关于使用IngressTime，window过小的问题

Flink 1.10，windows 10 flink api验证

代码如下
```

import org.apache.flink.api.java.functions.KeySelector;
import org.apache.flink.api.java.tuple.Tuple2;
import org.apache.flink.streaming.api.TimeCharacteristic;
import org.apache.flink.streaming.api.datastream.DataStreamSource;
import org.apache.flink.streaming.api.environment.StreamExecutionEnvironment;
import org.apache.flink.streaming.api.functions.IngestionTimeExtractor;
import org.apache.flink.streaming.api.windowing.assigners.TumblingEventTimeWindows;
import org.apache.flink.streaming.api.windowing.assigners.TumblingProcessingTimeWindows;
import org.apache.flink.streaming.api.windowing.time.Time;

import java.util.ArrayList;
import java.util.List;

public class KeyedStreamJob {
    public static void main(String[] args) throws Exception {
        StreamExecutionEnvironment env = StreamExecutionEnvironment.getExecutionEnvironment();
        env.setStreamTimeCharacteristic(TimeCharacteristic.IngestionTime);
//        env.setParallelism(3);

        Tuple2<String, Integer> item = null;
        List<Tuple2<String, Integer>> items = new ArrayList<>();
        item = new Tuple2<>("k1", 1);
        items.add(item);
        item = new Tuple2<>("k3", 10);
        items.add(item);
        item = new Tuple2<>("k1", 10);
        items.add(item);
        item = new Tuple2<>("k2", 2);
        items.add(item);
        item = new Tuple2<>("k1", 11);
        items.add(item);
        item = new Tuple2<>("k2", 20);
        items.add(item);
        DataStreamSource<Tuple2<String, Integer>> streamSource = env.fromCollection(items);
        streamSource
                //by 1
                //.assignTimestampsAndWatermarks(new IngestionTimeExtractor())
                .keyBy(new KeySelector<Tuple2<String, Integer>, String>() {
            @Override
            public String getKey(Tuple2<String, Integer> value) throws Exception {
                return value.f0;
            }
        })
                .window(TumblingEventTimeWindows.of(Time.milliseconds(10L)))
                .sum(1)
                .print("+++++++++++++++++++++++++++");

        env.execute("keyedSteamJob");
    }
}

```
输出
```
+++++++++++++++++++++++++++:1> (k3,10)
+++++++++++++++++++++++++++:2> (k1,1)
+++++++++++++++++++++++++++:8> (k2,22)
+++++++++++++++++++++++++++:2> (k1,21)
```
如果把

window(TumblingEventTimeWindows.of(Time.milliseconds(10L)))
改成

.window(TumblingEventTimeWindows.of(Time.seconds(10L)))
输出
```
+++++++++++++++++++++++++++:8> (k2,22)
+++++++++++++++++++++++++++:1> (k3,10)
+++++++++++++++++++++++++++:2> (k1,22)
```
两次不同的windows窗口，第一次输出对于key=‘k1‘不聚集，第二次输出聚集

为什么会这样，如何验证怎么样的过程处理流程导致这样的结果区别

如果k1=1已经在ValueState中（2>(k1,1)），
那么再次输出时currentKey=k1时，这个时候ValueState的value是1，那么输出应该是10+11+1，而不是10+11；


如果window改成1秒也是按照正常结果输出





发送自 Windows 10 版邮件<https://go.microsoft.com/fwlink/?LinkId=550986>应用

如果不使用window，那么输出会按照ValueState的存量的key的ValueState聚合
输出
```
+++++++++++++++++++++++++++:2> (k1,1)
+++++++++++++++++++++++++++:1> (k3,10)
+++++++++++++++++++++++++++:2> (k1,11)
+++++++++++++++++++++++++++:8> (k2,2)
+++++++++++++++++++++++++++:2> (k1,22)
+++++++++++++++++++++++++++:8> (k2,22)
```

发送自 Windows 10 版邮件<https://go.microsoft.com/fwlink/?LinkId=550986>应用

发件人: [hidden email]<mailto:[hidden email]>
发送时间: 2020年6月1日 22:22
收件人: user-zh<mailto:[hidden email]>
主题: 关于使用IngressTime，window过小的问题

Flink 1.10，windows 10 flink api验证

代码如下
```

import org.apache.flink.api.java.functions.KeySelector;
import org.apache.flink.api.java.tuple.Tuple2;
import org.apache.flink.streaming.api.TimeCharacteristic;
import org.apache.flink.streaming.api.datastream.DataStreamSource;
import org.apache.flink.streaming.api.environment.StreamExecutionEnvironment;
import org.apache.flink.streaming.api.functions.IngestionTimeExtractor;
import org.apache.flink.streaming.api.windowing.assigners.TumblingEventTimeWindows;
import org.apache.flink.streaming.api.windowing.assigners.TumblingProcessingTimeWindows;
import org.apache.flink.streaming.api.windowing.time.Time;

import java.util.ArrayList;
import java.util.List;

public class KeyedStreamJob {
    public static void main(String[] args) throws Exception {
        StreamExecutionEnvironment env = StreamExecutionEnvironment.getExecutionEnvironment();
        env.setStreamTimeCharacteristic(TimeCharacteristic.IngestionTime);
//        env.setParallelism(3);

        Tuple2<String, Integer> item = null;
        List<Tuple2<String, Integer>> items = new ArrayList<>();
        item = new Tuple2<>("k1", 1);
        items.add(item);
        item = new Tuple2<>("k3", 10);
        items.add(item);
        item = new Tuple2<>("k1", 10);
        items.add(item);
        item = new Tuple2<>("k2", 2);
        items.add(item);
        item = new Tuple2<>("k1", 11);
        items.add(item);
        item = new Tuple2<>("k2", 20);
        items.add(item);
        DataStreamSource<Tuple2<String, Integer>> streamSource = env.fromCollection(items);
        streamSource
                //by 1
                //.assignTimestampsAndWatermarks(new IngestionTimeExtractor())
                .keyBy(new KeySelector<Tuple2<String, Integer>, String>() {
            @Override
            public String getKey(Tuple2<String, Integer> value) throws Exception {
                return value.f0;
            }
        })
                .window(TumblingEventTimeWindows.of(Time.milliseconds(10L)))
                .sum(1)
                .print("+++++++++++++++++++++++++++");

        env.execute("keyedSteamJob");
    }
}

```
输出
```
+++++++++++++++++++++++++++:1> (k3,10)
+++++++++++++++++++++++++++:2> (k1,1)
+++++++++++++++++++++++++++:8> (k2,22)
+++++++++++++++++++++++++++:2> (k1,21)
```
如果把

window(TumblingEventTimeWindows.of(Time.milliseconds(10L)))
改成

.window(TumblingEventTimeWindows.of(Time.seconds(10L)))
输出
```
+++++++++++++++++++++++++++:8> (k2,22)
+++++++++++++++++++++++++++:1> (k3,10)
+++++++++++++++++++++++++++:2> (k1,22)
```
两次不同的windows窗口，第一次输出对于key=‘k1‘不聚集，第二次输出聚集

为什么会这样，如何验证怎么样的过程处理流程导致这样的结果区别

如果k1=1已经在ValueState中（2>(k1,1)），
那么再次输出时currentKey=k1时，这个时候ValueState的value是1，那么输出应该是10+11+1，而不是10+11；


如果window改成1秒也是按照正常结果输出





发送自 Windows 10 版邮件<https://go.microsoft.com/fwlink/?LinkId=550986>应用

如果说window的10毫秒的状态ValueState被超时逐出了，可以理解。但不带window的聚合操作是否意味着所有的key的
ValueState都存在与StateBackend中，是否会无限制增长，超过集群的一些限制，比如内存、slot等会怎么样。
即使ValueState在window中被逐出，但代码中未明确指定TimeService和逐出器。

如果假设相同的key在被处理时在时空上间隔足够远，不带windows和带windows的输出结果是否还会不同？


发送自 Windows 10 版邮件<https://go.microsoft.com/fwlink/?LinkId=550986>应用

发件人: [hidden email]<mailto:[hidden email]>
发送时间: 2020年6月1日 22:27
收件人: [hidden email]<mailto:[hidden email]>
主题: 回复: 关于使用IngressTime，window过小的问题

如果不使用window，那么输出会按照ValueState的存量的key的ValueState聚合
输出
```
+++++++++++++++++++++++++++:2> (k1,1)
+++++++++++++++++++++++++++:1> (k3,10)
+++++++++++++++++++++++++++:2> (k1,11)
+++++++++++++++++++++++++++:8> (k2,2)
+++++++++++++++++++++++++++:2> (k1,22)
+++++++++++++++++++++++++++:8> (k2,22)
```

发送自 Windows 10 版邮件<https://go.microsoft.com/fwlink/?LinkId=550986>应用

发件人: [hidden email]<mailto:[hidden email]>
发送时间: 2020年6月1日 22:22
收件人: user-zh<mailto:[hidden email]>
主题: 关于使用IngressTime，window过小的问题

Flink 1.10，windows 10 flink api验证

代码如下
```

import org.apache.flink.api.java.functions.KeySelector;
import org.apache.flink.api.java.tuple.Tuple2;
import org.apache.flink.streaming.api.TimeCharacteristic;
import org.apache.flink.streaming.api.datastream.DataStreamSource;
import org.apache.flink.streaming.api.environment.StreamExecutionEnvironment;
import org.apache.flink.streaming.api.functions.IngestionTimeExtractor;
import org.apache.flink.streaming.api.windowing.assigners.TumblingEventTimeWindows;
import org.apache.flink.streaming.api.windowing.assigners.TumblingProcessingTimeWindows;
import org.apache.flink.streaming.api.windowing.time.Time;

import java.util.ArrayList;
import java.util.List;

public class KeyedStreamJob {
    public static void main(String[] args) throws Exception {
        StreamExecutionEnvironment env = StreamExecutionEnvironment.getExecutionEnvironment();
        env.setStreamTimeCharacteristic(TimeCharacteristic.IngestionTime);
//        env.setParallelism(3);

        Tuple2<String, Integer> item = null;
        List<Tuple2<String, Integer>> items = new ArrayList<>();
        item = new Tuple2<>("k1", 1);
        items.add(item);
        item = new Tuple2<>("k3", 10);
        items.add(item);
        item = new Tuple2<>("k1", 10);
        items.add(item);
        item = new Tuple2<>("k2", 2);
        items.add(item);
        item = new Tuple2<>("k1", 11);
        items.add(item);
        item = new Tuple2<>("k2", 20);
        items.add(item);
        DataStreamSource<Tuple2<String, Integer>> streamSource = env.fromCollection(items);
        streamSource
                //by 1
                //.assignTimestampsAndWatermarks(new IngestionTimeExtractor())
                .keyBy(new KeySelector<Tuple2<String, Integer>, String>() {
            @Override
            public String getKey(Tuple2<String, Integer> value) throws Exception {
                return value.f0;
            }
        })
                .window(TumblingEventTimeWindows.of(Time.milliseconds(10L)))
                .sum(1)
                .print("+++++++++++++++++++++++++++");

        env.execute("keyedSteamJob");
    }
}

```
输出
```
+++++++++++++++++++++++++++:1> (k3,10)
+++++++++++++++++++++++++++:2> (k1,1)
+++++++++++++++++++++++++++:8> (k2,22)
+++++++++++++++++++++++++++:2> (k1,21)
```
如果把

window(TumblingEventTimeWindows.of(Time.milliseconds(10L)))
改成

.window(TumblingEventTimeWindows.of(Time.seconds(10L)))
输出
```
+++++++++++++++++++++++++++:8> (k2,22)
+++++++++++++++++++++++++++:1> (k3,10)
+++++++++++++++++++++++++++:2> (k1,22)
```
两次不同的windows窗口，第一次输出对于key=‘k1‘不聚集，第二次输出聚集

为什么会这样，如何验证怎么样的过程处理流程导致这样的结果区别

如果k1=1已经在ValueState中（2>(k1,1)），
那么再次输出时currentKey=k1时，这个时候ValueState的value是1，那么输出应该是10+11+1，而不是10+11；


如果window改成1秒也是按照正常结果输出





发送自 Windows 10 版邮件<https://go.microsoft.com/fwlink/?LinkId=550986>应用

[chaojianok]

把 Time.milliseconds(10L) 改成 Time.seconds(10L) 后，其实是改变了时间窗口的大小，这会使以前在同一个窗口的数据现在被分在了两个窗口里，而聚合的时候是按照窗口进行聚合的，所以结果变了。















在 2020-06-01 22:41:14，"[hidden email]" <[hidden email]> 写道：

>如果说window的10毫秒的状态ValueState被超时逐出了，可以理解。但不带window的聚合操作是否意味着所有的key的
>ValueState都存在与StateBackend中，是否会无限制增长，超过集群的一些限制，比如内存、slot等会怎么样。
>即使ValueState在window中被逐出，但代码中未明确指定TimeService和逐出器。
>
>如果假设相同的key在被处理时在时空上间隔足够远，不带windows和带windows的输出结果是否还会不同？
>
>
>发送自 Windows 10 版邮件<https://go.microsoft.com/fwlink/?LinkId=550986>应用
>
>发件人: [hidden email]<mailto:[hidden email]>
>发送时间: 2020年6月1日 22:27
>收件人: [hidden email]<mailto:[hidden email]>
>主题: 回复: 关于使用IngressTime，window过小的问题
>
>如果不使用window，那么输出会按照ValueState的存量的key的ValueState聚合
>输出
>```
>+++++++++++++++++++++++++++:2> (k1,1)
>+++++++++++++++++++++++++++:1> (k3,10)
>+++++++++++++++++++++++++++:2> (k1,11)
>+++++++++++++++++++++++++++:8> (k2,2)
>+++++++++++++++++++++++++++:2> (k1,22)
>+++++++++++++++++++++++++++:8> (k2,22)
>```
>
>发送自 Windows 10 版邮件<https://go.microsoft.com/fwlink/?LinkId=550986>应用
>
>发件人: [hidden email]<mailto:[hidden email]>
>发送时间: 2020年6月1日 22:22
>收件人: user-zh<mailto:[hidden email]>
>主题: 关于使用IngressTime，window过小的问题
>
>Flink 1.10，windows 10 flink api验证
>
>代码如下
>```
>
>import org.apache.flink.api.java.functions.KeySelector;
>import org.apache.flink.api.java.tuple.Tuple2;
>import org.apache.flink.streaming.api.TimeCharacteristic;
>import org.apache.flink.streaming.api.datastream.DataStreamSource;
>import org.apache.flink.streaming.api.environment.StreamExecutionEnvironment;
>import org.apache.flink.streaming.api.functions.IngestionTimeExtractor;
>import org.apache.flink.streaming.api.windowing.assigners.TumblingEventTimeWindows;
>import org.apache.flink.streaming.api.windowing.assigners.TumblingProcessingTimeWindows;
>import org.apache.flink.streaming.api.windowing.time.Time;
>
>import java.util.ArrayList;
>import java.util.List;
>
>public class KeyedStreamJob {
>    public static void main(String[] args) throws Exception {
>        StreamExecutionEnvironment env = StreamExecutionEnvironment.getExecutionEnvironment();
>        env.setStreamTimeCharacteristic(TimeCharacteristic.IngestionTime);
>//        env.setParallelism(3);
>
>        Tuple2<String, Integer> item = null;
>        List<Tuple2<String, Integer>> items = new ArrayList<>();
>        item = new Tuple2<>("k1", 1);
>        items.add(item);
>        item = new Tuple2<>("k3", 10);
>        items.add(item);
>        item = new Tuple2<>("k1", 10);
>        items.add(item);
>        item = new Tuple2<>("k2", 2);
>        items.add(item);
>        item = new Tuple2<>("k1", 11);
>        items.add(item);
>        item = new Tuple2<>("k2", 20);
>        items.add(item);
>        DataStreamSource<Tuple2<String, Integer>> streamSource = env.fromCollection(items);
>        streamSource
>                //by 1
>                //.assignTimestampsAndWatermarks(new IngestionTimeExtractor())
>                .keyBy(new KeySelector<Tuple2<String, Integer>, String>() {
>            @Override
>            public String getKey(Tuple2<String, Integer> value) throws Exception {
>                return value.f0;
>            }
>        })
>                .window(TumblingEventTimeWindows.of(Time.milliseconds(10L)))
>                .sum(1)
>                .print("+++++++++++++++++++++++++++");
>
>        env.execute("keyedSteamJob");
>    }
>}
>
>```
>输出
>```
>+++++++++++++++++++++++++++:1> (k3,10)
>+++++++++++++++++++++++++++:2> (k1,1)
>+++++++++++++++++++++++++++:8> (k2,22)
>+++++++++++++++++++++++++++:2> (k1,21)
>```
>如果把
>
>window(TumblingEventTimeWindows.of(Time.milliseconds(10L)))
>改成
>
>.window(TumblingEventTimeWindows.of(Time.seconds(10L)))
>输出
>```
>+++++++++++++++++++++++++++:8> (k2,22)
>+++++++++++++++++++++++++++:1> (k3,10)
>+++++++++++++++++++++++++++:2> (k1,22)
>```
>两次不同的windows窗口，第一次输出对于key=‘k1‘不聚集，第二次输出聚集
>
>为什么会这样，如何验证怎么样的过程处理流程导致这样的结果区别
>
>如果k1=1已经在ValueState中（2>(k1,1)），
>那么再次输出时currentKey=k1时，这个时候ValueState的value是1，那么输出应该是10+11+1，而不是10+11；
>
>
>如果window改成1秒也是按照正常结果输出
>
>
>
>
>
>发送自 Windows 10 版邮件<https://go.microsoft.com/fwlink/?LinkId=550986>应用
>
>