SQL Server loop through a table for every 5 rows - sql

Question

I need to write a stored procedure or table function to return a new data table as a new data source.
I wish to loop through the original table for every 5 rows base on the invoice ID column (it's possible not start from 1), the first 5 rows add to the left of the new table and the second 5 rows add to the right of the new table, the third 5 rows to the left and so on.
For example, Here is the original table:
Here is the expect table:
Thanks in advance!

Answer 1

declare #rowCount int = 5;
with cte as (
select *,( (IN_InvoiceID-1) / #rowCount ) % 2 group1
,( (IN_InvoiceID-1) / #rowCount ) group2
,IN_InvoiceID % #rowCount group3
from T
)
select * from cte
select T1.INID,T1.IN_InvoiceID,T1.IN_InvoiceAmount,T2.INID,T2.IN_InvoiceID,T2.IN_InvoiceAmount
from CTE T1
left join CTE T2 on T2.group1 = 1 and T1.group2 = T2.group2-1 and T1.group3 = T2.group3
where T1.group1 = 0
Test DDL
CREATE TABLE T
([INID] varchar(38), [IN_InvoiceID] int, [IN_InvoiceAmount] int)
;
INSERT INTO T
([INID], [IN_InvoiceID], [IN_InvoiceAmount])
VALUES
('DB3E17E6-35C5-41:121-93B1-F809BF6B2972', 1, 2999),
('3212F048-8213-4FCC-AB64-121485B77D4E43', 2, 3737),
('E3526373-A204-40F5-801C-7F8302A4E5E2', 3, 3175),
('76CC9C19-BF79-4E8A-8034-A33805AD3390', 4, 391),
('EC7A2FBC-B62D-4865-88DE-A8097975F125', 5, 1206),
('52AD3046-21331-4F0A-BD1D-67F232C54244', 6, 402),
('CA48F132-A9F5-4516-9E58-CDEE6644AAD1', 7, 1996),
('02E10C31-CAB2-4220-B66A-CEE5E67A9378', 8, 3906),
('98F1EEFF-B07A-4B65-87F4-E165264284DD', 9, 2575),
('91EBDD8B-B73C-470C-8900-DD66078483DB', 10, 2965),
('6E2490E5-C4DE-4833-877F-1590F7BDC1B8', 11, 1603),
('00985921-AC3C-4E3E-BAE1-7F58302F831A', 12, 1302)
;
Result:

Answer 2

Could you please check article Display Data in Multiple Columns using SQL showing with example case how a database developer can show the list of data rows in a columnar mode using Row_Number() function and mode arithmetic expression
You need to add additional columns from the same row that is different in the sample

Answer 3

Seems as if you want to split the table into 2 tables with alternating 5 rows. An easy way to do this would be:
Take data into a temp table having an extra column (lets say
grouping_id)
Update the grouping id so that each 5 rows have the same id. You can
use in_invoiceId % 5 (the nod function). After this step the first 5
rows will have grouping_id 0, next 5 will have 1, next will have 2
(assuming your invoice id is incremented +1 for all rows).
You can just do a normal select with where clause for odd and even grouping_id

Answer 4

Ideally, you can manage with the 2 tables Master and detail table.
But due to my curiosity, I am able to solve and give the answer as
Declare #table table(id int identity, invoice_id int)
; WITH Numbers AS
(
SELECT n = 1
UNION ALL
SELECT n + 1
FROM Numbers
WHERE n+1 <= 50
)
insert into #table SELECT n
FROM Numbers
Select (a.id )%5 ,* from #table a join #table b on a.id+5 = b.id and a.id != b.id
;WITH Numbers AS
(
SELECT n = 1, o = 5
UNION ALL
SELECT n + 10, o = o+10
FROM Numbers
WHERE n+1 <= 50
)
select a.id ParentId,a.invoice_id ParentInvoiceId, --b.n, b.o,
c.invoice_id childInvoiceID from #table a
join Numbers b on a.id between b.n and b.o
left join #table c on a.id + 5 = c.id

Answer 5

Here is my solution
First i create grps based on whether the in_invoiceid is divisible by 5 or not.(Ignore the remainders)
After that i create a category to indicate between alternative groups(ie by checking if the remainder is 0 or otherise)
Then its a matter of dense_ranking the records on the basis of the category field ordered by in_invoiceid
Lastly a join with category=1 rows with same dense_rank as those records in category=0
create table Invoicetable(IN_ID varchar(100), IN_InvoiceID int)
INSERT INTO Invoicetable (IN_ID, IN_InvoiceID)
VALUES
('2345-BCDE-6645-1DDF', 1),
('2345-BCDE-6645-3DDF', 2),
('2345-BCDE-6645-4DDF', 3),
('2345-BCDE-6645-5DDF', 4),
('2345-BCDE-6645-6DDF', 5),
('2345-BCDE-6645-7DDF', 6),
('2345-BCDE-6645-aDDF', 7),
('2345-BCDE-6645-sDDF', 8),
('2345-BCDE-6645-dDDF', 9),
('2345-BCDE-6645-dDDF', 10),
('2345-BCDE-6645-dDDF', 11),
('2345-BCDE-6645-dDDF', 12);
with data
as (
select *
,(in_invoiceid-1)/5 as grp
,case when ((in_invoiceid-1)/5)%2=0 then '1' else '0' end as category
,dense_rank() over(partition by case when ((in_invoiceid-1)/5)%2=0 then '1' else '0' end
order by in_invoiceid) as rnk
from invoicetable a
)
select *
from data a
left join data b
on a.rnk=b.rnk
and b.category=0
where a.category=1
Here is db fiddle link.
https://dbfiddle.uk/?rdbms=sqlserver_2017&fiddle=287f101737c580ca271940764b2536ae

Answer 6

You may try with the following approach. Dividing the table is done with (((ROW_NUMBER() OVER (ORDER BY IN_InvoiceID) - 1) / 5) % 2 = 0) which groups records in left and right groups.
CREATE TABLE #InvoiceTable(
IN_ID varchar(24),
IN_InvoiceID int
)
INSERT INTO #InvoiceTable (IN_ID, IN_InvoiceID)
VALUES
('2345-BCDE-6645-1DDF', 1),
('2345-BCDE-6645-3DDF', 2),
('2345-BCDE-6645-4DDF', 3),
('2345-BCDE-6645-5DDF', 4),
('2345-BCDE-6645-6DDF', 5),
('2345-BCDE-6645-7DDF', 6),
('2345-BCDE-6645-aDDF', 7),
('2345-BCDE-6645-sDDF', 8),
('2345-BCDE-6645-dDDF', 9),
('2345-BCDE-6645-dDDF', 10),
('2345-BCDE-6645-dDDF', 11),
('2345-BCDE-6645-dDDF', 12);
WITH cte AS (
SELECT
IN_ID,
IN_InvoiceID,
CASE
WHEN (((ROW_NUMBER() OVER (ORDER BY IN_InvoiceID) - 1) / 5) % 2 = 0) THEN 'L'
ELSE 'R'
END AS IN_Position
FROM #InvoiceTable
),
cteL AS (
SELECT IN_ID, IN_InvoiceID, ROW_NUMBER() OVER (ORDER BY IN_InvoiceID) AS IN_RowNumber
FROM cte
WHERE IN_Position = 'L'
),
cteR AS (
SELECT IN_ID, IN_InvoiceID, ROW_NUMBER() OVER (ORDER BY IN_InvoiceID) AS IN_RowNumber
FROM cte
WHERE IN_Position = 'R'
)
SELECT cteL.IN_ID, cteL.IN_InvoiceID, cteR.IN_ID, cteR.IN_InvoiceID
FROM cteL
LEFT JOIN cteR ON (cteL.IN_RowNumber = cteR.IN_RowNumber)
Output:
IN_ID IN_InvoiceID IN_ID IN_InvoiceID
2345-BCDE-6645-1DDF 1 2345-BCDE-6645-7DDF 6
2345-BCDE-6645-3DDF 2 2345-BCDE-6645-aDDF 7
2345-BCDE-6645-4DDF 3 2345-BCDE-6645-sDDF 8
2345-BCDE-6645-5DDF 4 2345-BCDE-6645-dDDF 9
2345-BCDE-6645-6DDF 5 2345-BCDE-6645-dDDF 10
2345-BCDE-6645-dDDF 11 NULL NULL
2345-BCDE-6645-dDDF 12 NULL NULL