下面sql 语句 怎么用group by将相同的数据合并？
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悬赏园豆：30
[已解决问题]
解决于 2016-06-02 18:18
select B.EQ_IMPORTANTPARTLCMID, D.SORT_NAME,
c.equipment_name,
A.KEYSPARENAME ,
decode(A.ENGINELECTRICSORTCODE,'1','机械','2','电气','3','其他')ENGINELECTRICSORTCODE,
A.LIFECYCLE,
C.site_code, E.zcs,RWWCL,
E.YWC,E.WWC,E.YYH,WYH,WJCYH,
decode( A.taskexecstatus,'0','未完成','1','已完成') as taskexecstatus
from T_EQ_IMPORTANTPARTLCM A
left join T_EQ_IMPORTANTPARTLCMTASK B on A.EQ_IMPORTANTPARTLCMID=b.eq_importantpartlcmid
left join T_EQ_BA_EQUIPMENT C on A.eq_ba_equipment_id=C.t_eq_ba_equipment_id
left join T_EQ_BA_EQUIPMENT_SORT D on D.t_eq_ba_equipment_sort_id=C.section_id
left join (select t.eq_importantpartlcmid ,
count(t.eq_importantpartlcmid) zcs,
round((count(t.CHECKTIME)/count(t.eq_importantpartlcmid))*100,2) RWWCL,
count(t.CHECKTIME) YWC,
m.WWC, count(t.eq_importantpartlcmid)-E.WYH as WJCYH ,
count(t.eq_importantpartlcmtaskid) YYH,
E.WYH
from T_EQ_IMPORTANTPARTLCMTASK t,
(select eq_importantpartlcmid, count(*) WWC from T_EQ_IMPORTANTPARTLCMTASK where CHECKTIME is null group by eq_importantpartlcmid ) m,
(select eq_importantpartlcmid, count(*) WYH from T_EQ_IMPORTANTPARTLCMTASK where FOUNDHIDDENTIME is null group by eq_importantpartlcmid ) E
where t.eq_importantpartlcmid=e.eq_importantpartlcmid and e.eq_importantpartlcmid=t.eq_importantpartlcmid
group by t.eq_importantpartlcmid,m.WWC,E.WYH) E on e.eq_importantpartlcmid=A.eq_importantpartlcmid where B.EQ_IMPORTANTPARTLCMID is not null
以上是sql语句，现在需要按照c.equipment_name和A.KEYSPARENAME 两个字段分组要怎么写
最佳答案
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如果是Oracle的话，你可以用分析函数试试
提示一下：row_number() over(partition by xxx )
举例，根据部门分组,并按工资排序
select e.* , row_number() over(partition by deptno order by sal ) rn from emp e;
得到每个部门工资最高的两个员工信息：
select * from (select e.*, row_number() over(partition by deptno order by sal desc)rn from emp e) where rn <=2
希望这么举例，能帮到你。
收获园豆：30
虽然已经解决了，不过很谢谢你的回答。分给你了
其他回答(1)
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你的意思是根据c.equipment_name和A.KEYSPARENAME两个字段 去重？
差不多这意思，假如以上sql查出的数据为第一条1 2 3 4 5 6，第二条1 2 3 4 5 6。1 2为c.equipment_name、A.KEYSPARENAME，现在要将两条数据合为一条 用 group by 怎么写
