Question

我有一个简单的片段来处理我的应用程序的登录。由于我正在处理登录请求，我不想缓存它们。此策略正常工作，直到我在请求中间引入暂停或方向更改。当用户单击登录按钮时，我显示ProgressDialog。当我得到响应（成功或失败）时，这就消失了。如果我进入主屏幕，然后在登录请求中间返回应用程序，我的监听器永远不会得到通知，因此我的ProgressDialog不会被解雇，我的应用程序被冻结。我尝试在我的onStart中添加spiceManager.getFromCache。这有帮助，但是当应用程序尝试恢复时结果始终为null ...这是有道理的，因为结果不会被缓存。配置我的侦听器以在此方案中得到通知的正确方法是什么？

// 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。使用一些缓存键执行请求

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

并且在侦听器中，如果在切换到另一个活动之前成功返回，则会从缓存中删除对此请求的响应，因为您不希望根据您的要求缓存帐户信息。

@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尝试获取缓存响应，如果您切换到另一个活动，现在回到上一个活动。这个返回响应是在调用spiceManager.shouldStop（）之后到达的。否则返回null。

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

如何使用RoboSpice从中断的未缓存请求中正确恢复

1 个答案: