高分，在oracle中 sql转hql

高分请教高手，在oracle中 sql转hql

Java code

select tt.date1, count(tt.id)  from (select t.user_id id,               to_char(t.login_time, 'yyyy-mm') date1          from v2_user_login_log t         group by to_char(t.login_time, 'yyyy-mm'), t.user_id        having count(*) > 5) tt group by date1 order by date1 desc

[解决办法]
你这句话没用到特殊语法，应该没区别的。直接写就行了，

我自己是这么用的，不知道理解错了没。

<!-- 查询统计数据：属性值 -->
<select id="getStatisticsDataByOrgIndprocode2" parameterType="java.lang.String" resultMap="dataViewMap">
select * from 
(
${selectClause}
where indCode = #{indCode} and RangeMinDate &gt;= to_date(#{beginDate},'yyyymmdd') and RangeMinDate &lt;= to_date(#{endDate} ,'yyyymmdd')
<if test="indProCode != null"> and IndProCode=#{indProCode}</if>
<if test="province != null"> and Province =#{province}</if>
<if test="cityp != null"> and Cityp =#{cityp}</if>
<if test="region != null"> and Region =#{region}</if>
<if test="orgCode != null"> and OrgCode =#{orgCode}</if>
<if test="deptCode != null"> and DeptCode =#{deptCode}</if>
${groupByClause}
) ${pivotClause}
</select>
[解决办法]
是HIVE-QL吗？

SQL code

select tt.date1, count(tt.id)  from (select t.user_id id,               SUBSTR(t.login_time,1,7) date1          from v2_user_login_log t         group by SUBSTR(t.login_time,1,7), t.user_id        having count(*) > 5) tt group by date1 order by date1 desc 

 

查看更多 下一篇