SQL排序：前面的排序结果对后面的排序结果有影响的排序方法

andy_alexander 2018-08-06 08:10:07

将最近遇到的一个业务场景进行抽象，希望能有高手指点。（可以通过存储过程实现，这里求教非循环的实现方式）排序规则如下：
对于发生时间相同的几组事件，与这个时间前一时间属于相同方的事件，排在前面。
一组简单的事例：
时间队
A A
B B
B A
C A
C B
排序的结果应该为：
时间队
A A
B A
B B
C B
C A

...全文

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andy_alexander 2018-08-08

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谢谢您的回复，看您用了些非标准SQL的内容，我好好消化一下。不管能不能理解吧，感谢您的回复

引用 12 楼 nayi_224 的回复:

[quote=引用 11 楼 weixin_42503575 的回复:]
[quote=引用 10 楼 nayi_224 的回复:]
[quote=引用 9 楼 weixin_42503575 的回复:]
理解了下您的思路，还是通过取当前组的前一行值来实现。那我这个需求的难点在于，如果上一组的排序结果发生变动了，会对下一组排序结果产生影响。比如，如果第一组的值，队对应的变为B，则相应的，下面的排序都需要跟着变动。而我理解，分析函数，其分析的依据，应该还是变化前的表格吧
[quote=引用 8 楼 nayi_224 的回复:]
实际数据看着好乱，我就直接拿开始的例子写了。
with tab1 as (

select 'A' a, 'A' b from dual union all

select 'B' a, 'B' b from dual union all

select 'B' a, 'A' b from dual union all

select 'C' a, 'A' b from dual union all

select 'C' a, 'B' b from dual 

)

, tab2 as (

select a, b, dense_rank() over(order by a) dr from tab1)

select a, b

  from tab2

 order by a, decode(b, first_value(a) over(order by dr range between 1 preceding and 0 following), 1, 2), b

;
oracle中能直接用分析函数解决的还是不要自连接比较好，效率一般会快很多

[/quote]

看不懂你想要干什么。
“组”是怎么划分的，与前一组的“距离”是什么，“事件”是什么，“组”内的“事件”是什么规则排序，第一组组内是什么规则排序，“同一队”指的又是什么。x y end_x end_y都是什么东西。
挑出或者手写出不超过10条可以涵盖所有特征的数据，包括“同一时间有三个时间的”，尤其是可以让人一眼看出你所说的“难点”的数据，并给出预期结果。[/quote]

我近一步简化问题吧：按tour_id、second列升序排列，对tour_id、second相同的，根据该记录对应的team_id与上一组(即与该记录tour_id,second上一条记录）相同的排在前面。示例数据如下所示：

tour_id team_id second
1 26 9
1 96 10
1 26 10
1 26 12
1 96 12
1 26 19
1 96 22
1 26 22
2 26 9
2 16 10
2 26 10
2 26 12
2 16 12
2 26 19
2 16 22
2 26 22

************************************
想要的结果：
tour_id team_id second
1 26 9
1 26 10
1 96 10
1 96 12
1 26 12
1 26 19
1 26 22
1 96 22
2 26 9
2 26 10
2 16 10
2 16 12
2 26 12
2 26 19
2 26 22
2 16 22
[/quote]

with tab1 as (

select 1 tour_id, 26 team_id, 9 sec from dual union all

select 1 , 96 , 10  from dual union all

select 1 , 26 , 10  from dual union all

select 1 , 26 , 12  from dual union all

select 1 , 96 , 12  from dual union all

select 1 , 26 , 19  from dual union all

select 1 , 96 , 22  from dual union all

select 2 , 26 , 9  from dual union all

select 2 , 10 , 10  from dual 

)

select *

  from tab1 t1

model

dimension by(

       dense_rank() over(order by t1.tour_id, sec) dr,

       row_number() over(partition by t1.tour_id, sec order by team_id desc) rn)

measures(-1 ord, tour_id, team_id, sec)

rules automatic order(

 ord[dr, rn] order by dr, rn = decode(team_id[cv(), cv()], team_id[cv() - 1, 1], 1, 2) 

) 

order by dr, ord, rn

;

[/quote]

nayi_224 2018-08-08

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引用 11 楼 weixin_42503575 的回复:

[quote=引用 10 楼 nayi_224 的回复:]
[quote=引用 9 楼 weixin_42503575 的回复:]
理解了下您的思路，还是通过取当前组的前一行值来实现。那我这个需求的难点在于，如果上一组的排序结果发生变动了，会对下一组排序结果产生影响。比如，如果第一组的值，队对应的变为B，则相应的，下面的排序都需要跟着变动。而我理解，分析函数，其分析的依据，应该还是变化前的表格吧
[quote=引用 8 楼 nayi_224 的回复:]
实际数据看着好乱，我就直接拿开始的例子写了。
with tab1 as (

select 'A' a, 'A' b from dual union all

select 'B' a, 'B' b from dual union all

select 'B' a, 'A' b from dual union all

select 'C' a, 'A' b from dual union all

select 'C' a, 'B' b from dual 

)

, tab2 as (

select a, b, dense_rank() over(order by a) dr from tab1)

select a, b

  from tab2

 order by a, decode(b, first_value(a) over(order by dr range between 1 preceding and 0 following), 1, 2), b

;
oracle中能直接用分析函数解决的还是不要自连接比较好，效率一般会快很多

with tab1 as (

select 1 tour_id, 26 team_id, 9 sec from dual union all

select 1 , 96 , 10  from dual union all

select 1 , 26 , 10  from dual union all

select 1 , 26 , 12  from dual union all

select 1 , 96 , 12  from dual union all

select 1 , 26 , 19  from dual union all

select 1 , 96 , 22  from dual union all

select 2 , 26 , 9  from dual union all

select 2 , 10 , 10  from dual 

)

select *

  from tab1 t1

model

dimension by(

       dense_rank() over(order by t1.tour_id, sec) dr,

       row_number() over(partition by t1.tour_id, sec order by team_id desc) rn)

measures(-1 ord, tour_id, team_id, sec)

rules automatic order(

 ord[dr, rn] order by dr, rn = decode(team_id[cv(), cv()], team_id[cv() - 1, 1], 1, 2) 

) 

order by dr, ord, rn

;

nayi_224 2018-08-08

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引用 13 楼 weixin_42503575 的回复:

谢谢您的回复，看您用了些非标准SQL的内容，我好好消化一下。不管能不能理解吧，感谢您的回复

[quote=引用 12 楼 nayi_224 的回复:]
[quote=引用 11 楼 weixin_42503575 的回复:]
[quote=引用 10 楼 nayi_224 的回复:]
[quote=引用 9 楼 weixin_42503575 的回复:]
理解了下您的思路，还是通过取当前组的前一行值来实现。那我这个需求的难点在于，如果上一组的排序结果发生变动了，会对下一组排序结果产生影响。比如，如果第一组的值，队对应的变为B，则相应的，下面的排序都需要跟着变动。而我理解，分析函数，其分析的依据，应该还是变化前的表格吧
[quote=引用 8 楼 nayi_224 的回复:]
实际数据看着好乱，我就直接拿开始的例子写了。
with tab1 as (

select 'A' a, 'A' b from dual union all

select 'B' a, 'B' b from dual union all

select 'B' a, 'A' b from dual union all

select 'C' a, 'A' b from dual union all

select 'C' a, 'B' b from dual 

)

, tab2 as (

select a, b, dense_rank() over(order by a) dr from tab1)

select a, b

  from tab2

 order by a, decode(b, first_value(a) over(order by dr range between 1 preceding and 0 following), 1, 2), b

;
oracle中能直接用分析函数解决的还是不要自连接比较好，效率一般会快很多

with tab1 as (

select 1 tour_id, 26 team_id, 9 sec from dual union all

select 1 , 96 , 10  from dual union all

select 1 , 26 , 10  from dual union all

select 1 , 26 , 12  from dual union all

select 1 , 96 , 12  from dual union all

select 1 , 26 , 19  from dual union all

select 1 , 96 , 22  from dual union all

select 2 , 26 , 9  from dual union all

select 2 , 10 , 10  from dual 

)

select *

  from tab1 t1

model

dimension by(

       dense_rank() over(order by t1.tour_id, sec) dr,

       row_number() over(partition by t1.tour_id, sec order by team_id desc) rn)

measures(-1 ord, tour_id, team_id, sec)

rules automatic order(

 ord[dr, rn] order by dr, rn = decode(team_id[cv(), cv()], team_id[cv() - 1, 1], 1, 2) 

) 

order by dr, ord, rn

;

[/quote][/quote]

又仔细看了一下，有点bug，应该这么写

with tab1 as (

select 1 tour_id, 13 team_id, 9 sec from dual union all

select 1 , 26 , 9  from dual union all

select 1 , 96 , 10  from dual union all

select 1 , 26 , 10  from dual union all

select 1 , 26 , 12  from dual union all

select 1 , 96 , 12  from dual union all

select 1 , 13 , 19  from dual union all

select 1 , 26 , 19  from dual union all

--select 1 , 96 , 20  from dual union all

select 1 , 133 , 22  from dual union all

select 1 , 96 , 22  from dual union all

select 2 , 26 , 9  from dual union all

select 2 , 10 , 10  from dual 

)

select *

  from tab1 t1

model

dimension by(

       dense_rank() over(order by t1.tour_id, sec) dr,

       row_number() over(partition by t1.tour_id, sec order by team_id ) rn)

measures(-1 ord, tour_id, team_id, sec)

rules automatic order(

 --ord[dr, rn] order by dr, rn = decode(team_id[cv(), cv()], team_id[cv() - 1, 1], 1, 2) 

 ord[dr, rn] order by dr, rn = 

         case when team_id[cv(), cv()] = max(decode(ord, 1, -99999999, team_id))[cv() - 1, rn]

         then 1 else 2 end

) 

order by dr, ord, rn

;

这种真递归的需求要是能用标准sql的select语句写出来，请一定要@我。

andy_alexander 2018-08-07

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好的，我截取了三分钟以内的数据。主要是解决对同时发生的事件进行排序的问题，对同时的事件，与前一组事件距离近的在前；距离相同的，与前一组事件的最后一个事件属于同一队的在前。难点在于后一组排序会依赖于前一组的排序结果，比如对0分10秒、0分12秒的结果。（对于同一时间有三个时间的，需要结合距离之类的计算，我觉得思路一致，）
现只麻烦您看一下，能否解决时间相同的，与前一组事件最后一个属于同一队的，排在前面这个难点吧。麻烦了
CREATE TABLE table_order (
"team_id" varchar(255) COLLATE "default",
"minute" varchar(255) COLLATE "default",
"second" varchar(255) COLLATE "default",
"x" varchar(255) COLLATE "default",
"y" varchar(255) COLLATE "default",
"end_x" varchar(255) COLLATE "default",
"end_y" varchar(255) COLLATE "default"
)WITH (OIDS=FALSE);

INSERT INTO table_order VALUES ('26', '0', '1', '52.605', '34.68', '55.755', '33.116');
INSERT INTO table_order VALUES ('26', '0', '10', '41.58', '52.496', NULL, NULL);
INSERT INTO table_order VALUES ('26', '0', '15', '24.465', '43.112', NULL, NULL);
INSERT INTO table_order VALUES ('26', '0', '16', '24.465', '43.112', '26.46', '18.02');
INSERT INTO table_order VALUES ('26', '0', '19', '28.77', '17.136', '44.52', '5.576');
INSERT INTO table_order VALUES ('26', '0', '2', '55.755', '33.116', '48.51', '37.06');
INSERT INTO table_order VALUES ('26', '0', '21', '44.415', '5.644', '48.3', '16.796');
INSERT INTO table_order VALUES ('26', '0', '22', '48.3', '16.796', NULL, NULL);
INSERT INTO table_order VALUES ('26', '0', '29', '49.35', '40.936', NULL, NULL);
INSERT INTO table_order VALUES ('26', '0', '3', '48.51', '37.06', '33.81', '57.392');
INSERT INTO table_order VALUES ('26', '0', '4', '33.81', '57.392', '23.73', '41.548');
INSERT INTO table_order VALUES ('26', '0', '5', '23.73', '41.548', '33.6', '41.548');
INSERT INTO table_order VALUES ('26', '0', '53', '47.04', '16.184', NULL, NULL);
INSERT INTO table_order VALUES ('26', '0', '54', '45.045', '19.244', '73.92', '17.816');
INSERT INTO table_order VALUES ('26', '0', '58', '62.37', '16.116', '76.125', '9.724');
INSERT INTO table_order VALUES ('26', '0', '6', '33.81', '42.024', '23.31', '44.88');
INSERT INTO table_order VALUES ('26', '0', '9', '23.94', '46.308', '42.735', '50.184');
INSERT INTO table_order VALUES ('26', '1', '11', '80.22', '0', '76.44', '3.196');
INSERT INTO table_order VALUES ('26', '1', '13', '76.44', '3.196', '79.905', '1.564');
INSERT INTO table_order VALUES ('26', '1', '14', '79.905', '1.564', '69.93', '1.972');
INSERT INTO table_order VALUES ('26', '1', '16', '69.93', '1.972', '56.91', '4.76');
INSERT INTO table_order VALUES ('26', '1', '19', '56.28', '11.832', '46.83', '33.184');
INSERT INTO table_order VALUES ('26', '1', '2', '82.635', '13.668', '82.845', '15.3');
INSERT INTO table_order VALUES ('26', '1', '22', '55.755', '41.004', '63.315', '43.452');
INSERT INTO table_order VALUES ('26', '1', '26', '64.575', '48.756', NULL, NULL);
INSERT INTO table_order VALUES ('26', '1', '52', '70.035', '39.984', '94.5', '35.496');
INSERT INTO table_order VALUES ('26', '1', '58', '87.465', '65.688', NULL, NULL);
INSERT INTO table_order VALUES ('26', '2', '13', '76.23', '50.796', '74.445', '37.264');
INSERT INTO table_order VALUES ('26', '2', '16', '73.71', '35.292', '73.395', '18.632');
INSERT INTO table_order VALUES ('26', '2', '18', '73.29', '17.816', '78.54', '1.904');
INSERT INTO table_order VALUES ('26', '2', '22', '79.59', '4.828', '79.59', '6.596');
INSERT INTO table_order VALUES ('26', '2', '27', '81.48', '0', '92.4', '3.604');
INSERT INTO table_order VALUES ('26', '2', '28', '92.4', '3.604', '85.575', '0');
INSERT INTO table_order VALUES ('26', '2', '3', '87.255', '68', '79.065', '63.376');
INSERT INTO table_order VALUES ('26', '2', '4', '79.065', '63.376', '85.365', '65.416');
INSERT INTO table_order VALUES ('26', '2', '45', '69.825', '32.776', NULL, NULL);
INSERT INTO table_order VALUES ('26', '2', '47', '74.235', '31.348', '77.28', '29.92');
INSERT INTO table_order VALUES ('26', '2', '51', '74.655', '18.36', '68.565', '25.568');
INSERT INTO table_order VALUES ('26', '2', '53', '68.565', '25.568', '80.01', '25.16');
INSERT INTO table_order VALUES ('26', '2', '54', '80.01', '25.16', '78.855', '29.104');
INSERT INTO table_order VALUES ('26', '2', '55', '78.96', '29.308', '79.8', '24.48');
INSERT INTO table_order VALUES ('26', '2', '56', '79.485', '27.676', '75.81', '19.244');
INSERT INTO table_order VALUES ('26', '2', '58', '75.81', '19.04', '101.535', '12.376');
INSERT INTO table_order VALUES ('26', '2', '7', '83.16', '59.84', '77.595', '52.496');
INSERT INTO table_order VALUES ('26', '2', '9', '77.595', '52.564', '84.525', '63.58');
INSERT INTO table_order VALUES ('26', '3', '14', '85.365', '0', '85.995', '9.384');
INSERT INTO table_order VALUES ('26', '3', '19', '91.035', '15.708', '90.3', '30.736');
INSERT INTO table_order VALUES ('26', '3', '2', '101.43', '12.376', '90.825', '28.016');
INSERT INTO table_order VALUES ('26', '3', '23', '74.865', '46.308', '70.875', '37.876');
INSERT INTO table_order VALUES ('26', '3', '25', '70.875', '39.168', '98.385', '56.576');
INSERT INTO table_order VALUES ('26', '3', '30', '99.75', '49.368', NULL, NULL);
INSERT INTO table_order VALUES ('26', '3', '44', '40.95', '68', '51.765', '66.028');
INSERT INTO table_order VALUES ('26', '3', '45', '51.765', '66.028', '38.01', '63.444');
INSERT INTO table_order VALUES ('26', '3', '46', '38.01', '63.444', NULL, NULL);
INSERT INTO table_order VALUES ('26', '3', '47', '50.505', '55.624', '69.3', '48.348');
INSERT INTO table_order VALUES ('26', '3', '49', '56.805', '45.56', '62.055', '68');
INSERT INTO table_order VALUES ('26', '3', '7', '85.785', '6.46', NULL, NULL);
INSERT INTO table_order VALUES ('96', '0', '10', '41.58', '52.496', NULL, NULL);
INSERT INTO table_order VALUES ('96', '0', '12', '32.76', '53.788', '27.72', '51.204');
INSERT INTO table_order VALUES ('96', '0', '12', '32.76', '53.788', NULL, NULL);
INSERT INTO table_order VALUES ('96', '0', '14', '27.93', '50.864', NULL, NULL);
INSERT INTO table_order VALUES ('96', '0', '22', '49.14', '18.36', NULL, NULL);
INSERT INTO table_order VALUES ('96', '0', '23', '45.15', '19.788', NULL, NULL);
INSERT INTO table_order VALUES ('96', '0', '24', '45.045', '22.508', '46.2', '29.648');
INSERT INTO table_order VALUES ('96', '0', '25', '46.2', '29.648', '49.245', '27.676');
INSERT INTO table_order VALUES ('96', '0', '26', '49.35', '27.88', '47.145', '37.672');
INSERT INTO table_order VALUES ('96', '0', '29', '49.35', '40.936', NULL, NULL);
INSERT INTO table_order VALUES ('96', '0', '38', '50.19', '40.392', '57.855', '23.528');
INSERT INTO table_order VALUES ('96', '0', '40', '58.065', '23.528', '62.37', '8.976');
INSERT INTO table_order VALUES ('96', '0', '43', '64.785', '14.28', '99.96', '17.204');
INSERT INTO table_order VALUES ('96', '0', '49', '97.65', '13.056', '49.035', '13.192');
INSERT INTO table_order VALUES ('96', '0', '53', '47.04', '16.184', NULL, NULL);
INSERT INTO table_order VALUES ('96', '0', '56', '78.54', '15.232', '68.775', '11.764');
INSERT INTO table_order VALUES ('96', '0', '58', '68.775', '11.764', '63.945', '13.056');
INSERT INTO table_order VALUES ('96', '1', '26', '64.575', '48.756', NULL, NULL);
INSERT INTO table_order VALUES ('96', '1', '3', '85.05', '15.028', NULL, NULL);
INSERT INTO table_order VALUES ('96', '1', '53', '95.025', '41.344', NULL, NULL);
INSERT INTO table_order VALUES ('96', '1', '57', '90.93', '52.768', NULL, NULL);
INSERT INTO table_order VALUES ('96', '1', '58', '87.465', '65.688', NULL, NULL);
INSERT INTO table_order VALUES ('96', '2', '22', '78.225', '8.296', NULL, NULL);
INSERT INTO table_order VALUES ('96', '2', '40', '83.16', '0', '66.99', '12.172');
INSERT INTO table_order VALUES ('96', '2', '41', '66.99', '12.172', '75.285', '2.38');
INSERT INTO table_order VALUES ('96', '2', '42', '74.76', '2.38', '69.3', '28.084');
INSERT INTO table_order VALUES ('96', '2', '48', '79.695', '26.588', NULL, NULL);
INSERT INTO table_order VALUES ('96', '2', '52', '80.22', '16.796', NULL, NULL);
INSERT INTO table_order VALUES ('96', '2', '55', '80.955', '21.012', NULL, NULL);
INSERT INTO table_order VALUES ('96', '3', '20', '89.145', '35.224', NULL, NULL);
INSERT INTO table_order VALUES ('96', '3', '3', '89.985', '30.124', NULL, NULL);
INSERT INTO table_order VALUES ('96', '3', '30', '97.125', '49.776', NULL, NULL);
INSERT INTO table_order VALUES ('96', '3', '30', '99.75', '49.368', NULL, NULL);
INSERT INTO table_order VALUES ('96', '3', '32', '91.56', '51.748', '67.2', '68');
INSERT INTO table_order VALUES ('96', '3', '46', '41.055', '63.512', NULL, NULL);
INSERT INTO table_order VALUES ('96', '3', '47', '61.11', '51.408', NULL, NULL);
INSERT INTO table_order VALUES ('96', '3', '8', '84', '11.968', NULL, NULL);

吉普赛的歌 2018-08-07

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麻烦贴一下完整可行的测试数据脚本吧，可以保证按你的脚本数据做出来，你正式的就可以用的

andy_alexander 2018-08-07

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您好，可能我在具体抽象中表述的不太准确了。我看您在最后排序时，tt列与seq列进行了关联。而实际中，tt列是一系列的复杂情况，比如具体的时间（我实际的业务需求是就是解决同时发生的事件排序问题），没有这类简单的对应关系。您看是否可以解决？

引用 1 楼 yenange 的回复:

没 oracle 环境，用 sqlserver 做了一个，不过思路一样，你参照着改一下就好。

USE tempdb

GO

IF OBJECT_ID('t') IS NOT NULL DROP TABLE t

GO

CREATE TABLE t(

	tt VARCHAR(10),

	seq     VARCHAR(10)	

)

GO

INSERT INTO t VALUES ('A','A')

INSERT INTO t VALUES ('B','B')

INSERT INTO t VALUES ('B','A')

INSERT INTO t VALUES ('C','A')

INSERT INTO t VALUES ('C','B')

---------- 以上为测试表及测试数据 ----------------



;WITH cte AS (

	SELECT ROW_NUMBER() OVER (ORDER BY tt) AS rid,tt FROM t GROUP BY tt

),cte2 AS (

	SELECT a.*,b.tt AS upTT FROM cte AS a LEFT JOIN cte AS b ON a.rid=b.rid+1

)

SELECT * 

FROM t 

ORDER BY 

tt

,CASE WHEN EXISTS(

	SELECT 1 FROM cte2 WHERE cte2.tt=t.tt AND cte2.upTT=t.seq  	

) THEN 0 ELSE 1 END,seq

andy_alexander 2018-08-07

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看了一上思路，是在CTE语句中进行关联查询，怎么没想到呢。谢谢啦，马上尝试一下，有消息了告诉您。谢谢谢谢

引用 1 楼 yenange 的回复:

没 oracle 环境，用 sqlserver 做了一个，不过思路一样，你参照着改一下就好。

USE tempdb

GO

IF OBJECT_ID('t') IS NOT NULL DROP TABLE t

GO

CREATE TABLE t(

	tt VARCHAR(10),

	seq     VARCHAR(10)	

)

GO

INSERT INTO t VALUES ('A','A')

INSERT INTO t VALUES ('B','B')

INSERT INTO t VALUES ('B','A')

INSERT INTO t VALUES ('C','A')

INSERT INTO t VALUES ('C','B')

---------- 以上为测试表及测试数据 ----------------



;WITH cte AS (

	SELECT ROW_NUMBER() OVER (ORDER BY tt) AS rid,tt FROM t GROUP BY tt

),cte2 AS (

	SELECT a.*,b.tt AS upTT FROM cte AS a LEFT JOIN cte AS b ON a.rid=b.rid+1

)

SELECT * 

FROM t 

ORDER BY 

tt

,CASE WHEN EXISTS(

	SELECT 1 FROM cte2 WHERE cte2.tt=t.tt AND cte2.upTT=t.seq  	

) THEN 0 ELSE 1 END,seq

andy_alexander 2018-08-07

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队那一列只有两种可能，我抽象出了最原始的情况。抱歉哈

引用 2 楼 yaiger 的回复:

没说清楚啊，
如果数据是
时间队
A D
A B
A C
B A
B C
B D
该如何排序？

yaiger 2018-08-07

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没说清楚啊，
如果数据是
时间队
A D
A B
A C
B A
B C
B D
该如何排序？

吉普赛的歌 2018-08-07

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没 oracle 环境，用 sqlserver 做了一个，不过思路一样，你参照着改一下就好。

USE tempdb

GO

IF OBJECT_ID('t') IS NOT NULL DROP TABLE t

GO

CREATE TABLE t(

	tt VARCHAR(10),

	seq     VARCHAR(10)	

)

GO

INSERT INTO t VALUES ('A','A')

INSERT INTO t VALUES ('B','B')

INSERT INTO t VALUES ('B','A')

INSERT INTO t VALUES ('C','A')

INSERT INTO t VALUES ('C','B')

---------- 以上为测试表及测试数据 ----------------



;WITH cte AS (

	SELECT ROW_NUMBER() OVER (ORDER BY tt) AS rid,tt FROM t GROUP BY tt

),cte2 AS (

	SELECT a.*,b.tt AS upTT FROM cte AS a LEFT JOIN cte AS b ON a.rid=b.rid+1

)

SELECT * 

FROM t 

ORDER BY 

tt

,CASE WHEN EXISTS(

	SELECT 1 FROM cte2 WHERE cte2.tt=t.tt AND cte2.upTT=t.seq  	

) THEN 0 ELSE 1 END,seq

andy_alexander 2018-08-07

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引用 10 楼 nayi_224 的回复:

[quote=引用 9 楼 weixin_42503575 的回复:]
理解了下您的思路，还是通过取当前组的前一行值来实现。那我这个需求的难点在于，如果上一组的排序结果发生变动了，会对下一组排序结果产生影响。比如，如果第一组的值，队对应的变为B，则相应的，下面的排序都需要跟着变动。而我理解，分析函数，其分析的依据，应该还是变化前的表格吧
[quote=引用 8 楼 nayi_224 的回复:]
实际数据看着好乱，我就直接拿开始的例子写了。
with tab1 as (

select 'A' a, 'A' b from dual union all

select 'B' a, 'B' b from dual union all

select 'B' a, 'A' b from dual union all

select 'C' a, 'A' b from dual union all

select 'C' a, 'B' b from dual 

)

, tab2 as (

select a, b, dense_rank() over(order by a) dr from tab1)

select a, b

  from tab2

 order by a, decode(b, first_value(a) over(order by dr range between 1 preceding and 0 following), 1, 2), b

;
oracle中能直接用分析函数解决的还是不要自连接比较好，效率一般会快很多

nayi_224 2018-08-07

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引用 9 楼 weixin_42503575 的回复:

理解了下您的思路，还是通过取当前组的前一行值来实现。那我这个需求的难点在于，如果上一组的排序结果发生变动了，会对下一组排序结果产生影响。比如，如果第一组的值，队对应的变为B，则相应的，下面的排序都需要跟着变动。而我理解，分析函数，其分析的依据，应该还是变化前的表格吧
[quote=引用 8 楼 nayi_224 的回复:]
实际数据看着好乱，我就直接拿开始的例子写了。
with tab1 as (

select 'A' a, 'A' b from dual union all

select 'B' a, 'B' b from dual union all

select 'B' a, 'A' b from dual union all

select 'C' a, 'A' b from dual union all

select 'C' a, 'B' b from dual 

)

, tab2 as (

select a, b, dense_rank() over(order by a) dr from tab1)

select a, b

  from tab2

 order by a, decode(b, first_value(a) over(order by dr range between 1 preceding and 0 following), 1, 2), b

;
oracle中能直接用分析函数解决的还是不要自连接比较好，效率一般会快很多

andy_alexander 2018-08-07

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理解了下您的思路，还是通过取当前组的前一行值来实现。那我这个需求的难点在于，如果上一组的排序结果发生变动了，会对下一组排序结果产生影响。比如，如果第一组的值，队对应的变为B，则相应的，下面的排序都需要跟着变动。而我理解，分析函数，其分析的依据，应该还是变化前的表格吧

引用 8 楼 nayi_224 的回复:

实际数据看着好乱，我就直接拿开始的例子写了。

with tab1 as (

select 'A' a, 'A' b from dual union all

select 'B' a, 'B' b from dual union all

select 'B' a, 'A' b from dual union all

select 'C' a, 'A' b from dual union all

select 'C' a, 'B' b from dual 

)

, tab2 as (

select a, b, dense_rank() over(order by a) dr from tab1)

select a, b

  from tab2

 order by a, decode(b, first_value(a) over(order by dr range between 1 preceding and 0 following), 1, 2), b

;

oracle中能直接用分析函数解决的还是不要自连接比较好，效率一般会快很多

nayi_224 2018-08-07

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实际数据看着好乱，我就直接拿开始的例子写了。

with tab1 as (

select 'A' a, 'A' b from dual union all

select 'B' a, 'B' b from dual union all

select 'B' a, 'A' b from dual union all

select 'C' a, 'A' b from dual union all

select 'C' a, 'B' b from dual 

)

, tab2 as (

select a, b, dense_rank() over(order by a) dr from tab1)

select a, b

  from tab2

 order by a, decode(b, first_value(a) over(order by dr range between 1 preceding and 0 following), 1, 2), b

;

oracle中能直接用分析函数解决的还是不要自连接比较好，效率一般会快很多