195
votes

Can I run a select statement and get the row number if the items are sorted?

I have a table like this:

mysql> describe orders;
+-------------+---------------------+------+-----+---------+----------------+
| Field       | Type                | Null | Key | Default | Extra          |
+-------------+---------------------+------+-----+---------+----------------+
| orderID     | bigint(20) unsigned | NO   | PRI | NULL    | auto_increment |
| itemID      | bigint(20) unsigned | NO   |     | NULL    |                |
+-------------+---------------------+------+-----+---------+----------------+

I can then run this query to get the number of orders by ID:

SELECT itemID, COUNT(*) as ordercount
FROM orders
GROUP BY itemID ORDER BY ordercount DESC;

This gives me a count of each itemID in the table like this:

+--------+------------+
| itemID | ordercount |
+--------+------------+
|    388 |          3 |
|    234 |          2 |
|   3432 |          1 |
|    693 |          1 |
|   3459 |          1 |
+--------+------------+

I want to get the row number as well, so I could tell that itemID=388 is the first row, 234 is second, etc (essentially the ranking of the orders, not just a raw count). I know I can do this in Java when I get the result set back, but I was wondering if there was a way to handle it purely in SQL.

Update

Setting the rank adds it to the result set, but not properly ordered:

mysql> SET @rank=0;
Query OK, 0 rows affected (0.00 sec)

mysql> SELECT @rank:=@rank+1 AS rank, itemID, COUNT(*) as ordercount
    -> FROM orders
    -> GROUP BY itemID ORDER BY rank DESC;
+------+--------+------------+
| rank | itemID | ordercount |
+------+--------+------------+
|    5 |   3459 |          1 |
|    4 |    234 |          2 |
|    3 |    693 |          1 |
|    2 |   3432 |          1 |
|    1 |    388 |          3 |
+------+--------+------------+
5 rows in set (0.00 sec)
5
For future reference: If you want to order from rank 1 to rank 5, use ORDER BY rank ASC (ordering by rank in ASCending order). I guess that is what you mean by but not properly orderedBlueCacti

5 Answers

192
votes

Take a look at this.

Change your query to:

SET @rank=0;
SELECT @rank:=@rank+1 AS rank, itemID, COUNT(*) as ordercount
  FROM orders
  GROUP BY itemID
  ORDER BY ordercount DESC;
SELECT @rank;

The last select is your count.

181
votes
SELECT @rn:=@rn+1 AS rank, itemID, ordercount
FROM (
  SELECT itemID, COUNT(*) AS ordercount
  FROM orders
  GROUP BY itemID
  ORDER BY ordercount DESC
) t1, (SELECT @rn:=0) t2;
36
votes

Swamibebop's solution works, but by taking advantage of table.* syntax, we can avoid repeating the column names of the inner select and get a simpler/shorter result:

SELECT @r := @r+1 , 
       z.* 
FROM(/* your original select statement goes in here */)z, 
(SELECT @r:=0)y;

So that will give you:

SELECT @r := @r+1 , 
       z.* 
FROM(
     SELECT itemID, 
     count(*) AS ordercount
     FROM orders
     GROUP BY itemID
     ORDER BY ordercount DESC
    )z,
    (SELECT @r:=0)y;
12
votes

You can use MySQL variables to do it. Something like this should work (though, it consists of two queries).

SELECT 0 INTO @x;

SELECT itemID, 
       COUNT(*) AS ordercount, 
       (@x:=@x+1) AS rownumber 
FROM orders 
GROUP BY itemID 
ORDER BY ordercount DESC; 
6
votes

It's now builtin in MySQL 8.0 and MariaDB 10.2:

SELECT
  itemID, COUNT(*) as ordercount,
  ROW_NUMBER OVER (PARTITION BY itemID ORDER BY rank DESC) as rank
FROM orders
GROUP BY itemID ORDER BY rank DESC