SQL从表中选择匹配最低ID的所有行

Question

I have table like so:

    id  |  order_id  |  product_id
------------------------------------
    1   |  356       |  80
    2   |  356       |  81
    3   |  357       |  77
    4   |  357       |  80
    5   |  357       |  78
    6   |  358       |  90

What I would like to be able to do is to grab the 2 lowest order_id values without knowing how many there are to begin with, but I am having trouble thinking of how to create the query to select these rows...

SELECT * FROM table ORDER BY order_id ASC

Is there a way to do this within MySQL, that will always return the lowest order_id value rows, so in this case, I should get the first 2 rows returned. If the first 2 rows didn't exist, than I should get returned rows 3 - 5 (totaling 3 rows for order id 357).

How to do this?

Answer 1

Please try this:

---

SELECT 
*
FROM 
table INNER JOIN 
(
    SELECT MIN(order_id) AS min_order_id
    FROM table
) AS t
ON t.order_id = table.order_id;

• First get the minimum order id and give it an alias t.
• Now Make an inner join between your table and this alias t ON order_id.