Selecting only the first two items from and order - mysql

Question

I need you help regarding something, i have 3 tables ORDERS, ORDER_ITEM, ORDER_ITEM_LINE.
CREATE TABLE orders
(`id` int, `date` datetime)
;
INSERT INTO orders
(`id`, `date`)
VALUES
(78, '2017-01-03 00:00:00'),
(79, '2017-02-03 00:00:00'),
(80, '2017-03-03 00:00:00'),
(81, '2017-04-03 00:00:00'),
(82, '2017-05-03 00:00:00'),
(83, '2017-06-03 00:00:00'),
(84, '2017-07-03 00:00:00')
;
CREATE TABLE order_item
(`id` int, `fk_o_id` int, `sku` int)
;
INSERT INTO order_item
(`id`, `fk_o_id`, `sku`)
VALUES
(10, 78, 123),
(11, 79, 124),
(12, 79, 125),
(13, 80, 126),
(14, 82, 127),
(15, 82, 128),
(16, 82, 129)
;
CREATE TABLE order_item_line
(`id` int, `fk_oi_id` int, `line_id` int)
;
INSERT INTO order_item_line
(`id`, `fk_oi_id`, `line_id`)
VALUES
(33, 10, 1),
(34, 11, 1),
(35, 12, 2),
(36, 13, 1),
(37, 14, 1),
(38, 15, 2),
(39, 16, 3)
;
I would like to display all orders with 2 or more than 2 items but only first two so it will be line_id - 1 and 2.
The outcome should look like:
Outcome
If you have any ideas, thank you in advance.

Answer 1

To get the result you require, you will need to create another table. In this example I created a table called TESTQUERY and inserted data to count how many times the orders id appeared
Table creation
CREATE TABLE TESTQUERY
(`id` int, `count` int)
Data into the test table
INSERT INTO TESTQUERY
(
SELECT o.id, COUNT(o.id) as count FROM orders o
JOIN order_item oi ON oi.fk_o_id = o.id
JOIN order_item_line oil ON oil.fk_oi_id = oi.id
GROUP BY o.id
)
I then queried against all for databases using the query below and it returned your desired outcome
SELECT o.id, oi.sku, oil.line_id FROM orders o
JOIN order_item oi ON oi.fk_o_id = o.id
JOIN order_item_line oil ON oil.fk_oi_id = oi.id
JOIN TESTQUERY t ON t.id = o.id
WHERE t.count > 1 AND oil.line_id < 3
I hope this helps