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我有一个使用EclipseLink 2.7.5的带有构造函数表达式的 JPQL 查询。

它失败了：

Caused by: Exception [EclipseLink-0] (Eclipse Persistence Services - 2.7.5.v20191016-ea124dd158): org.eclipse.persistence.exceptions.JPQLException
Exception Description: Syntax error parsing [
            SELECT NEW net.bbstats.dto.PlayerStatDto(
              gl.playerId,
              gl.seasonStartYear,
              gl.lastName,
              gl.firstName,
              gl.incognito,
              COUNT(DISTINCT gl.gameId) AS games,
              SUM(CASE WHEN gl.hasPlayed = FALSE THEN 1 ELSE 0 END) AS gamesplayed,
              SUM(CASE WHEN gl.starter = TRUE THEN 1 ELSE 0 END) AS gamesstarted,
              SUM(gl.threePointersMade),
              SUM(gl.freeThrowsMade),
              SUM(gl.freeThrowsAttempted),
              SUM(gl.personalFouls),
              SUM(CASE WHEN gl.personalFouls >= 5 THEN 1 ELSE 0 END),
              SUM(gl.points) AS points,
              MAX(gl.points),
              MIN(gl.points)
            )
            FROM GameLog gl
            WHERE gl.roundId = :roundId AND gl.groupCode = :groupCode AND gl.rosterId IN :rosterIds
            GROUP BY
              gl.playerId,
              gl.seasonStartYear,
              gl.lastName,
              gl.firstName,
              gl.incognito
            ORDER BY gamesstarted DESC, games DESC, points DESC, gl.lastName, gl.firstName
        ]. 
[140, 141] The constructor expression has two constructor items ('COUNT(DISTINCT gl.gameId)' and 'AS games') that are not separated by a comma.
[204, 205] The constructor expression has two constructor items ('SUM(CASE WHEN gl.hasPlayed = FALSE THEN 1 ELSE 0 END)' and 'AS gamesplayed') that are not separated by a comma.
[271, 272] The constructor expression has two constructor items ('SUM(CASE WHEN gl.starter = TRUE THEN 1 ELSE 0 END)' and 'AS gamesstarted') that are not separated by a comma.
[462, 463] The constructor expression has two constructor items ('SUM(gl.points)' and 'AS points') that are not separated by a comma.
[141, 149]  The expression is invalid, which means it does not follow the JPQL grammar.
[205, 219]  The expression is invalid, which means it does not follow the JPQL grammar.
[272, 287]  The expression is invalid, which means it does not follow the JPQL grammar.
[463, 472]  The expression is invalid, which means it does not follow the JPQL grammar.
    at org.eclipse.persistence.internal.jpa.jpql.HermesParser.buildException(HermesParser.java:157)
    at org.eclipse.persistence.internal.jpa.jpql.HermesParser.validate(HermesParser.java:336)
    at org.eclipse.persistence.internal.jpa.jpql.HermesParser.populateQueryImp(HermesParser.java:280)
    at org.eclipse.persistence.internal.jpa.jpql.HermesParser.buildQuery(HermesParser.java:165)
    at org.eclipse.persistence.internal.jpa.EJBQueryImpl.buildEJBQLDatabaseQuery(EJBQueryImpl.java:142)
    at org.eclipse.persistence.internal.jpa.JPAQuery.processJPQLQuery(JPAQuery.java:225)
    at org.eclipse.persistence.internal.jpa.JPAQuery.prepare(JPAQuery.java:186)
    at org.eclipse.persistence.queries.DatabaseQuery.prepareInternal(DatabaseQuery.java:631)
    at org.eclipse.persistence.internal.sessions.AbstractSession.processJPAQuery(AbstractSession.java:4502)
    at org.eclipse.persistence.internal.sessions.AbstractSession.processJPAQueries(AbstractSession.java:4462)
    at org.eclipse.persistence.internal.sessions.DatabaseSessionImpl.initializeDescriptors(DatabaseSessionImpl.java:645)
    at org.eclipse.persistence.internal.sessions.DatabaseSessionImpl.postConnectDatasource(DatabaseSessionImpl.java:868)
    at org.eclipse.persistence.internal.sessions.DatabaseSessionImpl.loginAndDetectDatasource(DatabaseSessionImpl.java:811)
    at org.eclipse.persistence.internal.jpa.EntityManagerFactoryProvider.login(EntityManagerFactoryProvider.java:256)
    at org.eclipse.persistence.internal.jpa.EntityManagerSetupImpl.deploy(EntityManagerSetupImpl.java:772)
    ... 201 more

这适用于休眠。我认为这将是标准的 JPA 2.2 甚至 JPA 2.1 在AS之后放置一个子句SELECT ...，以便您可以重用ORDER BYetc. 子句的标识符。

问：

怎么了？它是 EclipseLink 错误/缺失的功能吗？它不应该那样工作吗，因为 Hibernate 在这里有一些额外的肉？

如何在构造函数表达式中解决这个问题（无需重复ORDER BY子句中的表达式）？

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1 回答 1

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那确实是无效的 JPQL，因为select_item不是constructor_item. 以下是取自 JPA 2.2 规范的 SELECT 子句的语法：

select_clause ::= SELECT [DISTINCT] select_item {, select_item}* 
select_item ::= select_expression [ [AS] result_variable]
select_expression ::=
  single_valued_path_expression | 
  scalar_expression |
  aggregate_expression |
  identification_variable | 
  OBJECT(identification_variable) |
  constructor_expression
constructor_expression ::= 
  NEW constructor_name ( constructor_item {, constructor_item}* )

constructor_item ::=
  single_valued_path_expression |
  scalar_expression |
  aggregate_expression |
  identification_variable
aggregate_expression ::= 
  { AVG | MAX | MIN | SUM} ([DISTINCT] state_valued_path_expression) |
  COUNT ([DISTINCT] identification_variable | state_valued_path_expression
    |single_valued_object_path_expression) |
  function_invocation

恐怕在这种情况下没有避免重复的解决方法。

于 2020-01-11T08:53:03.987 回答