How to get column value from next record - mysql

Question

I have a following result set:
POST | DATE
--------------------------------------
Senior Software Engg. | 2018-04-18
Software Engg. | 2017-04-18
Assoc. Software Engg. | 2016-04-18
SQL query:
SELECT DISTINCT designation_id as id, d.title as POST, DATE(dt_datetime) as DATE
FROM users_history_check u
INNER JOIN
designations d
ON d.id = u.designation_id
WHERE u.id = $userID
ORDER BY DATE DESC
I want to fetch next record and perform date difference calculation in months, and display records.
Expected Output :
POST | Start DATE | End DATE | MONTHS
---------------------------------------------------------------
Senior Software Engg. | 2018-04-18 | - |
Software Engg. | 2017-04-18 | 2018-04-18 | 12
Assoc. Software Engg. | 2016-04-18 | 2017-04-18 | 12
Something like :
SELECT DISTINCT designation_id as id, d.title as POST, DATE(dt_datetime) as Start DATE, NEXT_RECORD(DATE(dt_datetime)) as End DATE, DATEDIFF(Start DATE, End DATE) as MONTHS....
Any help is very much appreciated. Thanks.

Answer 1

SELECT `POST`,
`DATE`,
IFNULL(END_DATE,'') AS END_DATE,
IFNULL(MONTH,'') AS MONTH
FROM
(SELECT `POST`,
`DATE`,
#prev AS END_DATE,
TIMESTAMPDIFF(month,DATE,#prev) AS MONTH,
#prev := T.DATE AS VarDate
FROM Table1 T,
(SELECT #prev:=null)R
) T1
OUTPUT
POST DATE END_DATE MONTH
Senior Software Engg. 2018-04-18
Software Engg. 2017-04-18 2018-04-18 12
Assoc. Software Engg. 2016-04-18 2017-04-18 12
Demo Link
http://sqlfiddle.com/#!9/33260/15
EXPLANATION:
In Sub query, I am saving the Date value in #prev variable and in each row using that variable to calculate the END_DATE before assigning the current date value from Column Date.
Then using the sub query to present the data in a proper way.

Answer 2

You can get the previous date using variables:
SELECT id, post, date,
(CASE WHEN (#tmp_prevd := #prevd) = NULL THEN NULL -- never happens
WHEN (#prevd := date) = NULL THEN NULL -- never happens
ELSE #tmp_prevd
END) as prev_date
FROM (SELECT DISTINCT designation_id as id, d.title as POST, DATE(dt_datetime) as DATE
FROM users_history_check u INNER JOIN
designations d
ON d.id = u.designation_id
WHERE u.id = $userID
ORDER BY DATE DESC
) ud CROSS JOIN
(SELECT #prevd := NULL) params;
This is tricky, because all references to a variable need to be in the same expression. That is why this uses CASE in a rather arcane way.
In MySQL 8.0 and basically all other databases, you could use LEAD() instead.

Answer 3

Try This....
SELECT T1.POST,T1.DATE ,T2.DATE,DATEDIFF(MONTH,T1.DATE,T2.DATE)
FROM(
SELECT ROW_NUMBER()OVER(ORDER BY DATE DESC) AS SlNo,*
FROM Mytable)T1
LEFT JOIN (SELECT ROW_NUMBER()OVER(ORDER BY DATE DESC)+1 AS SlNo,*
FROM Mytable)T2
ON(T1.SlNo = T2.SlNo )

Answer 4

I suggest using self-join like this
select d1.post,
d1.d `start DATE`,
min(d2.d) `end DATE`,
timestampdiff(month, d1.d, min(d2.d)) `MONTHS`
from data d1
left join data d2 on d1.d < d2.d
group by d1.post, d1.d
dbfiddle demo
The data table is the result of your SQL. It can be added using WITH or you may use subquery as well.

Answer 5

SELECT * ,Datediff(Month,[Date],endate)
FROM
(
SELECT *,Lead( [Date], 1, Null) OVER (
ORDER BY [Date]) AS Endate --INTO SourceTable
FROM
(
SELECT 'Senior Software Engg.' POST , '2018-04-18' DATE UNION ALL
SELECT 'Software Engg.' POST , '2017-04-18' DATE UNION ALL
SELECT 'Assoc. Software Engg.' POST , '2016-04-18' DATE
)A
)B
ORDER BY [Date] desc