Order By 2 columns - sql

Question

I have this table and .
id | item_id | created_at
-----+-----------+------------
1 | Apple | 2017-03-21
2 | Grape | 2017-03-23
3 | Grape | 2017-03-24
4 | Apple | 2017-03-25
I want to order by created_at and also at the same time order by item_id like this:
id | item_id | created_at
-----+-----------+------------
4 | Apple | 2017-03-25
1 | Apple | 2017-03-21
3 | Grape | 2017-03-24
2 | Grape | 2017-03-23
So if i add a new row for item_id: Grape my new results should be like this:
id | item_id | created_at
-----+-----------+------------
5 | Grape | 2017-03-28 (NEW)
3 | Grape | 2017-03-24
2 | Grape | 2017-03-23
4 | Apple | 2017-03-25
1 | Apple | 2017-03-21
and then if i add new row for item_id: Apple it should be like this:
id | item_id | created_at
-----+-----------+------------
6 | Apple | 2017-03-28 (NEW)
4 | Apple | 2017-03-25
1 | Apple | 2017-03-21
5 | Grape | 2017-03-27
3 | Grape | 2017-03-24
2 | Grape | 2017-03-23
...So it orders by the latest created_at and show the other rows with the same item_id below it
I have tried ORDER BY created_at, item_id DESC but it does not works and give me this instead:
id | item_id | created_at
-----+-----------+------------
6 | Apple | 2017-03-28
5 | Grape | 2017-03-27
4 | Apple | 2017-03-25
3 | Grape | 2017-03-24
2 | Grape | 2017-03-23
1 | Apple | 2017-03-21

Answer 1

SQLFiddle: PostgreSQL
WITH grouped AS (
select item_id, max(created_at) as max_dt
from tbl
group by item_id
)
SELECT tbl.*
FROM grouped
LEFT JOIN tbl USING (item_id)
ORDER BY grouped.max_dt desc,
grouped.item_id, -- important if apple and grape both have same max dt
tbl.created_at desc;
subquery to get the most recent dates by item (the item grouping)
left join those results on the actual table to get desired records

Answer 2

You need to first identify your first rule which is get the max created_at per item_id, therefore the subquery with the item_id grouped and a row_number() added so it can know which fruit cames first. Then just join with your table.
SELECT tt.id, tt.item_id, tt.created_at
FROM test_table tt
INNER JOIN
(SELECT item_id,
MAX(created_at),
ROW_NUMBER() OVER (ORDER BY MAX(created_at) DESC) rn
FROM test_table
GROUP BY item_id
ORDER BY 3) ord
ON tt.item_id = ord.item_id
ORDER BY ord.rn, tt.created_at DESC;

Answer 3

For mysql, try this:
select item.*
from item
left join (
select item_id, max(created_at) maxdate
from item
group by item_id
) t on item.item_id = t.item_id
order by t.maxdate desc, item.created_at desc
Demo1 in SQLFiddle
Demo2 in SQLFiddle

Answer 4

Here is a mysql version
set #ordr = 0;
select t.*
from tbl t
join (
select o.item_id, o.created_at, #ordr := #ordr + 1 as `order`
from (
select tbl.item_id, max(tbl.created_at) as created_at
from tbl
group by tbl.item_id
order by created_at desc
) as o
) sub
on t.item_id = sub.item_id
order by sub.`order`, sub.created_at desc

Answer 5

Please try -
SELECT id,
item_ID,
created_at
FROM tblOrders
ORDER BY item_ID,
item_ID = ( SELECT item_ID
FROM tblOrders
WHERE id = ( SELECT MAX( id ) FROM tblOrders ) ),
created_at;