java - 如何将多条消息转换为一条消息？

Question

每天我们都会在 Student-Topic-In 中收到以下消息

Message 1: {"StudentID": "1", "StudentName":"aaa","fatherName":"aaa1",  "class":"1"}
Message 2: {"StudentID": "2", "StudentName":"bbb","fatherName":"bbb1",  "class":"1"}
Message 3: {"StudentID": "3", "StudentName":"ccc","fatherName":"ccc1",  "class":"2"}
Message 4: {"StudentID": "4", "StudentName":"ddd","fatherName":"ddd1",  "class":"2"}
Message 5: {"StudentID": "5", "StudentName":"eee","fatherName":"eee1",  "class":"2"}

一天结束时(一天一次)基于每个类(class)，我们必须合并所有消息并以以下格式发布到“Student-Topic-Out”。

Message 1:{"Class":"1"
          {"StudentID": "1", "StudentName":"aaa","fatherName":"aaa1"},
          {"StudentID": "2", "StudentName":"bbb","fatherName":"bbb1"}
       }
Message 2:{"Class":"2" 
          {"StudentID": "3", "StudentName":"ccc","fatherName":"ccc1"},
          {"StudentID": "4", "StudentName":"ddd","fatherName":"ddd1"},
          {"StudentID": "5", "StudentName":"eee","fatherName":"eee1"}
       }

我尝试了以下方法，但不知道如何创建没有类(class)名称的学生列表？

KStream<String, Object> sampleStream = builder.stream("Student-Topic-in");
    sampleStream
            .filter((k, v) -> v != null)
            .mapValues(v -> (Student) v)
            .groupBy((k, v) -> KeyValue.pair(v.getClass_name(), v))
            .windowedBy(TimeWindows.of(5000))
            //I am not sure how to create a student list without Classname
            .aggregate(Student::new, (k, v, list) -> (Student)list.add((Student)v)

能否请您告诉我，如何构造输出 JSON 消息 Kafka Streams？

Answer 1

你可以做一个 KStream.groupBy(...).windowedBy().aggregate().mapValues使用“类”属性进行分组。

在Aggregator()你可以组装一个List你变成的学生 JSON在 mapValues()