign_Up extends AppCompatActivity {

  ....

  register();


  public void register()
  {
  final String url = "someURL";

        new Json().checkJsonFile(url,getApplicationContext());
  }

  //Now in an non-activity

  public class Json {

    public void checkJsonFile(final String url, final Context context) {

        new Thread(new Runnable() {
            public void run() {
                String result;
                String line;

                try {

                    URL obj = new URL(url);
                    HttpURLConnection conn = (HttpURLConnection) obj.openConnection();
                    conn.setReadTimeout(5000);
                    conn.addRequestProperty("Accept-Language", "en-US,en;q=0.8");
                    conn.addRequestProperty("User-Agent", "Mozilla");
                    conn.addRequestProperty("Referer", "google.com");


                    boolean redirect = false;

                    // normally, 3xx is redirect
                    int status = conn.getResponseCode();
                    if (status != HttpURLConnection.HTTP_OK) {
                        if (status == HttpURLConnection.HTTP_MOVED_TEMP
                                || status == HttpURLConnection.HTTP_MOVED_PERM
                                || status == HttpURLConnection.HTTP_SEE_OTHER)
                            redirect = true;
                    }

                    if (redirect) {

                        // get redirect url from "location" header field
                        String newUrl = conn.getHeaderField("Location");

                        // get the cookie if need, for login
                        String cookies = conn.getHeaderField("Set-Cookie");

                        // open the new connnection again
                        conn = (HttpURLConnection) new URL(newUrl).openConnection();
                        conn.setRequestProperty("Cookie", cookies);
                        conn.addRequestProperty("Accept-Language", "en-US,en;q=0.8");
                        conn.addRequestProperty("User-Agent", "Mozilla");
                        conn.addRequestProperty("Referer", "google.com");


                    }

                    BufferedReader in = new BufferedReader(
                            new InputStreamReader(conn.getInputStream()));

                    StringBuilder sb = new StringBuilder();

                    while ((line = in.readLine()) != null) {
                        sb.append(line);

                    }
                    in.close();

                    result = sb.toString();

                    new Sign_Up().parseJSON(result);

                } catch (Exception e) {
                    e.printStackTrace();
                }
            }
        }).start();
    }
}


        //Back in Sign_Up class

        public void parseJSON(String JSON){
        int error = -1;
        try {
                JSONObject jsonObject = new JSONObject(JSON);

                error = Integer.parseInt(jsonObject.getString("error_code"));

        } catch (JSONException e) {
            //Retry
            register();

            e.printStackTrace();
        }

        progressDialog.dismiss();

        switch (error) {
            case 0: //Successful
                finish();
                startActivity(new Intent(Sign_Up.this, Loading.class));
                break;
            case 1:
                Toast.makeText(getApplicationContext(), "blabla", Toast.LENGTH_SHORT).show();
                break;
            case 2:
                Toast.makeText(getApplicationContext(), "other blabla", Toast.LENGTH_SHORT).show();
                break;
            default:
                register();
                break;
        }
    }

但是，当我将 Json 内容放入 Sign_Up class 时，至少 startActivity 有效，但 Toast 无效。同样，我想将 Json 的内容放在单独的 class.

中

非常感谢您！

Answer 1

您不能通过调用

创建 Activity

new MyActivity()

并期望一切正常。行

new Sign_Up().parseJSON(result);

正在错误地创建 Sign_Up activity 的新实例，因此它未绑定到上下文，并且将执行其生命周期调用的 none。您无法在该状态下显示 Toast。

更好的解决方案是将接口从 Sign_Up 传递到 Json class 以让它回调到原始 Sign_Up 实例，而不是尝试创建一个新实例。例如：

class Json {
    public interface Callback {
        public void run(String result);    
    }

    private Callback callback

    Json(Callback c) {
        callback = c;
    }

    // later, replace new Sign_Up().parseJSON(result) with callback.run(result)
}

并创建它替换

new Json().checkJsonFile(url,getApplicationContext());

和

new Json(new Json.Callback() {
    @Override
    public void run(String result) {
        parseJSON(result);
    }
}).checkJsonFile(url,getApplicationContext());

您还需要确保所有 UI 操作（toast 生成）都在 UI 线程上运行。

class 切换后 startActivity 不工作

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