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Get Last Value of previous row partition in SQL

Time:01-31

In my data set, each customer has some orders on different dates. For each customer each month, I want to check his/her last order in the previous month in which city.

For example, it is my data for one of the customers.

customer year month day order id city id
1544 2022 2 6 413 9
1544 2022 2 17 39 10
1544 2022 3 5 115 21
1544 2022 5 29 2153 4
1544 2022 5 30 955 9

the result should be the same as this:

customer year month city of last order of prev month(prevCity)
1544 2022 2 null or 9
1544 2022 3 10
1544 2022 5 21

(the first row of the above table is not my question now. ) I write my query using last_value the same as this:

select customer,
   year,
   month,
   last_value(City) over (partition by customer, year, month order by created_at desc
       ROWS BETWEEN CURRENT ROW AND UNBOUNDED FOLLOWING) as prevCity 
 from table1

but the result is false!

How can I correct this?

CodePudding user response:

Using the window function lag() over() in concert with the WITH TIES clause

Select top 1 with ties 
       customer
      ,year
      ,month
      ,LastCityID = lag([city id],1) over (partition by customer order by year, month,day) 
 From YourTable
 order by row_number() over (partition by customer,year,month order by year, month,day) 

Or an Nudge More Perforamt

with cte as (
Select *
      ,LastCityID = lag([city id],1) over (partition by customer order by year, month,day) 
      ,RN = row_number() over (partition by customer,year,month order by year, month,day) 
 From YourTable
)
Select customer
      ,year
      ,month
      ,LastCityID 
 From  cte
 Where RN =1 

Results

customer    year    month   LastCityID
1544        2022    2       NULL
1544        2022    3       10
1544        2022    5       21
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