Let's consider that there are three material types such as ('COTTON', 'LEATHER', 'SILK') and I want to fetch the dress_id's which has all theses three material types. I want to rank them as well.
Can someone explain step by step on how to do this ? I came through few examples and none of them seems to be clear to me.
The output should look something like
DRESS_ID MATERIAL LAST_UPDATED_DATE RANK
111 COTTON 2019-08-29 1
111 SILK 2019-08-30 2
111 LEATHER 2019-08-31 3
222 COTTON 2019-08-29 1
222 SILK 2019-08-30 2
222 LEATHER 2019-08-31 3
222 LEATHER 2019-09-02 4
I get an error in MYSQL work bench while executing this query. Error Code: 1305. FUNCTION rank does not exist.
SELECT dress_id,
rank() over(PARTITION BY dress_id, material ORDER by LAST_UPDATED_DATE asc) as rank
FROM dress_types;
In earlier versions of MySQL, you can either use variables or a correlated subquery.
Because you have only a handful of materials for each dress, a correlated subquery is reasonable, particularly with the right index. The code looks like:
SELECT d.dress_id, d.material,
(SELECT COUNT(*)
FROM dress_types d2
WHERE d2.dress_id = d.dress_id AND
d2.last_updated_date <= d.last_updated_date
) as rank
FROM dress_types d;
Note that this implements the logic based on your data not the query. The corresponding query would be:
SELECT dress_id,
rank() over (PARTITION BY dress_id ORDER by LAST_UPDATED_DATE asc) as rank
FROM dress_types;
The index that you want is on dress_types(dress_id, last_updated_date)
.
Actually, these are the same so long as there are no duplicates (by date). The logic may be different if there are duplicates.
SELECT dress_id
, material
, LAST_UPDATED_DATE
rank() over(PARTITION BY dress_id ORDER by LAST_UPDATED_DATE asc) as rank
FROM dress_types
SELECT T.*,
CASE WHEN @prev_dress_id != T.dress_id THEN @rank:=1
ELSE @rank:=@rank+1
END as rank,
@prev_dress_id := T.dress_id as set_prev_dress_id
FROM
(SELECT dress_id,material,last_updated_date
FROM dress_types T1
WHERE EXISTS (SELECT 1 FROM dress_types E1 WHERE E1.dress_id = T1.dress_ID AND E1.material = 'COTTON')
AND EXISTS (SELECT 1 FROM dress_types E2 WHERE E2.dress_id = T1.dress_ID AND E2.material = 'SILK')
AND EXISTS (SELECT 1 FROM dress_types E3 WHERE E3.dress_id = T1.dress_ID AND E3.material = 'LEATHER')
ORDER BY dress_id asc,last_updated_date asc
)T,(SELECT @prev_dress_id:=-1)V
The inner select selects dresses that have existence of all 3 materials and ordered by dress_id, last_updated_date. The outer joins it with a prev_dress_id variable that can be set at the end of each row. The the logics in case statement to calculate rank based on @prev_dress_id != or = T.dress_id. sqlfiddle
For previous versions of MySQL 8.0 you must use variables to simulate the ranking:
SET @rownum := 0;
SET @group_number := 0;
SELECT dress_id, material, last_updated_date, rank FROM (
SELECT @rownum := case
when @group_number = dress_id then @rownum + 1
else 1
end AS rank, dress_id, material, last_updated_date,
@group_number := dress_id
FROM dress_types
ORDER BY
dress_id,
FIELD(material, 'COTTON', 'SILK', 'LEATHER'),
last_updated_date
) t
See the demo .
Results:
| dress_id | material | last_updated_date | rank |
| -------- | -------- | ------------------- | ---- |
| 111 | COTTON | 2019-08-29 00:00:00 | 1 |
| 111 | SILK | 2019-08-30 00:00:00 | 2 |
| 111 | LEATHER | 2019-08-31 00:00:00 | 3 |
| 222 | COTTON | 2019-08-29 00:00:00 | 1 |
| 222 | SILK | 2019-08-30 00:00:00 | 2 |
| 222 | LEATHER | 2019-08-31 00:00:00 | 3 |
| 222 | LEATHER | 2019-09-02 00:00:00 | 4 |
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