利用rownumber ,关键字partition进行小范围分页
方法一：
--所有供应商对该产品最近的一次报价
with oa as
(
select a.SupplierId ,UnitPrice,ProductBaseId, detail.LastModified,detail.Id from Latent_Export.dbo.bjQuotationForm a
inner join bjQuotationFormDetail detail
on detail.QuotationFormId=a.Id
where ProductBaseId='4CEA04EE-8B70-477D-AC26-0098D13DC2B2' and UnitPrice>0
order by SupplierId,LastModified desc
),
ob as
(
select a.SupplierId,MAX(a.LastModified) as LastModified from oa a group by a.SupplierId
)
select *from ob inner join bjQuotationForm b on ob.SupplierId=b.SupplierId inner join bjQuotationFormDetail c
on ob.LastModified=c.LastModified and b.Id=c.QuotationFormId
--方法二：（改进）
SELECT SupplierId,LastModified,ProductBaseId,UnitPrice FROM (
select ROW_NUMBER() OVER (partition by form.SupplierId order by detail.LastModified desc) as row, form.SupplierId ,
UnitPrice,ProductBaseId, detail.LastModified from Latent_Export.dbo.bjQuotationForm form
inner join bjQuotationFormDetail detail
on detail.QuotationFormId=form.Id
where ProductBaseId='4CEA04EE-8B70-477D-AC26-0098D13DC2B2' and UnitPrice>0
) S WHERE S.row=1
--某个供应商对该产品的所有报价
select top 50 * from bjQuotationForm form inner join bjQuotationFormDetail detail
on form.Id =detail.QuotationFormId
where ProductBaseId='4CEA04EE-8B70-477D-AC26-0098D13DC2B2' and SupplierId='E2F18AB9-0468-4CA9-BB4C-500D59BEC958'
order by detail.LastModified desc
参考：http://www.cnblogs.com/lanzi/archive/2010/10/26/1861338.html
OVER(PARTITION BY)函数介绍
Oracle从8.1.6开始提供分析函数，分析函数用于计算基于组的某种聚合值，它和聚合函数的不同之处是：对于每个组返回多行，而聚合函数对于每个组只返回一行。
开窗函数指定了分析函数工作的数据窗口大小，这个数据窗口大小可能会随着行的变化而变化，举例如下：
1：over后的写法：
over（order by salary） 按照salary排序进行累计，order by是个默认的开窗函数
over（partition by deptno）按照部门分区
2：开窗的窗口范围：
over（order by salary range between 5 preceding and 5 following）：窗口范围为当前行数据幅度减5加5后的范围内的。
举例：
--sum(s)over(order by s range between 2 preceding and 2 following) 表示加2或2的范围内的求和
select name,class,s, sum(s)over(order by s range between 2 preceding and 2 following) mm from t2adf 3 45 45 --45加2减2即43到47，但是s在这个范围内只有45
asdf 3 55 55
cfe 2 74 74
3dd 3 78 158 --78在76到80范围内有78，80，求和得158
fda 1 80 158
gds 2 92 92
ffd 1 95 190
dss 1 95 190
ddd 3 99 198
gf 3 99 198
举例：
select name,class,s, sum(s)over(order by s rows between 2 preceding and 2 following) mm from t2
adf 3 45 174 （45+55+74=174）
asdf 3 55 252 （45+55+74+78=252）
cfe 2 74 332 （74+55+45+78+80=332）
3dd 3 78 379 （78+74+55+80+92=379）
fda 1 80 419
gds 2 92 440
ffd 1 95 461
dss 1 95 480
ddd 3 99 388
gf 3 99 293
3、与over函数结合的几个函数介绍
下面以班级成绩表t2来说明其应用
t2表信息如下：
cfe 2 74
dss 1 95
ffd 1 95
fda 1 80
gds 2 92
gf 3 99
ddd 3 99
adf 3 45
asdf 3 55
3dd 3 78
select * from
(
select name,class,s,rank()over(partition by class order by s desc) mm from t2
)
where mm=1；
得到的结果是:
dss 1 95 1
ffd 1 95 1
gds 2 92 1
gf 3 99 1
ddd 3 99 1
注意：
1.在求第一名成绩的时候，不能用row_number()，因为如果同班有两个并列第一，row_number()只返回一个结果;
select * from
(
select name,class,s,row_number()over(partition by class order by s desc) mm from t2
)
where mm=1；
1 95 1 --95有两名但是只显示一个
2 92 1
3 99 1 --99有两名但也只显示一个
2.rank()和dense_rank()可以将所有的都查找出来：
如上可以看到采用rank可以将并列第一名的都查找出来；
rank()和dense_rank()区别：
--rank()是跳跃排序，有两个第二名时接下来就是第四名；
select name,class,s,rank()over(partition by class order by s desc) mm from t2
dss 1 95 1
ffd 1 95 1
fda 1 80 3 --直接就跳到了第三
gds 2 92 1
cfe 2 74 2
gf 3 99 1
ddd 3 99 1
3dd 3 78 3
asdf 3 55 4
adf 3 45 5
--dense_rank()l是连续排序，有两个第二名时仍然跟着第三名
select name,class,s,dense_rank()over(partition by class order by s desc) mm from t2
dss 1 95 1
ffd 1 95 1
fda 1 80 2 --连续排序（仍为2）
gds 2 92 1
cfe 2 74 2
gf 3 99 1
ddd 3 99 1
3dd 3 78 2
asdf 3 55 3
adf 3 45 4
--sum()over（）的使用
select name,class,s, sum(s)over(partition by class order by s desc) mm from t2 --根据班级进行分数求和
dss 1 95 190 --由于两个95都是第一名，所以累加时是两个第一名的相加
ffd 1 95 190
fda 1 80 270 --第一名加上第二名的
gds 2 92 92
cfe 2 74 166
gf 3 99 198
ddd 3 99 198
3dd 3 78 276
asdf 3 55 331
adf 3 45 376
first_value() over()和last_value() over()的使用
--找出这三条电路每条电路的第一条记录类型和最后一条记录类型
first_value(res_type) over(PARTITION BY opr_id ORDER BY res_type) low,
last_value(res_type) over(PARTITION BY opr_id ORDER BY res_type rows BETWEEN unbounded preceding AND unbounded following) high
FROM rm_circuit_route
WHERE opr_id IN ('000100190000000000021311','000100190000000000021355','000100190000000000021339')
ORDER BY opr_id;
注：rows BETWEEN unbounded preceding AND unbounded following 的使用
--取last_value时不使用rows BETWEEN unbounded preceding AND unbounded following的结果
first_value(res_type) over(PARTITION BY opr_id ORDER BY res_type) low,
last_value(res_type) over(PARTITION BY opr_id ORDER BY res_type) high
FROM rm_circuit_route
WHERE opr_id IN ('000100190000000000021311','000100190000000000021355','000100190000000000021339')
ORDER BY opr_id;
如下图可以看到，如果不使用
数据如下：
取出该电路的第一条记录，加上ignore nulls后，如果第一条是判断的那个字段是空的，则默认取下一条，结果如下所示：
lag(expresstion,<offset>,<default>)
with a as
(select 1 id,'a' name from dual
union
select 2 id,'b' name from dual
union
select 3 id,'c' name from dual
union
select 4 id,'d' name from dual
union
select 5 id,'e' name from dual
)
select id,name,lag(id,1,'')over(order by name) from a;
--lead() over()函数用法（取出后N行数据）
lead(expresstion,<offset>,<default>)
with a as
(select 1 id,'a' name from dual
union
select 2 id,'b' name from dual
union
select 3 id,'c' name from dual
union
select 4 id,'d' name from dual
union
select 5 id,'e' name from dual
)
select id,name,lead(id,1,'')over(order by name) from a;
--ratio_to_report(a)函数用法 Ratio_to_report() 括号中就是分子，over() 括号中就是分母
with a as (select 1 a from dual
union all
select 1 a from dual
union all
select 1 a from dual
union all
select 2 a from dual
union all
select 3 a from dual
union all
select 4 a from dual
union all
select 4 a from dual
union all
select 5 a from dual
)
select a, ratio_to_report(a)over(partition by a) b from a
order by a;
with a as (select 1 a from dual
union all
select 1 a from dual
union all
select 1 a from dual
union all
select 2 a from dual
union all
select 3 a from dual
union all
select 4 a from dual
union all
select 4 a from dual
union all
select 5 a from dual
)
select a, ratio_to_report(a)over() b from a --分母缺省就是整个占比
order by a;
with a as (select 1 a from dual
union all
select 1 a from dual
union all
select 1 a from dual
union all
select 2 a from dual
union all
select 3 a from dual
union all
select 4 a from dual
union all
select 4 a from dual
union all
select 5 a from dual
)
select a, ratio_to_report(a)over() b from a
group by a order by a;--分组后的占比
SELECT a.deptno,
a.ename,
a.sal,
a.r,
b.n,
(a.r-1)/(n-1) pr1,
percent_rank() over(PARTITION BY a.deptno ORDER BY a.sal) pr2
FROM (SELECT deptno,
ename,
sal,
rank() over(PARTITION BY deptno ORDER BY sal) r --计算出在组中的排名序号
FROM emp
ORDER BY deptno, sal) a,
(SELECT deptno, COUNT(1) n FROM emp GROUP BY deptno) b --按部门计算每个部门的所有成员数
WHERE a.deptno = b.deptno;
如下所示自己计算的pr1与通过percent_rank函数得到的值是一样的：
SELECT a.deptno,
a.ename,
a.sal,
a.r,
b.n,
c.rn,
(a.r + c.rn - 1) / n pr1,
cume_dist() over(PARTITION BY a.deptno ORDER BY a.sal) pr2
FROM (SELECT deptno,
ename,
sal,
rank() over(PARTITION BY deptno ORDER BY sal) r
FROM emp
ORDER BY deptno, sal) a,
(SELECT deptno, COUNT(1) n FROM emp GROUP BY deptno) b,
(SELECT deptno, r, COUNT(1) rn,sal
FROM (SELECT deptno,sal,
rank() over(PARTITION BY deptno ORDER BY sal) r
FROM emp)
GROUP BY deptno, r,sal
ORDER BY deptno) c --c表就是为了得到每个部门员工工资的一样的个数
WHERE a.deptno = b.deptno
AND a.deptno = c.deptno(+)
AND a.sal = c.sal;
如下，输入百分比为0.7，因为0.7介于0.6和0.8之间，因此返回的结果就是0.6对应的sal的1500加上0.8对应的sal的1600平均
SELECT ename,
sal,
deptno,
percentile_cont(0.7) within GROUP(ORDER BY sal) over(PARTITION BY deptno) "Percentile_Cont",
percent_rank() over(PARTITION BY deptno ORDER BY sal) "Percent_Rank"
FROM emp
WHERE deptno IN (30, 60);
SELECT ename,
sal,
deptno,
percentile_cont(0.6) within GROUP(ORDER BY sal) over(PARTITION BY deptno) "Percentile_Cont",
percent_rank() over(PARTITION BY deptno ORDER BY sal) "Percent_Rank"
FROM emp
WHERE deptno IN (30, 60);
注意：本函数与PERCENTILE_CONT的区别在找不到对应的分布值时返回的替代值的计算方法不同
SAMPLE：下例中0.7的分布值在部门30中没有对应的Cume_Dist值，所以就取下一个分布值0.83333333所对应的SALARY来替代
SELECT ename,
sal,
deptno,
percentile_disc(0.7) within GROUP(ORDER BY sal) over(PARTITION BY deptno) "Percentile_Disc",
cume_dist() over(PARTITION BY deptno ORDER BY sal) "Cume_Dist"
FROM emp
WHERE deptno IN (30, 60);
