Get the previous amount of accounts? - sql

Question

I got a table like this:
===============================
| ID | Salary | Date |
===============================
| A1 | $1000 | 2020-01-03|
-------------------------------
| A1 | $1300 | 2020-02-03|
-------------------------------
| A1 | $1500 | 2020-03-01|
-------------------------------
| A2 | $1300 | 2020-01-13|
-------------------------------
| A2 | $1500 | 2020-02-11|
-------------------------------
Expected output:
==================================================
| ID | Salary | Previous Salary | Date |
==================================================
| A1 | $1500 | $1300 | 2020-03-01|
--------------------------------------------------
| A2 | $1500 | $1300 | 2020-02-03|
--------------------------------------------------
How could I query to always get their previous salary and to show in another column/table ?

Answer 1

You can combine both the row_number and the lag windows functions to locate the last salary for every id and to return their last and previous salary.
with cte as (
select id, salary,
row_number() over (partition by id order by date desc) as position,
lag(salary) over (partition by id order by date) as previous,
date
from payroll
)
select id, salary, previous, date
from cte
where position = 1 -- It's the first one because we ordered by date descendingly
Result :
ID Salary Previous Date
----- --------------------- --------------------- ----------
A1 1500,00 1300,00 2020-03-01
A2 1500,00 1300,00 2020-02-11
Online sample: http://sqlfiddle.com/#!18/770472/15/0

Answer 2

In SQL Server you can use the LAG Window Function to reference a field from the previous record in a partitioned set (within a specified data window)
with [Data] as
(
SELECT ID, Salary, Cast([Date] as Date) [Date]
FROM (VALUES
('A1', 1000, '2020-01-03'),
('A1',1300,'2020-02-03'),
('A1',1500,'2020-03-01'),
('A2',1300,'2020-01-13'),
('A2',1500,'2020-02-11')
) as t(ID,Salary,Date)
)
-- above is a simple demo dataset definition, your actual query is below
SELECT ID, Salary, LAG(Salary, 1)
OVER (
PARTITION BY [ID]
ORDER BY [Date]
) as [Previous_Salary], [Date]
FROM [Data]
ORDER BY [Data].[Date] DESC
Produces the following output:
ID Salary Previous_Salary Date
---- ----------- --------------- ----------
A1 1500 1300 2020-03-01
A2 1500 1300 2020-02-11
A1 1300 1000 2020-02-03
A2 1300 NULL 2020-01-13
A1 1000 NULL 2020-01-03
(5 rows affected)
Experiment with the ordering, note here in the window we are using ascending order, and in the display, we can show descending order.
Window functions create a virtual dataset outside of your current query, think of windows functions as a way to execute correlated queries in parallel and merge in the result.
In many simple implementations window functions like this should provide better or the same performance as writing your own logic to self-join or to sub query, for example this is an equivalent query using CROSS APPLY
SELECT ID, [Data].Salary, previous.Salary as [Previous_Salary], [Data].[Date]
FROM [Data]
CROSS APPLY (SELECT TOP 1 x.Salary
FROM [Data] x
WHERE x.[ID] = [Data].ID AND x.[Date] > [Data].[Date]
ORDER BY x.[Date] DESC) as previous
ORDER BY [Data].[Date] DESC
The LAG syntax requires less code, clearly defines your intent and allows set-based execution and optimizations.
Other JOIN style queries will still be blocking queries as they will require the reversal of the original data-set (by forcing the entire set to be loaded in reverse order) and so will not offer a truly set-based or optimal approach.
SQL Server developers realized that there is genuine need for these types of queries, and that in general when left to our own devices we create inefficient lookup queries, Window Functions were designed to offer a best-practice solution to these types of analytical queries.

Answer 3

This query could work for you
select *, LAG(salary) OVER (partition by id ORDER BY id) as previous from A

Answer 4

Try this: (Assumption: Table name 'PAYROLL' in 'PLS' schema)
SELECT R1.ID,R1.SALARY AS SALARY,R2.SALARY AS PREVIOUS_SALARY
FROM PLS.PAYROLL R1
LEFT OUTER JOIN PLS.PAYROLL R2 ON R1.ID=R2.ID AND R2.SALARY_DATE = (SELECT MAX(SALARY_DATE) FROM PLS.PAYROLL WHERE ID=R1.ID AND SALARY_DATE<R1.SALARY_DATE)
WHERE R1.SALARY_DATE=(SELECT MAX(SALARY_DATE) FROM PLS.PAYROLL WHERE ID=R1.ID)
ORDER BY R1.ID

Answer 5

You can use window functions and pivot to make this.
DECLARE #SampleData TABLE (ID VARCHAR(5), Salary MONEY ,[Date] DATE)
INSERT INTO #SampleData VALUES
('A1', $1000 , '2020-01-03'),
('A1', $1300 , '2020-02-03'),
('A1', $1500 , '2020-03-01'),
('A2', $1300 , '2020-01-13'),
('A2', $1500 , '2020-02-11')
SELECT ID, [1] Salary, [2] [Previous Salary], [Date]
FROM (SELECT ID, Salary, MAX([Date]) OVER(PARTITION BY ID) AS [Date],
ROW_NUMBER() OVER(PARTITION BY ID ORDER BY [Date] DESC) RN
FROM #SampleData ) SRC
PIVOT(MAX(Salary) FOR RN IN ([1],[2])) PVT
ORDER BY ID
Result:
ID Salary Previous Salary Date
----- --------------------- --------------------- ----------
A1 1500,00 1300,00 2020-03-01
A2 1500,00 1300,00 2020-02-11