1、 生成连续整数序列
MySQL8: with recursive t(n) as (
select 1
union all
select n+1 from t where n<7
)
select * from t;
Oracle：select level n
from dual connect by level<=7;
集算器 SPL：
A |
|
1 |
=to(1,7) |
A1：构造从 1 到 7 的整数序列
示例 1：百鸡问题，鸡翁一值钱五，鸡母一值钱三，鸡雏三值钱一。百钱买百鸡，问鸡翁、母、雏各几
MySQL8:
with recursive jg(n) as (select 1 union all select n+1 from jg where n<100/5),
jm(n) as (select 1 union all select n+1 from jm where n<100/3),
jc(n) as (select 3 union all select n+3 from jc where n<98)
select jg.n jw, jm.n jm, jc.n jc
from jg cross join jm cross join jc
where jg.n*5+jm.n*3+jc.n/3=100 and jg.n+jm.n+jc.n=100
集算器 SPL：
A |
|
1 |
=to(100/5) |
2 |
=to(100/3) |
3 |
=33.(~*3) |
4 |
=create(jw,jm,jc) |
5 |
>A1.run(A2.run(A3.run(if(A1.~+A2.~+A3.~==100 && A1.~*5+A2.~*3+A3.~/3==100,A4.insert(0,A1.~,A2.~,A3.~))))) |
A1：构造1到20的整数序列
A2：构造1到33的整数序列
A3：构造1到99且步长为3的整数序列
A4：创建数据结构为(jw,jm,jc)的序表
A5：对A1、A2、A3的数据进行嵌套循环，若满足于A1成员+A2成员+A3成员==100且A1成员*5+A2成员*3+A3成员/3==100则追加到A4序表中
示例2：将指定列中冒号分隔的串划分成多行
Oracle:
with t(k,f) as (select 1 , 'a1:a2:a3' from dual
union all select 2, 'b1:b2' from dual),
t1 as (select k,f, length(f)-length(replace(f,':',''))+1 cnt from t),
t2 as (select level n from dual connect by level<=(select max(cnt) from t1)),
t3 as (select t1.k, t1.f, n, cnt,
case when n=1 then 1 else instr(f,':',1,n-1)+1 end p1,
case when n=cnt then length(f)+1 else instr(f,':',1,n) end p2
from t1 join t2 on t2.n<=t1.cnt)
select k,substr(f,p1,p2-p1) f from t3 order by k;
集算器 SPL：
A |
|
1 |
=create(k,f).record([1,"a1:a2:a3",2,"b1:b2"]) |
2 |
>A1.run(f=f.split(":")) |
3 |
=A1.(f.new(A1.k:k, ~:f)) |
4 |
=A3.conj() |
A1：创建数据结构为(k,f)的序表，并追加2条记录(1, “a1:a2:a3)和(2,”b1:b2”)
A2：将A1的字段f用冒号划分成序列并重新赋值给字段f
A3：针对A1每条记录构造数据结构为(k,f)的序表，并根据字段f中成员构造记录(A1.k，f成员)追加到此序表中
2、 生成连续日期序列
MySQL8：
with recursive
t(d) as (select date'2018-10-03'
union all
select d+1 from t where d select d,dayofweek(d) w from t; 集算器 SPL： A 1 =periods("2018-10-03", "2018-10-09") A1：生成2018-10-03到2018-10-09的日期序列 示例：列出2015-01-03到2015-01-07每天的销量汇总 MySQL8： with recursive t(d,v) as (select date'2015-01-04',30 union all select date'2015-01-06',50 union all select date'2015-01-07',50 union all select date'2015-01-03',40 union all select date'2015-01-04', 80), s(d) as (select date'2015-01-03' union all select d+1 from s where d select s.d, sum(t.v) v from s left join t on s.d=t.d group by s.d; 集算器 SPL： A 1 [2015-01-04, 30, 2015-01-06,50, 2015-01-07,50, 2015-01-03,40, 2015-01-04,80] 2 =create(d,v).record(A1) 3 =periods("2015-01-03", "2015-01-07") 4 =A2.align@a(A3,d) 5 =A4.new(A3(#):d, ~.sum(v):v) A4：A2中记录按字段d的值对齐到A3 A5：根据A4和A3对位构造统计后的序表 3、 生成连续的工作日(不包含周六周日)序列 MySQL8: with recursive t(d) as (select date'2018-10-03' union all select d+1 from t where d select d,dayofweek(d) w from t where dayofweek(d)<=5; 集算器 SPL： A 1 =workdays(date("2018-10-03"), date("2018-10-09")) 2 =A1.new(~:d,day@w(~)-1:w) A1：构造从2018-10-03到2018-10-09不包含周六周日的日期序列 A2：根据A1构造日期及相应周几的序表 4、 根据序列生成表 MySQL8： with recursive t1(n) as (select 1 union all select n+1 from t1 where n<14), t2(n, name) as (select n, concat('a',n) name from t1) select max(if(n%4=1, name, null)) f1, max(if(n%4=2, name, null)) f2, max(if(n%4=3, name, null)) f3, max(if(n%4=0, name, null)) f4 from t2 group by floor((n+3)/4); 集算器 SPL： A 1 =to(14).("a"/~) 2 =create(f1,f2,f3,f4).record(A1)
