Overcome SQL Headache – Generating a Sequence

1. Generating a continuous integer sequence

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;

SPL:

A
1 =to(1,7)

A1: Create a sequence consisting integers from 1 to 7

001png

Example 1: As Hundred Fowls Problem asks: If a rooster is worth 5 coins, a hen 3 coins, and 3 chicks together are worth 1 coin. How many roosters, hens and chicks totaling 100 can be bought for 100 coins?

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 Rooster, jm.n Hen, jc.n Chick
  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

SPL:

A
1 =to(100/5)
2 =to(100/3)
3 =33.(~*3)
4 =create(Rooster,Hen,Chick)
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: Create an integer sequence from 1 to 20;
A2: Create an integer sequence from 1 to 33;
A3: Create an integer sequence from 1 to 99 with the step being 3;
A4: Create a table sequence consisting of fields Rooster, Hen, and Chick;
A5: Perform nested loop through members in A1, A2 and A3. If A1’s member + A2’s member + A3’s member ==100 & A1’s member *5+A2’s member *3+A3’s member/3==100, insert them to A4’s table sequence.

002png

Example 2: Convert colon-delimited strings in certain columns into multiple rows.

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;

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: Create a table sequence whose data structure is (k,f) and add two records - (1, “a1:a2:a3) and (2,”b1:b2”) to it;
A2: Separate each value of A1’s f field into a sequence by colon and reassign the sequence members to the field;
A3: Create a table sequence whose data structure is (k,f) over each of A1’s records, generate records in the form of (A1.k,member of f field) according to members of each f field value, and insert them into the newly-created sequence.

003png

2. Generating a continuous date sequence

MySQL8:
with recursive
  t(d) as (select date’2018-10-03’
      union all
      select d+1 from t where d<date’2018-10-09’)
select d,dayofweek(d) w from t;

SPL:

A
1 =periods("2018-10-03", "2018-10-09")
A1: Create a date sequence from date 2018-10-03 to date 2018-10-09;

004png

Example: List sales of each day from 2015-01-03 to 2015-01-07.

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<date’2015-01-07’)
select s.d, sum(t.v) v
from s left join t on s.d=t.d
group by s.d;

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: Align A2’s records to A3 according to d field values;
A5: Create a table sequence according to the correspondence between A4 and A3.

005png

3. Generating a continuous sequence of workdays(excluding Saturdays and Sundays)

MySQL8:
with recursive
  t(d) as (select date’2018-10-03’
    union all
    select d+1 from t where d<date’2018-10-09’)
select d,dayofweek(d) w from t
where dayofweek(d)<=5;

SPL:

A
1 =workdays(date("2018-10-03"), date("2018-10-09"))
2 =A1.new(~:d,day@w(~)-1:w)

A1: Create a date sequence without Saturdays and Sundays from 2018-10-03 to 2018-10-09;
A2: Create a table sequence consisting of date and the corresponding day according to A1.

006png

4. Generating a table from a sequence

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);

SPL:

A
1 =to(14).("a"/~)
2 =create(f1,f2,f3,f4).record(A1)

007png