1

假设我有这个字符串（它是一个 JPQL 查询 - 使用实体类正确替换了表名和列名）

//The actual string is auto-generated, but this is just an example:
String sql =
 "select la.laNo,la.status " +
 "from LA la " +
 "where (la.cc,la.laNo) in (" +
 "select lap.cc,lap.laNo " +
 "from LAP lap " +
 "where lap.paNo = '145'" +
 ")";

当我尝试这样做时：

Query q = org.eclipse.persistence.internal.jpa.EntityManagerImpl.createQuery(sql);

我得到了例外（为简洁起见，一些行已被删除）：

Exception Description: Syntax error parsing the query [select la.laNo,la.status from LA la where (la.cc,la.laNo) in (select lap.cc,lap.laNo from LAP lap where lap.paNo = '145')], line 1, column 48: syntax error at [,].
Internal Exception: MismatchedTokenException(81!=84)
        at org.eclipse.persistence.internal.jpa.EntityManagerImpl.createQuery(EntityManagerImpl.java:1375)
Caused by: Exception [EclipseLink-8024] (Eclipse Persistence Services - 2.1.2.v20101206-r8635): org.eclipse.persistence.exceptions.JPQLException
Internal Exception: MismatchedTokenException(81!=84)
        at org.eclipse.persistence.exceptions.JPQLException.syntaxErrorAt(JPQLException.java:362)
        at org.eclipse.persistence.internal.jpa.parsing.jpql.JPQLParser.handleRecognitionException(JPQLParser.java:301)
     ...at org.eclipse.persistence.internal.jpa.parsing.jpql.antlr.JPQLParser.arithmeticPrimary(JPQLParser.java:17303)
     ...at org.eclipse.persistence.internal.jpa.parsing.jpql.JPQLParser.parse(JPQLParser.java:130)
     ...at org.eclipse.persistence.internal.jpa.EJBQueryImpl.buildEJBQLDatabaseQuery(EJBQueryImpl.java:207)
     ...at org.eclipse.persistence.internal.jpa.EJBQueryImpl.<init>(EJBQueryImpl.java:134)
     ...at org.eclipse.persistence.internal.jpa.EntityManagerImpl.createQuery(EntityManagerImpl.java:1373)
        ... 13 more
Caused by: MismatchedTokenException(81!=84)
        at org.eclipse.persistence.internal.libraries.antlr.runtime.BaseRecognizer.mismatch(Unknown Source)
        at org.eclipse.persistence.internal.libraries.antlr.runtime.BaseRecognizer.match(Unknown Source)
        at org.eclipse.persistence.internal.jpa.parsing.jpql.antlr.JPQLParser.arithmeticPrimary(JPQLParser.java:17279)
        ... 32 more

SQL 在语法上是正确的，并且当我在数据库管理器中运行它时可以正确返回结果。那么问题可能出在哪里？

更新：
做了一些测试，显然，这有效：

String sql =
 "select la.laNo,la.status " +
 "from LA la " +
 "where la.cc in (" +
 "select lap.cc " +
 "from LAP lap " +
 "where lap.paNo = '145'" +
 ") " +
 "and la.laNo in (" +
 "select lap1.laNo " +
 "from LAP lap1 " +
 "where lap1.paNo = '145'" +
 ")";

为什么不选择多列？

4

2 回答 2

1

JPQL 不支持带有 IN 的数组，但现在 EclipseLink 2.5 支持这一点。

见， http://java-persistence-performance.blogspot.com/2013/06/eclipselink-supports-hql-and-several.html

于 2013-07-15T14:25:17.207 回答
0

createQuery()期望 JPQL 查询作为参数，而不是 SQL 查询。用于createNativeQuery()传递和执行 SQL 查询。

于 2013-07-14T08:30:39.753 回答