Question

相关：How to countDistinct on multiple columns

我有一个包含许多字段的实体类，其中三个是longitude，latitude和update_time。我正在尝试添加一个@Formula字段来连接这三个字段：

@Formula("concat(longitude, latitude, update_time)")
public String fix;

然后我想将该字段用作countDistinct查询的一部分：

@SuppressWarnings( {"unchecked", "rawtypes"} )
public long getCountDistinctPositions() {
    Session session = sessionFactory.openSession();

    CriteriaBuilder criteriaBuilder = session.getCriteriaBuilder();

    CriteriaQuery criteriaQuery = criteriaBuilder.createQuery();
    Root<Position> position = criteriaQuery.from(Position.class);
    Expression fix = position.get("fix");
    Expression countDistinct = criteriaBuilder.countDistinct(fix);
    criteriaQuery.select(countDistinct);
    Query query = session.createQuery(criteriaQuery);
    Long result = (Long)query.getSingleResult();

    session.close();

    return result;
}

但我一直有例外：

java.lang.IllegalArgumentException: Unable to locate Attribute  with the the given name [fix] on this ManagedType [aaa.Position]

Answer 1

无法查询@Formula字段，它基本上是仅在执行选择查询时Hibernate将填充的合成字段。

对于您的用例，您需要执行：

SELECT DISTINCT longitude, latitude, update_time FROM Entity

从JPA查询的角度来看，您需要在所有3个单独的列上执行#countDistinct，而不是在公式列上执行@Transient public String getFix() { return this.longitude + this.latitude + this.update_time; }。

考虑公式列的最佳方法是，您已在实体上有效地翻译了以下方法

CriteriaQuery<Long> countQuery = cb.createQuery( Long.class );
Root<TheEntity> root = countQuery.from( TheEntity.class );

countQuery.select( cb.count( root.get( "id" ) ) );

Subquery<Integer> subQuery = countQuery.subquery( Integer.class );
Root<TheEntity> subRoot = subQuery.from( TheEntity.class );
subQuery.select( cb.min( subRoot.get( "id" ) ) );
subQuery.groupBy( subRoot.get( "longitude" ), 
   subRoot.get( "latitude" ), 
   subRoot.get( "updateTime" ) );

countQuery.where( root.get( "id" ).in( subQuery ) );

Long count = entityManager.createQuery( countQuery ).getSingleResult();

到查询时数据库连接的属性。

更新

# Get list of numbers
with open("/path/to/numbers.txt") as f:
    content = f.read()
    numbers = content.split()

# Handle each URL in a loop
for number in numbers:
    url = 'http://www.example.com/%s' % number
    # Do something with url

客户端代码无法识别@Formula字段

1 个答案: