java-如何使用RoboSpice从中断的未缓存请求中正确恢复

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,则我的监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);
        }

    }
}

解决方法:

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());

并且在侦听器中,如果您在切换到另一个活动之前成功返回了对该请求的响应,则从缓存中删除对该请求的响应,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());

标签：robospice,java,android
来源： https://codeday.me/bug/20191121/2048550.html