【问题标题】:Error: org.hibernate.exception.SQLGrammarException: could not extract ResultSet and dateAccessed is unknown column错误:org.hibernate.exception.SQLGrammarException:无法提取 ResultSet 并且 dateAccessed 是未知列
【发布时间】:2016-12-26 10:44:41
【问题描述】:

我在 Spring MVC JAVA 中执行下面的查询

if (currentFireTime != null && previousFireTime != null) {
        timePart = " sal.dateAccessed between '" + previousFireTime + "' and '" + currentFireTime + "' and ";
    }
    String queryString = " select count(*) as COUNT FROM LogData sal where " + timePart + " sal.userSigned = false and sal.batchMetaDataId = "+batchId;
    SQLQuery sq =sessionFactory.getCurrentSession().createSQLQuery(queryString);
    sq.addScalar("COUNT", IntegerType.INSTANCE);
    sq.setResultTransformer(AliasToEntityMapResultTransformer.INSTANCE);
    List<Map<String,Object>> resultSet=sq.list();
    return (Integer)resultSet.get(0).get("COUNT");

当执行看起来像这样并将它放在 Mysql 中它工作正常。但我得到如下所示的错误。

select count(*) as COUNT FROM LogData sal where sal.dateAccessed between '2016-12-26 00:00:00.618' and '2016-12-26 01:00:00.618' and  sal.userSigned = false and sal.batchMetaDataId = 86

显示错误

    org.hibernate.exception.SQLGrammarException: could not extract ResultSet
at org.hibernate.exception.internal.SQLExceptionTypeDelegate.convert(SQLExceptionTypeDelegate.java:82)
at org.hibernate.exception.internal.StandardSQLExceptionConverter.convert(StandardSQLExceptionConverter.java:49)
at org.hibernate.engine.jdbc.spi.SqlExceptionHelper.convert(SqlExceptionHelper.java:125)
at org.hibernate.engine.jdbc.spi.SqlExceptionHelper.convert(SqlExceptionHelper.java:110)

我需要在显示的代码中进行哪些更改?

【问题讨论】:

    标签: java mysql spring hibernate


    【解决方案1】:

    您以错误的方式获取计数。试试这个:

    resultSet.getInt("COUNT");
    

    【讨论】:

    • 我在第二行出现错误并给出错误 sal.dateAccessed is unknown column。
    • @daisy 是第 2 行吗??
    • @SachinSarawgi 第二行是 timePart = "sal.dateAccessed between '" + previousFireTime + "' and '" + currentFireTime + "' and ";
    • @daisy 你能在没有sal. 的情况下尝试相同的查询吗?因为它只有一张桌子,所以他们不会混淆。
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