【问题标题】:Cassandra select by timeuuidCassandra 按 timeuuid 选择
【发布时间】:2018-05-13 02:56:29
【问题描述】:

Spring-boot-data-cassandra

CQL:

CREATE TABLE device_data (
    device_id uuid,
    time timeuuid,
    unit text,
    value double,
    PRIMARY KEY (device_id, time)
)

存储库:

public interface DeviceDataRepository extends CassandraRepository<DeviceData> {

    @Query("SELECT * FROM device_data WHERE device_id = ?0 AND time > ?1")
    List<DeviceData> findByDeviceIdAndTime(UUID deviceId, Date from);
}

用法:

    Calendar calendar = new GregorianCalendar(1990, 1, 1);
    List<DeviceData> pump1Data = deviceDataService.findByDeviceIdAndFrom(UUID.fromString(pumpid), calendar.getTime());

这给了我以下错误:

Invalid INTEGER constant (633826800000) for "time" of type timeuuid; nested exception is com.datastax.driver.core.exceptions.InvalidQueryException: Invalid INTEGER constant (633826800000) for "time" of type timeuuid

我做错了什么?

【问题讨论】:

    标签: spring-boot cassandra timeuuid


    【解决方案1】:

    修复了问题:

    @Query("SELECT * FROM device_data WHERE device_id = ?0 AND time > ?1")
    List<DeviceData> findByDeviceIdAndTime(UUID deviceId, UUID from);
    

    设备数据:

    ...
    private UUID time;
    ...
    

    控制器:

        List<DeviceData> pumpData = deviceDataService.findByDeviceIdAndFrom(UUID.fromString(id), calendar.getTime());
    

    【讨论】:

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