2012-07-09 58 views
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任何人都可以解释为什么会抛出JPQL异常吗?jpql按尺寸排序错误

Query q = em.createQuery("select object(o) from SiteMember as o order by size(o.memberThread) desc"); 

memberThread refer a list. 

回答

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很可能是因为您使用的JPA实现(也是数据库供应商可能很重要),它没有针对此查询的供应商扩展。根据JPA 2.0以下类型的值可以在ORDER BY

1.A state_field_path_expression that evaluates to an orderable state field of an 
    entity or embeddable class abstract schema type designated in the SELECT clause 
    by one of the following: 
• a general_identification_variable 
• a single_valued_object_path_expression 

2.A state_field_path_expression that evaluates to the same state field of the same 
    entity or embeddable abstract schema type as a state_field_path_expression in the 
    SELECT clause 

3.A result_variable that refers to an orderable item in the SELECT clause for which 
the same result_variable has been specified. This may be the result of an 
aggregate_expression, a scalar_expression, or a state_field_path_expression in the 
SELECT clause. 

您的查询工作,例如与EclipseLink的(2.3.2)和MySQL。

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我使用Toplink和这段代码工作正常“从MemberThread选择对象(o)作为o其中size(o.replys)='0'order by o.id desc”.any解决方案请吗? – bews99 2012-07-09 07:37:23

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那一个是有效的JPQL查询。问题在于ORDER BY与非法价值相结合。很可能只是查询的排序结果很好,至少没有可移植性问题。如果这不是选项,您可以尝试遵循(不保证工作):将'size(o.memberThread)'作为ord'添加到SELECT,按ord排序,并按SiteMember中的字段进行分组。然后它会变得丑陋,因为结果还包含仅用于ORDER BY的属性。 – 2012-07-09 08:04:16

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感谢您的详细回复。 :D – bews99 2012-07-09 10:05:05