RocMarshal commented on a change in pull request #11979:
URL: https://github.com/apache/flink/pull/11979#discussion_r422002930
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File path: docs/training/event_driven.zh.md
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@@ -269,38 +259,35 @@ SingleOutputStreamOperator hourlyTips = fares
hourlyTips.getSideOutput(lateFares).print();
{% endhighlight %}
-Alternatively, you can use two OutputTags with the
-same name to refer to the same side output, but if you do, they must have the
same type.
+或者，可以使用两个同名的 OutputTag 来引用同一个旁路输出，但如果这样做，它们必须具有相同的类型。
{% top %}
-## Closing Remarks
+## 结语
-In this example you have seen how a `ProcessFunction` can be used to
reimplement a straightforward time
-window. Of course, if Flink's built-in windowing API meets your needs, by all
means, go ahead and
-use it. But if you find yourself considering doing something contorted with
Flink's windows, don't
-be afraid to roll your own.
+在本例中，你已经了解了如何使用 `ProcessFunction` 重新实现一个简单的时间窗口。
+当然，如果 Flink 内置的窗口 API 能够满足你的开发需求，那么一定要优先使用它。
+但如果你发现自己在考虑用 Flink 的窗口做些错综复杂的事情，不要害怕自己动手。
-Also, `ProcessFunctions` are useful for many other use cases beyond computing
analytics. The hands-on
-exercise below provides an example of something completely different.
+此外，`ProcessFunctions` 对于计算分析之外的许多其他用例也很有用。
+下面的实践练习提供了一个完全不同的例子。
-Another common use case for ProcessFunctions is for expiring stale state. If
you think back to the
-[Rides and Fares Exercise](https://github.com/apache/flink-training/tree/{% if
site.is_stable %}release-{{ site.version_title }}{% else %}master{% endif
%}/rides-and-fares),
-where a `RichCoFlatMapFunction` is used to compute a simple join, the sample
solution assumes that
-the TaxiRides and TaxiFares are perfectly matched, one-to-one for each
`rideId`. If an event is lost,
-the other event for the same `rideId` will be held in state forever. This
could instead be implemented
-as a `KeyedCoProcessFunction`, and a timer could be used to detect and clear
any stale state.
+ProcessFunctions 的另一个常见用例是过期过时 State。如果你回想一下
+[Rides and Fares Exercise](https://github.com/apache/flink-training/tree/{% if
site.is_stable %}release-{{ site.version_title }}{% else %}master{% endif
%}/rides-and-fares)，
+其中使用 `RichCoFlatMapFunction` 来计算简单 Join，那么示例解决方案假设 TaxiRides 和 TaxiFares
+完全匹配，每个 `rideId` 一对一。如果某个事件丢失，则同一 `rideId` 的另一个事件将永远保持 State。
Review comment:
@klion26
Hi, @klion26
I re-integrated the entire scene of the example to translate these sentences
more semantically and apply your suggestions. If you have any better
suggestions, please leave comments, I will continue to improve it. From the
bottom of my heart, thank you so much for your guidance and help !
Sincerely,
Roc
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