如何使用RoboSpice从中断的未缓存请求中正确恢复

Question

我有一个处理我的应用程序登录的简单片段。由于我正在处理登录请求，因此我不想缓存它们。在我在请求中间引入暂停或方向更改之前，该策略可以正常工作。当用户单击登录按钮时，我将显示一个ProgressDialog。当我得到响应（成功或失败）时，这消失了。如果我进入主屏幕，然后在登录请求过程中返回到应用程序，则我的侦听器将永远不会收到通知，因此我的ProgressDialog不会被关闭并且我的应用程序将被冻结。我尝试在onStart中添加spiceManager.getFromCache。这很有帮助，但是当应用程序尝试恢复时，结果始终为null
…这是有道理的，因为未缓存结果。在这种情况下，配置要通知我的侦听器的正确方法是什么？

// using Jackson2SpringAndroidSpiceService

public void onStart() {
    super.onStart();
    spiceManager.start(getActivity());
    spiceManager.addListenerIfPending(AccessTokenResponse.class, null,
            new AccessTokenResponseRequestListener());

    //spiceManager.getFromCache(AccessTokenResponse.class,
    //        null, DurationInMillis.ALWAYS_EXPIRED,
    //        new AccessTokenResponseRequestListener());
}


private void performRequest(String username, String password) {
    progressDialog = ProgressDialog.show(getActivity(), "", "Logging in...", true);
    LoginFragment.this.getActivity().setProgressBarIndeterminateVisibility(true);
    LoginRequest request = new LoginRequest(username, password);
    spiceManager.execute(request, null, DurationInMillis.ALWAYS_EXPIRED, new AccessTokenResponseRequestListener());
}


private class AccessTokenResponseRequestListener implements RequestListener<AccessTokenResponse> {

    @Override
    public void onRequestFailure(SpiceException e) {
        //update your UI
        if(progressDialog != null && progressDialog.isShowing()) {
            progressDialog.dismiss();
        }
        buttonLogin.setEnabled(true);
        Log.e(TAG, "Login unsuccessful");
        if(e.getCause() instanceof HttpClientErrorException)
        {
            HttpClientErrorException exception = (HttpClientErrorException)e.getCause();
            if(exception.getStatusCode().equals(HttpStatus.BAD_REQUEST))
            {
                Log.e(TAG, "Login unsuccessful");
                Toast.makeText(getActivity().getApplicationContext(),
                        "Wrong username/password combo!",
                        Toast.LENGTH_LONG).show();
            }
            else
            {
                Toast.makeText(getActivity().getApplicationContext(),
                        "Login unsuccessful! If the problem persists, please contact support.",
                        Toast.LENGTH_LONG).show();
            }
        } else {
            Toast.makeText(getActivity().getApplicationContext(),
                "Login unsuccessful! If the problem persists, please contact support.",
                Toast.LENGTH_LONG).show();
        }
    }

    @Override
    public void onRequestSuccess(AccessTokenResponse accessToken) {
        //update  UI
        if(progressDialog != null && progressDialog.isShowing()) {
            progressDialog.dismiss();
        }
        buttonLogin.setEnabled(true);

        if (accessToken != null) { 
            OnAuthenticatedListener listener = (OnAuthenticatedListener) getActivity();
            listener.userLoggedIn(editTextUsername.getText().toString(), accessToken);
        }

    }
}

Answer 1

使用缓存。使用某些缓存键执行请求

spiceManager.execute(request, "your_cache_key", DurationInMillis.ALWAYS_EXPIRED, new AccessTokenResponseRequestListener());

并在侦听器中删除该请求的响应，如果该请求在您切换到另一个活动之前成功返回，则从缓存中删除该响应，因为您不想根据需要缓存帐户信息。

@Override
public void onRequestFailure(SpiceException e) {
    ....
    spiceManager.removeDataFromCache(AccessTokenResponse.class);
    ....
}

@Override
public void onRequestSuccess(AccessTokenResponse accessToken) {
    if (accessToken == null) {
        return;
    }
    ....
    spiceManager.removeDataFromCache(AccessTokenResponse.class);
    ....
}

在onStart中，如果您切换到另一个活动，现在返回上一个活动，则尝试获取缓存的响应。该返回响应在您调用spiceManager.shouldStop（）之后到达。否则返回null。

spiceManager.getFromCache(AccessTokenResponse.class, "your_cache_key", DurationInMillis.ALWAYS_RETURNED, new AccessTokenResponseRequestListener());