Java Hibernate org.hibernate.exception.SQLGrammarException:无法在createSQLQuery上提取ResultSet

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.

这篇关于Java Hibernate org.hibernate.exception.SQLGrammarException:无法在createSQLQuery上提取ResultSet的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！