我有这个方法。

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);
}        

我想做一个更新并返回受影响的 IDs 这工作完美使用 rawSQL 以字符串方式返回 id 但我无法使用 Hibernate 任何提示使其工作!!!

在此先致谢并致以最诚挚的问候。

更新

我需要进行更新并返回受影响的 ID!!我不想进行简单的更新。

你可以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

最佳答案

you have to use HQL Query for bulk update. you are going write way only thing is that, you have to create HQL query for example


    Your Query Might be like this:-
    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.createQuery(sqlQuery)
                .setParameter("activeRecord",Boolean.FALSE);
        final Object idsList=query.executeUpdate();

    Example Query:
    final String tableName = ((Table)clazz.getAnnotation(Table.class)).name();
       Query qry = session.createQuery("update "+tableName+" p set p.proName=?
    where p.productId=111");
                qry.setParameter(0,"updated..");
                int res = qry.executeUpdate();

关于Java hibernate org.hibernate.exception.SQLGrammarException : could not extract ResultSet on createSQLQuery，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/44641851/