SQL 難點解決：直觀分組

一、   對位分組

示例 1：按順序分別列出使用 Chinese、English、French 做爲官方語言的國家數量

MySQL8:

with t(name,ord) as (select 'Chinese',1

union all select 'English',2

union all select 'French',3)

select t.name, count(countrycode) cnt

from t left join world.countrylanguage s on t.name=s.language

where s.isofficial='T'

group by name,ord

order by ord;

注意：表的字符集和數據庫會話的字符集要保持一致。

(1)  show variables like 'character_set_connection'查看當前會話字符集

(2)  show create table world.countrylanguage查看錶的字符集

(3)  set character_set_connection=[字符集]更新當前會話字符集

集算器SPL:

 A1: 鏈接數據庫

A2: 查詢出全部官方語言的記錄

A3: 須要列出的語言

A4: 將全部記錄按Language對位到A3相應位置

A5: 構造以語言和使用此語言爲官方語言的國家數量的序表

示例 2：按順序分別列出使用 Chinese、English、French 及其它語言做爲官方語言的國家數量

MySQL8:

with t(name,ord) as (select 'Chinese',1 union all select 'English',2

union all select 'French',3 union all select 'Other', 4),

s(name, cnt) as (

select language, count(countrycode) cnt

from world.countrylanguage s

where s.isofficial='T' and language in ('Chinese','English','French')

group by language

union all

select 'Other', count(distinct countrycode) cnt

from world.countrylanguage s

where isofficial='T' and language not in ('Chinese','English','French')

)

select t.name, s.cnt

from t left join s using (name)

order by t.ord;

集算器SPL:

 A4: 將全部記錄按Language對位到A3.to(3)相應位置，並追加一組用於存放不能對位的記錄

A5: 第4組計算不一樣CountryCode的數量

二、   枚舉分組

示例 1：按順序列出各種型城市的數量

MySQL8:

with t as (select * from world.city where CountryCode='CHN'),

segment(class,start,end) as (select 'tiny', 0, 200000

union all select 'small', 200000, 1000000

union all select 'medium', 1000000, 2000000

union all select 'big', 2000000, 100000000

)

select class, count(1) cnt

from segment s join t on t.population>=s.start and t.population<s.end

group by class, start

order by start;

集算器SPL：

A3: ${…}宏替換，以大括號內表達式的結果做爲新表達式進行計算，結果爲序列["?<200000","?<1000000","?<2000000","?<100000000"]

A5: 針對 A2 中每條記錄，尋找 A3 中第 1 個成立的條件，並追加到對應的組中

示例 2：列出華東地區大型城市數量、其它地區大型城市數量、非大型城市數量

MySQL8:

with t as (select * from world.city where CountryCode='CHN')

select 'East&Big' class, count(*) cnt

from t

where population>=2000000

and district in ('Shanghai','Jiangshu', 'Shandong','Zhejiang','Anhui','Jiangxi')

union all

select 'Other&Big', count(*)

from t

where population>=2000000

and district not in ('Shanghai','Jiangshu','Shandong','Zhejiang','Anhui','Jiangxi')

union all

select 'Not Big', count(*)

from t

where population<2000000;

集算器SPL:

A5: enum@n將不知足 A4 中全部條件的記錄存放到追加的最後一組中

示例 3：列出全部地區大型城市數量、華東地區大型城市數量、非大型城市數量

MySQL8:

with t as (select * from world.city where CountryCode='CHN')

select 'Big' class, count(*) cnt

from t

where population>=2000000

union all

select 'East&Big' class, count(*) cnt

from t

where population>=2000000

and district in ('Shanghai','Jiangshu','Shandong','Zhejiang','Anhui','Jiangxi')

union all

select 'Not Big' class, count(*) cnt

from t

where population<2000000;

集算器SPL:

 A6: 若A2中記錄知足A4中多個條件時，enum@r會將其追加到對應的每一個組中

三、    返回值直接做爲序號進行定位分組

示例 1: 按順序列出各種型城市的數量

MySQL8: 參見「枚舉分組」中 SQL

集算器SPL:

 A5: 先計算 A2.Population 在 A3 中段號，而後根據段號進行定位分組

四、   原序保持下的相鄰記錄分組

示例 1: 列出前 10 屆奧運金牌榜 (olympic 表中只有歷屆成績前 3 名的信息，且沒有獎牌徹底相同的狀況)

MySQL8:

with t1 as (select *,rank() over(partition by game order by gold*1000000+silver*1000+copper desc) rn from olympic where game<=10)

select game,nation,gold,silver,copper from t1 where rn=1;

集算器SPL:

 A3: 按原序分到各組，每組取第 1 條記錄組成新序表

示例 2: 求奧運會國家總成績蟬聯第 1 的最長屆數

MySQL8:

with t1 as (select *,rank() over(partition by game order by gold*1000000+silver*1000+copper desc) rn from olympic),

t2 as (select game,ifnull(nation<>lag(nation) over(order by game),0)neq from t1 where rn=1),

t3 as (select sum(neq) over(order by game) acc from t2),

t4 as (select count(acc) cnt from t3 group by acc)

select max(cnt) cnt from t4;

t1: 求出成績排名

t2: 列出歷屆第1名，並根據nation是否與上屆不一樣置標誌neq(不一樣置1，相同置0)

t3: 累積標誌neq到acc，能夠保證相鄰nation相同的acc相同，不相鄰nation的acc不相同

集算器SPL:

 A4: 將相鄰nation相同的記錄按原序分到同組

A5: 求各組長度的最大值即最大屆數

示例3：列出奧運會總成績排名第一最長蟬聯時的各屆信息

MySQL:

with t1 as (select *,rank() over(partition by game order by gold*1000000+silver*1000+copper desc) rn from olympic),

t2 as (select *,ifnull(nation<>lag(nation) over(order by game),0)neq from t1 where rn=1),

t3 as (select *, sum(neq) over(order by game) acc from t2),

t4 as (select acc,count(acc) cnt from t3 group by acc),

t5 as (select * from t4 where cnt=(select max(cnt) cnt from t4))

select game,nation,gold,silver,copper from t3 join t5 using (acc);

集算器SPL:

 A5: 求出長度最大組

示例 4：求奧運會前3名金牌總數連續增加的最大屆數

MySQL8:

with t1 as (select game,sum(gold) gold from olympic group by game),

t2 as (select game,gold, gold<=lag(gold,1,-1) over(order by game) lt from t1),

t3 as (select game, sum(lt) over(order by game) acc from t2),

t4 as (select count(*) cnt from t3 group by acc)

select max(cnt)-1 cnt from t4;

集算器SPL:

A3: 根據條件值按原序分組，若gold小於等於上一個gold則產生新分組