Rez*_*Rad 5 android rx-java retrofit android-workmanager
在我的聊天应用程序中,我想从本地数据库读取消息并由工作经理将它们发送到服务器,并在用户发送新消息时进行改造。
在父母中:
OneTimeWorkRequest workRequest = new OneTimeWorkRequest
.Builder(SendMsg_Work.class)
.setConstraints(new Constraints
.Builder()
.setRequiredNetworkType(NetworkType.CONNECTED)
.build())
.setInitialDelay(5, TimeUnit.SECONDS)
.setBackoffCriteria(BackoffPolicy.EXPONENTIAL, 30, TimeUnit.SECONDS)
.build();
WorkManager.getInstance().enqueue(workRequest);
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在工人:
@NonNull
@Override
public Result doWork() {
ChatRoom_Model_Local ll = database.get_unSendMsg();
if (ll != null) {
sendMessage(ll);
} else {
return Result.failure();
}
}
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和改造 - 我也使用 RxJava :
private void sendMessage(ChatRoom_Model_Local model) {
Single<Response<MSG_Response>> api = ApiClient.createService(ApiInterface.class, pr.getData(pr.SendMessage_url))
.sendMSG(pr.getData(pr.MY_TOKEN), model.getOther_user(), model.getMsg_text());
api.subscribeOn(Schedulers.io())
.observeOn(AndroidSchedulers.mainThread())
.subscribe(new SingleObserver<Response<MSG_Response>>() {
@Override
public void onSubscribe(Disposable d) { }
@Override
public void onSuccess(Response<MSG_Response> response) {
if (response.isSuccessful() && response.body() != null) {
String message = response.body().getMessage();
Intent bi = new Intent(new Config().getCHAT_SERVICE());
if (message != null && message.equals("xyx")) { // INSERT Successfully
String server_id = response.body().getId();
String localId = response.body().getIdLocal();
bi.putExtra("ChatRoomService_id", localId);
bi.putExtra("ChatRoomService_s_id",server_id);
int result = database.setMsgServerId(localId,server_id);
}
LocalBroadcastManager.getInstance(getApplicationContext()).sendBroadcast(bi);
}
ChatRoom_Model_Local ll = database.get_unSendMsg();
if (ll != null) {
sendMessage(ll);
}
}
@Override
public void onError(Throwable e) {
}
});
}
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现在的问题是:
如何在改造仍在进行时在 doWork 方法中返回一些结果?
如果工人已经被调用并且改造正在进行中并且工人被再次调用怎么办?
如何将消息一个接一个地发送到服务器?
谢谢
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