'MySQL difference between two rows of a SELECT Statement
I am trying to make the difference of two rows in an mysql database.
I have this table containing ID, kilometers, date, car_id, car_driver etc...
Since I don't always enter the information in the table in the correct order, I may end up with information like this:
ID | Kilometers | date | car_id | car_driver | ...
1 | 100 | 2012-05-04 | 1 | 1
2 | 200 | 2012-05-08 | 1 | 1
3 | 1000 | 2012-05-25 | 1 | 1
4 | 600 | 2012-05-16 | 1 | 1
With a select statement I am able to sort my table correctly:
SELECT * FROM mytable ORDER BY car_driver ASC, car_id ASC, date ASC
I will obtain this:
ID | Kilometers | date | car_id | car_driver | ...
1 | 100 | 2012-05-04 | 1 | 1
2 | 200 | 2012-05-08 | 1 | 1
4 | 600 | 2012-05-16 | 1 | 1
3 | 1000 | 2012-05-25 | 1 | 1
Now I would like to make a view where basically I have this extra information: Number of kilometers since last date and I would like to obtain something like this:
ID | Kilometers | date | car_id | car_driver | number_km_since_last_date
1 | 100 | 2012-05-04 | 1 | 1 | 0
2 | 200 | 2012-05-08 | 1 | 1 | 100
4 | 600 | 2012-05-16 | 1 | 1 | 400
3 | 1000 | 2012-05-25 | 1 | 1 | 400
I thought of doing an INNER JOIN to perform what I wanted, but I have the feeling I can't do the join on my ID since they are not sorted correctly.
Is there a way to achieve what I want?
Shall I create a view with a sort of row_number that I can then used in my INNER JOIN?
Solution 1:[1]
SELECT
mt1.ID,
mt1.Kilometers,
mt1.date,
mt1.Kilometers - IFNULL(mt2.Kilometers, 0) AS number_km_since_last_date
FROM
myTable mt1
LEFT JOIN myTable mt2
ON mt2.Date = (
SELECT MAX(Date)
FROM myTable mt3
WHERE mt3.Date < mt1.Date
)
ORDER BY mt1.date
Sql Fiddle
Or, by emulating a lag() function through MySql hackiness...
SET @kilo=0;
SELECT
mt1.ID,
mt1.Kilometers - @kilo AS number_km_since_last_date,
@kilo := mt1.Kilometers Kilometers,
mt1.date
FROM myTable mt1
ORDER BY mt1.date
Sql Fiddle
Solution 2:[2]
In Postgres, Oracle and SQL-Server 2012, this is plain simple, using the LAG() function:
SELECT
id, kilometers, date,
kilometers
- COALESCE( LAG(kilometers) OVER (ORDER BY date ASC, car_driver ASC, id ASC)
, kilometers)
AS number_km_since_last_date
FROM
mytable ;
In MySQL, we have to do some nasty constructions. Either an inline subquery (with probably not very good performance):
SELECT
id, kilometers, date,
kilometers - COALESCE(
( SELECT p.kilometers
FROM mytable AS p
WHERE ( p.date = m.date AND p.car_driver = m.car_driver
AND p.id < m.id
OR p.date = m.date AND p.car_driver < m.car_driver
OR p.date < m.date
)
ORDER BY p.date DESC, p.car_driver DESC
LIMIT 1
), kilometers)
AS number_km_since_last_date
FROM
mytable AS m ;
or a self-join (already provided by @Michael Fredrickson) or using MySQL variables (already provided as well).
If you want the counter to start again from 0 for every car_id, which would be done with PARTITION BY in many other DBMS:
SELECT
id, kilometers, date,
kilometers
- COALESCE( LAG(kilometers) OVER (PARTITION BY car_id
ORDER BY date ASC, car_driver ASC, id ASC)
, kilometers)
AS number_km_since_last_date
FROM
mytable ;
it could be done in MySQL like this:
SELECT
id, kilometers, date,
kilometers - COALESCE(
( SELECT p.kilometers
FROM mytable AS p
WHERE p.car_id = m.car_id
AND ( p.date = m.date AND p.car_driver = m.car_driver
AND p.id < m.id
OR p.date = m.date AND p.car_driver < m.car_driver
OR p.date < m.date
)
ORDER BY p.date DESC, p.car_driver DESC
LIMIT 1
), kilometers)
AS number_km_since_last_date
FROM
mytable AS m ;
Solution 3:[3]
With data unsorted I can only think of inline subquery (not a good idea on the large table):
select t1.*,
t1.Kilometers - (select top 1 kilometers from mytable t2 where t2.date < t1.date order by t2.date desc) as number_km_since_last_date
from mytable t1
If you get data sorted you can use left join
select t1.*
t1.Kilometers - t2.Kilometers as number_km_since_last_date
from mytable t1
left join mytable t2
on t1.id = t2.id + 1
You can probably tell that I'm more of a TSQL guy so you might need to adjust syntax for MySQL.
Solution 4:[4]
Here's an example of using CURSOR for this use case as well
CREATE TABLE TEMP1
(
MyDate DATETIME,
MyQty INT
)
INSERT INTO TEMP1 VALUES ('01/08/17', 100)
INSERT INTO TEMP1 VALUES ('01/09/17', 120)
INSERT INTO TEMP1 VALUES ('01/10/17', 180)
DECLARE @LastDate DATETIME = NULL
DECLARE @LastQty INT = NULL
DECLARE @MyDate DATETIME = NULL
DECLARE @MyQty INT = NULL
DECLARE mycursor CURSOR FOR
SELECT MyDate, MyQty FROM TEMP1 ORDER BY MyDate
OPEN mycursor
FETCH NEXT FROM mycursor INTO @MyDate, @MyQty
WHILE @@FETCH_STATUS = 0
BEGIN
SELECT @MyDate, @MyQty - @LastQty
SET @LastDate = @MyDate
SET @LastQty = @MyQty
FETCH NEXT FROM mycursor INTO @MyDate, @MyQty
END
CLOSE mycursor
DEALLOCATE mycursor
Solution 5:[5]
With MySQL 8 you can use CTE and ROW_NUMBER window function to make a more readable query
WITH cte_name AS (
SELECT
ROW_NUMBER() OVER (ORDER BY update_time) as row_num,
id,
other_data,
update_time
FROM table_name WHERE condition = 'some_condition'
)
SELECT t2.id, t2.other_data, TIMEDIFF(t2.update_time, t1.update_time) AS time_taken
FROM
cte_name t1
JOIN cte_name t2 ON t1.row_num = t2.row_num-1
ORDER BY time_taken;
In this example I'm trying get the difference between datetime values.
- The idea is to use ROW_NUMBER window function to assign an incremental number to each row after ordering by update_time.
- The CTE allows us to write a subquery without having to repeat writing the same code.
- We self join the CTE. The joining condition is basically - each n?? item of the second subquery joins with the n-1?? item of the first subquery (this also means the first row will disappear from the result set. if you need it you can use a UNION to add the first row to the start).
There are some good tutorials for: CTE (Common Table Expression), ROW_NUMBER and even window functions
Solution 6:[6]
It is a way to solve this using mySQL's window functions:
SELECT
id,
kilometers,
date,
car_id,
car_driver,
COALESCE(kilometers - LAST_VALUE(kilometers) OVER(PARTITION BY car_id, car_driver ORDER BY date ROWS BETWEEN UNBOUNDED PRECEDING AND 1 PRECEDING), 0) AS number_km_since_last_date
FROM mytable
ORDER BY car_driver, car_id, date
Sources
This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.
Source: Stack Overflow
| Solution | Source |
|---|---|
| Solution 1 | |
| Solution 2 | |
| Solution 3 | Yan Sklyarenko |
| Solution 4 | |
| Solution 5 | shaahiin |
| Solution 6 |
