Java Hibernate org.hibernate.exception.SQLGrammarException:无法在createSQLQuery上提取ResultSet [英] Java Hibernate org.hibernate.exception.SQLGrammarException: could not extract ResultSet on createSQLQuery
问题描述
我有这种方法.
private final void updateAllTableFields(final Class clazz){
final String tableName = ((Table)clazz.getAnnotation(Table.class)).name();
final String sqlQuery = new StringBuilder("SET @ids = NULL; ")
.append("UPDATE ")
.append(tableName)
.append(' ')
.append("set activeRecord=:activeRecord ")
.append("where activeRecord=true and updateable=true ")
.append("and (SELECT @ids \\:= CONCAT_WS(',', id, @ids)); ")
.append("select @ids;")
.toString();
final Query query = session.createSQLQuery(sqlQuery)
.setParameter("activeRecord",Boolean.FALSE);
final Object idsList=query.uniqueResult();
System.out.println("idsList = " + idsList);
}
我想进行更新并返回受影响的ID,这可以正常工作.使用rawSQL可以以字符串形式返回ID,但是我无法使用Hibernate使其正常工作!
I want to do a update and also return the affected Ids this works Perfect using a rawSQL returns the id in a string fashion but i couldn't make it work using Hibernate any tip!!!
提前致以最诚挚的问候.
Thanks in advance and best regards.
更新
我需要进行更新并返回受影响的ID !!我不想做一个简单的更新.
I need to do a update and return the affected id!! I dont want to make a simple UPDATE.
您可以在此处找到原始问题pal : https://stackoverflow.com/questions/44604763/java-hibernate-tips-about-update-all-table-fields-performance
you can check it out the original question here pal: https://stackoverflow.com/questions/44604763/java-hibernate-tips-about-update-all-table-fields-performance
更新 错误是
at org.hibernate.exception.internal.SQLExceptionTypeDelegate.convert(SQLExceptionTypeDelegate.java:80)
at org.hibernate.exception.internal.StandardSQLExceptionConverter.convert(StandardSQLExceptionConverter.java:49)
at org.hibernate.engine.jdbc.spi.SqlExceptionHelper.convert(SqlExceptionHelper.java:126)
at org.hibernate.engine.jdbc.spi.SqlExceptionHelper.convert(SqlExceptionHelper.java:112)
at org.hibernate.engine.jdbc.internal.ResultSetReturnImpl.extract(ResultSetReturnImpl.java:89)
at org.hibernate.loader.Loader.getResultSet(Loader.java:2065)
at org.hibernate.loader.Loader.executeQueryStatement(Loader.java:1862)
at org.hibernate.loader.Loader.executeQueryStatement(Loader.java:1838)
at org.hibernate.loader.Loader.doQuery(Loader.java:909)
at org.hibernate.loader.Loader.doQueryAndInitializeNonLazyCollections(Loader.java:354)
at org.hibernate.loader.Loader.doList(Loader.java:2553)
at org.hibernate.loader.Loader.doList(Loader.java:2539)
at org.hibernate.loader.Loader.listIgnoreQueryCache(Loader.java:2369)
at org.hibernate.loader.Loader.list(Loader.java:2364)
at org.hibernate.loader.custom.CustomLoader.list(CustomLoader.java:353)
at org.hibernate.internal.SessionImpl.listCustomQuery(SessionImpl.java:1873)
at org.hibernate.internal.AbstractSessionImpl.list(AbstractSessionImpl.java:311)
at org.hibernate.internal.SQLQueryImpl.list(SQLQueryImpl.java:141)
at org.hibernate.internal.AbstractQueryImpl.uniqueResult(AbstractQueryImpl.java:966)
at company.nuevemil.code.finalizarEntornoDePrueba(Test.java:56)
at company.nuevemil.code.main(Test.java:27)
Caused by: com.mysql.jdbc.exceptions.jdbc4.MySQLSyntaxErrorException: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'UPDATE student set activeRecord=false,uid=1 where activeRecord=true at line 1
推荐答案
我想,您将无法以Hibernate方式制作它.
I suppose, you won't be able to make it in Hibernate fashion.
Hibernate独立于数据库.但是查询中用于初始化变量的部分(我的意思是set @ids = null;
)不能在所有关系数据库中移植,因此我不希望它位于Hibernate API中.
Hibernate is independent from a database. But the part of the query that initializes a variable (I mean set @ids = null;
) is not portable across all the relational databases so I wouldn't expect it to be in Hibernate API somewhere.
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