Question

我尝试使用Flink的SQL API从地图访问密钥。它在线程＆＃34; main＆＃34;中的错误Exception失败。 org.apache.flink.table.api.TableException：不支持Type：ANY 请告诉我如何解决它。这是我的活动课

 public class EventHolder {

    private Map<String,String> event;

    public Map<String, String> getEvent() {
        return event;
    }

    public void setEvent(Map<String, String> event) {
        this.event = event;
    }

}

这是提交flink作业的主要类

public class MapTableSource {
public static void main(String[] args) throws Exception {
    final StreamExecutionEnvironment env = StreamExecutionEnvironment.getExecutionEnvironment();
    DataStream<EventHolder> mapEventStream = env.fromCollection(getMaps());

    // register a table and use SQL
    StreamTableEnvironment tableEnv = TableEnvironment.getTableEnvironment(env);
    tableEnv.registerDataStream("mapEvent", mapEventStream); 
    //tableEnv.registerFunction("orderSizeType", new OrderSizeType());

    Table alerts = tableEnv.sql(
            "select event['key'] from mapEvent ");

    DataStream<String> alertStream = tableEnv.toAppendStream(alerts, String.class);

    alertStream.filter(new FilterFunction<String>() {
        private static final long serialVersionUID = -2438621539037257735L;

        @Override
        public boolean filter(String value) throws Exception {
            System.out.println("Key value is:"+value);
            return value!=null;
        }
    });


    env.execute("map-tablsource-job");
}


private static List<EventHolder> getMaps(){
    List<EventHolder> list = new ArrayList<>();
    for(int i=0;i<5;i++){
        EventHolder holder = new EventHolder();
        Map<String,String> map = new HashMap<>();
        map.put("key", "value");
        holder.setEvent(map);
        list.add(holder);
    }
    return list;
}
}

当我运行它时，我得到了异常

Exception in thread "main" org.apache.flink.table.api.TableException: Type is not supported: ANY
at org.apache.flink.table.api.TableException$.apply(exceptions.scala:53)
at org.apache.flink.table.calcite.FlinkTypeFactory$.toTypeInfo(FlinkTypeFactory.scala:341)
at org.apache.flink.table.plan.logical.LogicalRelNode$$anonfun$12.apply(operators.scala:530)
at org.apache.flink.table.plan.logical.LogicalRelNode$$anonfun$12.apply(operators.scala:529)
at scala.collection.TraversableLike$$anonfun$map$1.apply(TraversableLike.scala:245)
at scala.collection.TraversableLike$$anonfun$map$1.apply(TraversableLike.scala:245)
at scala.collection.Iterator$class.foreach(Iterator.scala:742)
at scala.collection.AbstractIterator.foreach(Iterator.scala:1194)
at scala.collection.IterableLike$class.foreach(IterableLike.scala:72)
at scala.collection.AbstractIterable.foreach(Iterable.scala:54)
at scala.collection.TraversableLike$class.map(TraversableLike.scala:245)
at scala.collection.AbstractTraversable.map(Traversable.scala:104)
at org.apache.flink.table.plan.logical.LogicalRelNode.<init>(operators.scala:529)
at org.apache.flink.table.api.TableEnvironment.sql(TableEnvironment.scala:503)
at com.c.p.flink.MapTableSource.main(MapTableSource.java:25)

我使用的是flink 1.3.1

Answer 1

我认为问题在于fromCollection。由于Java限制（即类型擦除），Flink无法提取所需的类型信息。因此，您将映射视为具有SQL ANY类型的黑盒子。您可以使用tableEnv.scan("mapEvent").printSchema()验证表格的类型。您可以使用fromCollection在Types.MAP(Types.STRING, Types.STRING)中指定类型信息。

Answer 2

我通过以下方法解决了类似的问题：

//Should probably make MapVal more generic, but works for this example
public class MapVal extends ScalarFunction {
    public String eval(Map<String, String> obj, String key) {
        return obj.get(key);
    }
}

public class Car {
    private String make;
    private String model;
    private int year;
    private Map<String, String> attributes;
    //getters/setters...
}

//After registering Stream and TableEnv etc

tableEnv.registerFunction("mapval", new MapVal());

Table cars = tableEnv
                .scan("Cars")
                .select("make, model, year, attributes.mapval('name')");

Flink SQL是否支持Java Map类型？

2 个答案: