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分享问个算去年同期值的sql
表中数据如下：
日期 得分
2012-1 85
2012-2 98
2012-3 56
2012-4 66
2012-5 34
。。。
。。。
2013-1 82
2013-2 78
2013-3 29
2013-4 77
sql查4个值，日期、得分、去年同期得分、历史最高得分，希望取到的值如下：
日期 得分 去年同期得分 历史最高得分
2012-1 85
2012-2 98
2012-3 56
2012-4 66
2012-5 34
。。。
。。。
2013-1 82 85 85
2013-2 78 98 98
2013-3 29 56 29
2013-4 66 77 77
请问这个sql怎么写？
日期 得分
2012-1 85
2012-2 98
2012-3 56
2012-4 66
2012-5 34
。。。
。。。
2013-1 82
2013-2 78
2013-3 29
2013-4 77
sql查4个值，日期、得分、去年同期得分、历史最高得分，希望取到的值如下：
日期 得分 去年同期得分 历史最高得分
2012-1 85
2012-2 98
2012-3 56
2012-4 66
2012-5 34
。。。
。。。
2013-1 82 85 85
2013-2 78 98 98
2013-3 29 56 29
2013-4 66 77 77
请问这个sql怎么写？
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run-and-debug 2013-05-22
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这样可以查出来：
SELECT t1.dt,t1.score,
(SELECT t2.score FROM VALUE t2 WHERE SUBSTR(t2.dt,6,2)= SUBSTR(t1.dt,6,2)
AND SUBSTR(t2.dt,1,4)= (SUBSTR(t1.dt,1,4)-1)
) AS lastYear,
(SELECT MAX(t3.score) FROM VALUE t3 WHERE SUBSTR(t3.dt,6,2)= SUBSTR(t1.dt,6,2)
AND SUBSTR(t3.dt,1,4) < SUBSTR(t1.dt,1,4)
) AS lastYear1
FROM VALUE t1
大明湖畔_帅锅 2013-05-21
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这效果么
如果日期格式都是固定的 可以
如果是2011-1那可能要找到'-'的位置开始截取了 稍微麻烦了点
with t1 as
(
select '2011-01' dt,85 sc from dual union all
select '2011-02' dt,98 sc from dual union all
select '2011-03' dt,56 sc from dual union all
select '2011-04' dt,66 sc from dual union all
select '2012-01' dt,33 sc from dual union all
select '2012-02' dt,34 sc from dual union all
select '2012-03' dt,54 sc from dual union all
select '2012-04' dt,69 sc from dual union all
select '2013-01' dt,97 sc from dual union all
select '2013-02' dt,43 sc from dual union all
select '2013-03' dt,44 sc from dual union all
select '2013-04' dt,34 sc from dual
)
select dt,sc,s_sc,
(select max(sc) from t1 where t1.dt <= t.dt and substr(t1.dt,-2)=substr(t.dt,-2)) m_sc
from
(
select dt,sc,
case row_number() over(partition by substr(dt,-2) order by rownum)
when 1 then null
else lag(sc) over(order by substr(dt,-2),substr(dt,1,4)) end s_sc
from t1
) t
order by dt
dt sc s_sc m_sc
------------------------------------------
1 2011-01 85 85
2 2011-02 98 98
3 2011-03 56 56
4 2011-04 66 66
5 2012-01 33 85 85
6 2012-02 34 98 98
7 2012-03 54 56 56
8 2012-04 69 66 69
9 2013-01 97 33 97
10 2013-02 43 34 98
11 2013-03 44 54 56
12 2013-04 34 69 69
u010412956 2013-05-21
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历史最高得分 是怎么算的？
大明湖畔_帅锅 2013-05-21
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看lz也发了200多个贴了 基本的sql应该清楚吧 where过滤地区
然后要用上面那sql的话 可能这里要改改 转换后的月份是有0的
(select sum(sc)
from t1 t
where t.dt =
to_char(add_months(to_date(t1.dt, 'yyyy-mm'), -12), 'yyyy-mm')) 去年同期得分
u010412956 2013-05-21
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最后面加个where不就可以了么？
from t1
where 地区='北京'
charlesxu 2013-05-21
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我用你的sql执行时报错：ORA:01427-single-row subquery returns more than one row[/quote]
你相同的年月 会有多条记录，，那就这样：
with t1 as
(select '2011-01' dt, 85 sc
from dual
union all
select '2011-02' dt, 98 sc
from dual
union all
select '2011-03' dt, 56 sc
from dual
union all
select '2011-04' dt, 66 sc
from dual
union all
select '2012-01' dt, 33 sc
from dual
union all
select '2012-02' dt, 34 sc
from dual
union all
select '2012-03' dt, 54 sc
from dual
union all
select '2012-04' dt, 69 sc
from dual
union all
select '2013-01' dt, 97 sc
from dual
union all
select '2013-02' dt, 43 sc
from dual
union all
select '2013-03' dt, 44 sc
from dual
union all
select '2013-04' dt, 34 sc from dual)
select t1.dt 日期,
t1.sc 得分,
(select sum(sc)
from t1 t
where t.dt =
to_char(add_months(to_date(t1.dt, 'yyyy-mm'), -12), 'yyyy-mm')) 去年同期得分,
(select max(sc)
from t1 t
where t.dt < t1.dt
and substr(t1. dt, 6, 2) = substr(t.dt, 6, 2)) 历史最高得分
from t1;
大明湖畔_帅锅 2013-05-21
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可以直接截取 因为'-'的位置是固定的 后面最多2位
或者to_char(to_date(dt,'yyyy-mm'),'mm')就是麻烦点 效率也不行
select dt,sc,
case row_number() over(partition by substr(dt,6,2) order by rownum)
when 1 then null
else lag(sc) over(order by substr(dt,6,2)) end s_sc,
(select max(sc)
from t1
where t1.dt <= t.dt
and substr(t1.dt,6,2)=substr(t.dt,6,2)
) m_sc
from t1 t
order by dt
u010412956 2013-05-21
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我用你的sql执行时报错：ORA:01427-single-row subquery returns more than one row[/quote]
你相同的年月 会有多条记录，，那就这样：
with t1 as
(select '2011-01' dt, 85 sc
from dual
union all
select '2011-02' dt, 98 sc
from dual
union all
select '2011-03' dt, 56 sc
from dual
union all
select '2011-04' dt, 66 sc
from dual
union all
select '2012-01' dt, 33 sc
from dual
union all
select '2012-02' dt, 34 sc
from dual
union all
select '2012-03' dt, 54 sc
from dual
union all
select '2012-04' dt, 69 sc
from dual
union all
select '2013-01' dt, 97 sc
from dual
union all
select '2013-02' dt, 43 sc
from dual
union all
select '2013-03' dt, 44 sc
from dual
union all
select '2013-04' dt, 34 sc from dual)
select t1.dt 日期,
t1.sc 得分,
(select sum(sc)
from t1 t
where t.dt =
to_char(add_months(to_date(t1.dt, 'yyyy-mm'), -12), 'yyyy-mm')) 去年同期得分,
(select max(sc)
from t1 t
where t.dt < t1.dt
and substr(t1. dt, 6, 2) = substr(t.dt, 6, 2)) 历史最高得分
from t1;
大明湖畔_帅锅 2013-05-21
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写多了 第四行去掉lag后面2个参数 lag(sc) over()
charlesxu 2013-05-21
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我用你的sql执行时报错：ORA:01427-single-row subquery returns more than one row
大明湖畔_帅锅 2013-05-21
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为什么插入会选择这种格式 转换成日期也不方便 截取也会降低效率..
select dt,sc,
case row_number() over(partition by substr(dt,instr(dt,'-')+1,length(dt)) order by rownum)
when 1 then null
else lag(sc,1,null) over(order by substr(dt,instr(dt,'-')+1,length(dt))) end s_sc,
(select max(sc)
from t1
where t1.dt <= t.dt
and substr(t1.dt,instr(t1.dt,'-')+1,length(t1.dt))=substr(t.dt,instr(t.dt,'-')+1,length(t.dt))
) m_sc
from t1 t
order by dt
rabitsky 2013-05-21
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数据入库的时候控制一下格式不就可以了吗 to_char(XXX,'yyyy-mm')。
或者重新拼接一个to_char(to_date(xx||'-01','yyyy-mm-dd'),'yyyy-mm')
……
这种基本的东西，方法很多
charlesxu 2013-05-21
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现在有个问题，我的日期字段是字符型的，格式如下
2011-1
2011-2
。。。
2011-10
2011-12
请问怎么处理才能取出相同的结果？
rabitsky 2013-05-21
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借用楼上数据。
with t1 as (
select '2011-01' dt,85 sc from dual union all
select '2011-02' dt,98 sc from dual union all
select '2011-03' dt,56 sc from dual union all
select '2011-04' dt,66 sc from dual union all
select '2012-01' dt,33 sc from dual union all
select '2012-02' dt,34 sc from dual union all
select '2012-03' dt,54 sc from dual union all
select '2012-04' dt,69 sc from dual union all
select '2013-01' dt,97 sc from dual union all
select '2013-02' dt,43 sc from dual union all
select '2013-03' dt,44 sc from dual union all
select '2013-04' dt,34 sc from dual )
select dt, sc, lag(sc,1,null) over(partition by substr(dt,-2) order by dt) last_sc,
(select max(sc)
from t b
where b.dt <= a.dt
and substr(b.dt,-2) = substr(a.dt,-2)) max_sc
from t1 a
order by 1;
vanjayhsu 2013-05-21
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赞一个
u010412956 2013-05-21
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借下2#的数据：
with t1 as
(select '2011-01' dt, 85 sc
from dual
union all
select '2011-02' dt, 98 sc
from dual
union all
select '2011-03' dt, 56 sc
from dual
union all
select '2011-04' dt, 66 sc
from dual
union all
select '2012-01' dt, 33 sc
from dual
union all
select '2012-02' dt, 34 sc
from dual
union all
select '2012-03' dt, 54 sc
from dual
union all
select '2012-04' dt, 69 sc
from dual
union all
select '2013-01' dt, 97 sc
from dual
union all
select '2013-02' dt, 43 sc
from dual
union all
select '2013-03' dt, 44 sc
from dual
union all
select '2013-04' dt, 34 sc from dual)
select t1.dt 日期,
t1.sc 得分,
(select sc
from t1 t
where t.dt =
to_char(add_months(to_date(t1.dt, 'yyyy-mm'), -12), 'yyyy-mm')) 去年同期得分,
(select max(sc)
from t1 t
where t.dt < t1.dt
and substr(t1. dt, 6, 2) = substr(t.dt, 6, 2)) 历史最高得分
from t1;
大明湖畔_帅锅 2013-05-21
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如果是日期格式 里面的截取换成取值
substr(dt,-2) → to_char(dt,'mm')
substr(dt,1,4) → to_char(dt,'yyyy')
大明湖畔_帅锅 2013-05-21
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写麻烦了 不用嵌套 换成子查询就可以了
select dt,sc,
case row_number() over(partition by substr(dt,-2) order by rownum)
when 1 then null
else lag(sc) over(order by substr(dt,-2),substr(dt,1,4)) end s_sc,
(select max(sc) from t1 where t1.dt <= t.dt and substr(t1.dt,-2)=substr(t.dt,-2)) m_sc
from t1 t
order by dt
charlesxu 2013-05-21
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历史最高得分是之前所有年份当月的最高分，比如2013年7月的历史最高得分是2010、2011、2012年7月得分中最高的。
