这是使用 Rx 轮询另一个系统的相当经典的案例。大多数人都会用Observable.Interval
作为他们的首选运营商,对于大多数人来说这没什么问题。
但是,您对超时和重试有特定要求。在这种情况下,我认为你最好使用运算符组合:
-
Observable.Timer
允许您在指定时间执行查询
-
Timeout
识别已溢出的数据库查询
-
ToObservable()
映射你的Task
结果为可观察的序列。
-
Retry
让您在超时后恢复
-
Repeat
允许您在成功的数据库查询后继续。这也将保持上一个数据库查询完成与下一个数据库查询开始之间的初始周期/间隙。
这个工作LINQPad代码片段应该显示查询正常工作:
void Main()
{
var pollingPeriod = TimeSpan.FromSeconds(5);
var dbQueryTimeout = TimeSpan.FromSeconds(10);
//You will want to have your Rx query timeout after the expected silence of the timer, and then further maximum silence.
var rxQueryTimeOut = pollingPeriod + dbQueryTimeout;
var scheduler = new EventLoopScheduler(ts => new Thread(ts) { Name = "DatabasePoller" });
var query = Observable.Timer(pollingPeriod, scheduler)
.SelectMany(_ => DatabaseQuery().ToObservable())
.Timeout(rxQueryTimeOut, Observable.Return("Timeout"), scheduler)
.Retry() //Loop on errors
.Repeat(); //Loop on success
query.StartWith("Seed")
.TimeInterval(scheduler) //Just to debug, print the timing gaps.
.Dump();
}
// Define other methods and classes here
private static int delay = 9;
private static int delayModifier = 1;
public async Task<string> DatabaseQuery()
{
//Oscillate the delay between 3 and 12 seconds
delay += delayModifier;
var timespan = TimeSpan.FromSeconds(delay);
if (delay < 4 || delay > 11)
delayModifier *= -1;
timespan.Dump("delay");
await Task.Delay(timespan);
return "Value";
}
结果如下:
Seed 00:00:00.0125407
Timeout 00:00:15.0166379
Timeout 00:00:15.0124480
Timeout 00:00:15.0004520
Timeout 00:00:15.0013296
Timeout 00:00:15.0140864
Value 00:00:14.0251731
Value 00:00:13.0231958
Value 00:00:12.0162236
Value 00:00:11.0138606
样本的关键部分是......
var query = Observable.Timer(TimeSpan.FromSeconds(5), scheduler)
.SelectMany(_ => DatabaseQuery().ToObservable())
.Timeout(rxQueryTimeOut, Observable.Return("Timeout"), scheduler)
.Retry() //Loop on errors
.Repeat(); //Loop on success
EDIT:以下是如何得出此解决方案的进一步说明。https://github.com/LeeCampbell/RxCookbook/blob/master/Repository/Polling.md