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SQL難點(diǎn)解決:直觀分組

1、    對(duì)位分組

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示例 1:按順序分別列出使用 Chinese、English、French 作為官方語(yǔ)言的國(guó)家數(shù)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(shù)據(jù)庫(kù)會(huì)話的字符集要保持一致。

(1)   show variables like 'character_set_connection'查看當(dāng)前會(huì)話字符集

(2)   show create table world.countrylanguage查看表的字符集

(3)   set character_set_connection=[字符集]更新當(dāng)前會(huì)話字符集

 

集算器SPL:


A
1=connect("mysql")
2=A1.query@x("select   * from world.countrylanguage where isofficial='T'")
3[Chinese,English,French]
4=A2.align@a(A3,Language)
5=A4.new(A3(#):name, ~.len():cnt)

A1: 連接數(shù)據(jù)庫(kù)

A2: 查詢出所有官方語(yǔ)言的記錄

A3: 需要列出的語(yǔ)言

A4: 將所有記錄按Language對(duì)位到A3相應(yīng)位置

A5: 構(gòu)造以語(yǔ)言和使用此語(yǔ)言為官方語(yǔ)言的國(guó)家數(shù)量的序表

SQL 難點(diǎn)解決:直觀分組

 

示例 2:按順序分別列出使用 Chinese、English、French 及其它語(yǔ)言作為官方語(yǔ)言的國(guó)家數(shù)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:


A
1=connect("mysql")
2=A1.query@x("select   * from world.countrylanguage where isofficial='T'")
3[Chinese,English,French,Other]
4=A2.align@an(A3.to(3),Language)
5=A4.new(A3(#):name, if(#<=3,~.len(), ~.icount(CountryCode)):cnt)

A4: 將所有記錄按Language對(duì)位到A3.to(3)相應(yīng)位置,并追加一組用于存放不能對(duì)位的記錄

A5: 第4組計(jì)算不同CountryCode的數(shù)量

SQL 難點(diǎn)解決:直觀分組

 

2、    枚舉分組

示例 1:按順序列出各類型城市的數(shù)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

group by class, start

order by start;

 

集算器SPL:


A
1=connect("mysql")
2=A1.query@x("select * from world.city where   CountryCode='CHN'")
3=${string([20,100,200,10000].(~*10000).("?<"/~))}
4[tiny,small,medium,big]
5=A2.enum(A3,Population)
6=A5.new(A4(#):class, ~.len():cnt)

A3: ${…}宏替換,以大括號(hào)內(nèi)表達(dá)式的結(jié)果作為新表達(dá)式進(jìn)行計(jì)算,結(jié)果為序列["?<200000","?<1000000","?<2000000","?<100000000"]

A5: 針對(duì) A2 中每條記錄,尋找 A3 中第 1 個(gè)成立的條件,并追加到對(duì)應(yīng)的組中

SQL 難點(diǎn)解決:直觀分組

 

示例 2:列出華東地區(qū)大型城市數(shù)量、其它地區(qū)大型城市數(shù)量、非大型城市數(shù)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:


A
1=connect("mysql")
2=A1.query@x("select * from world.city where   CountryCode='CHN'")
3[Shanghai,Jiangshu, Shandong,Zhejiang,Anhui,Jiangxi]
4[?(1)>=2000000   && A3.contain(?(2)), ?(1)>=2000000 && !A3.contain(?(2))]
5[East&Big,Other&Big, Not Big]
6=A2.enum@n(A4, [Population,District])
7=A6.new(A5(#):class, A6(#).len():cnt)

A5: enum@n將不滿足 A4 中所有條件的記錄存放到追加的最后一組中

SQL 難點(diǎn)解決:直觀分組

 

示例 3:列出所有地區(qū)大型城市數(shù)量、華東地區(qū)大型城市數(shù)量、非大型城市數(shù)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:


A
1=connect("mysql")
2=A1.query@x("select * from world.city where   CountryCode='CHN'")
3[Shanghai,Jiangshu, Shandong,Zhejiang,Anhui,Jiangxi]
4[?(1)>=2000000, ?(1)>=2000000 && A3.contain(?(2))]
5[Big, East&Big, Not Big]
6=A2.enum@rn(A4, [Population,District])
7=A6.new(A5(#):class, A6(#).len():cnt)

A6: 若A2中記錄滿足A4中多個(gè)條件時(shí),enum@r會(huì)將其追加到對(duì)應(yīng)的每個(gè)組中

SQL 難點(diǎn)解決:直觀分組

 

3、    返回值直接作為序號(hào)進(jìn)行定位分組

示例 1: 按順序列出各類型城市的數(shù)量

MySQL8: 參見(jiàn)“枚舉分組”中 SQL

 

集算器SPL:


A
1=connect("mysql")
2=A1.query@x("select * from world.city where   CountryCode='CHN'")
3=[0,20,100,200].(~*10000)
4[tiny,small,medium,big]
5=A2.group@n(A3.pseg(Population))
6=A5.new(A4(#):class, ~.len():cnt)

A5: 先計(jì)算 A2.Population 在 A3 中段號(hào),然后根據(jù)段號(hào)進(jìn)行定位分組

 

4、    原序保持下的相鄰記錄分組

示例 1: 列出前 10 屆奧運(yùn)金牌榜 (olympic 表中只有歷屆成績(jī)前 3 名的信息,且沒(méi)有獎(jiǎng)牌完全相同的情況)

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:


A
1=connect("mysql")
2=A1.query("select   * from olympic where game<=10 order by game,   gold*1000000+silver*1000+copper desc")
3=A2.group@o1(game)

A3: 按原序分到各組,每組取第 1 條記錄組成新序表

SQL 難點(diǎn)解決:直觀分組

 

示例 2: 求奧運(yùn)會(huì)國(guó)家總成績(jī)蟬聯(lián)第 1 的最長(zhǎng)屆數(shù)

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(jī)排名

t2: 列出歷屆第1名,并根據(jù)nation是否與上屆不同置標(biāo)志neq(不同置1,相同置0)

t3: 累積標(biāo)志neq到acc,可以保證相鄰nation相同的acc相同,不相鄰nation的acc不相同

 

集算器SPL:


A
1=connect("mysql")
2=A1.query("select   * from olympic order by game, gold*1000000+silver*1000+copper desc")
3=A2.group@o1(game)
4=A3.group@o(nation)
5=A4.max(~.len())

A4: 將相鄰nation相同的記錄按原序分到同組

A5: 求各組長(zhǎng)度的最大值即最大屆數(shù)

SQL 難點(diǎn)解決:直觀分組

 

示例3:列出奧運(yùn)會(huì)總成績(jī)排名第一最長(zhǎng)蟬聯(lián)時(shí)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:


A
1=connect("mysql")
2=A1.query("select   * from olympic order by game, gold*1000000+silver*1000+copper desc")
3=A2.group@o1(game)
4=A3.group@o(nation)
5=A4.maxp(~.len())

A5: 求出長(zhǎng)度最大組

SQL 難點(diǎn)解決:直觀分組

 

示例 4:求奧運(yùn)會(huì)前3名金牌總數(shù)連續(xù)增長(zhǎng)的最大屆數(shù)

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:


A
1=connect("mysql")
2=A1.query("select game,sum(gold) gold from olympic group by   game order by game")
3=A2.group@i(gold<=gold[-1])
4=A3.max(~.len())-1

A3: 根據(jù)條件值按原序分組,若gold小于等于上一個(gè)gold則產(chǎn)生新分組

SQL 難點(diǎn)解決:直觀分組


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