Efficient SQL to find specific ID used with pagination - sql

Question

At work I need to implement a feature in the API that returns a specific page (20 entries) that contains the entry with a specified ID. That entry could be any of those 20 in the page.
Normally, a page is determined by taking the ID of the last element of the previous page, applying a filter to the elements after the previous ID and take the first 20 entries of the result.
But with the new feature, you‘re supposed to receive the page that CONTAINS the specified ID, rather than using it to determine the first element of a new page.
I‘m not working with databases that much so I‘m not sure, but since I‘m the only developer in this company, there‘s no one I can ask. If it helps, the database is MS SQL Server. If more info is needed, I can give it, as long as it‘s not against company policy.

Answer 1

Let's say we look for ID 123456. This can be anywhere from page 1 to page 6173 (or even missing completely). There is no way to tell other than to count the rows/pages until we get there. And we even must count on to get all following rows that are still on the same page. This is not difficult, but rather slow.
We cannot know which ID comes after another; after ID 5 the next may be ID 6 or 1234 or 1000000 or whatever. So the first step is to number all rows ordered by ID. Or rather assign them pages, for we know numbers 1 to 20 = page 1, numbers 21 to 40 = page 2, etc. Thus we get to know that our ID is on page X and we must select all rows marked with page X.
with rows_with_page as
(
select
t.*,
(row_number() over (order by id) - 1) / 20 + 1 as page
from mytable t
)
select *
from rows_with_page
where page = (select page from rows_with_page where id = #id)
order by id;
As a row number is an integer, the division with / is an integer division resulting in an integer in SQL Server (like in elementary school; 5 / 2 = 2).

Answer 2

If I understand your question, you have a fixed page size of 20 records per page. You will receive a ID value and that ID value can appear anywhere on a given page of 20 records. So, assuming the ID values start at 1, if you receive an ID value between 1 and 20 it should return all the records on page 1, if the ID value received is between 21 and 40 it should return all the records on page 2, etc.
We can figure out what page we are on by doing integer division - divide the ID by 20 and then add 1. So for ID 4, (4 / 20) is zero (remember integer division!) and then add 1 to get page 1. You can use the LIMIT...OFFSET feature of SQL Server (assuming you are using version that supports it).
I will create a simple script to show how this works. I don't know if you plan is to create a stored procedure or what, but you can adapt this technique. So let's make a simple table variable to mimic your table, add 100 records for some sample data, and then write a query to retrieve the records on the appropriate page.
-- create a table for the query
DECLARE #Records AS TABLE(
ID INT IDENTITY(1,1) NOT NULL,
[Value] INT NOT NULL
);
-- populate with 100 sample records
DECLARE #i INT;
SELECT #i = 1;
WHILE (#i <= 100)
BEGIN
INSERT INTO #Records([Value]) VALUES (#i);
SELECT #i = #i + 1;
END
-- now find the records on the correct page
DECLARE #id INT = 41; -- the record to find
DECLARE #pageSize INT = 20; -- the number of records per page
DECLARE #page INT; -- the page, counting from 1
SELECT #page = (#id / 20) + 1;
SELECT ID, [Value]
FROM #Records
ORDER BY ID
OFFSET (#page -1 ) * #pageSize ROWS
FETCH NEXT #pageSize ROWS ONLY;
That should work for you (if I understand what you were asking).