Calculate sum of entries in T-SQL

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Currently i'm starting a new project for a delivering service.

I have this table given:

CustID MealID Week  Weekday
1	   3	  1	Monday
1	   5	  1	Tuesday
1	   6	  1	Wednesday
1	   6	  1	Thursday
1	   3	  1	Friday
1	   5	  1	Saturday
1	   1	  2	Monday
1	   9	  2	Tuesday
1	  11	  2	Wednesday
1	  12	  2	Thursday
1	  12	  2	Friday
1	   1	  2	Saturday
2	   3	  1	Monday
2	   5	  1	Tuesday
2	   6	  1	Wednesday
2	   6	  1	Thursday
2	   3	  1	Friday
2	   5	  1	Saturday

In the first step i need all meals from Week 1. So a where clause works perfectly.
Now i have a second step. I need the count of each meal cumulated.

Let me say Meal 6 x 4, Meal 3 x 4 and so on.

Exists a good way to cumulate this?

What I have tried:

I tried out a count clause, but what i've seen, it doesn't works for that case.

Posted 4-Jun-19 23:50pm

Sascha Manns

Updated 5-Jun-19 1:49am

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Comments

Patrice T 5-Jun-19 7:29am

And you plan to show the query you used ?

2 solutions

Solution 2

Instead of using a WHERE clause PARTITION your data OVER Clause (Transact-SQL) - SQL Server | Microsoft Docs[^] and get all the information you need in a single query.

Here is the example from the link

SQL

USE AdventureWorks2012;  
GO  
SELECT SalesOrderID, ProductID, OrderQty  
    ,SUM(OrderQty) OVER(PARTITION BY SalesOrderID) AS Total  
    ,AVG(OrderQty) OVER(PARTITION BY SalesOrderID) AS "Avg"  
    ,COUNT(OrderQty) OVER(PARTITION BY SalesOrderID) AS "Count"  
    ,MIN(OrderQty) OVER(PARTITION BY SalesOrderID) AS "Min"  
    ,MAX(OrderQty) OVER(PARTITION BY SalesOrderID) AS "Max"  
FROM Sales.SalesOrderDetail   
WHERE SalesOrderID IN(43659,43664);  
GO

Posted 5-Jun-19 1:50am

CHill60

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OriginalGriff · Accepted Answer · 2019-06-05T01:48:00

Solution 1

You need to look at GROUP BY: SQL GROUP BY Statement[^]
Try:

SQL

SELECT MealID, COUNT(MealID) FROM MyTable
WHERE Week = 1
GROUP BY MealID

Posted 5-Jun-19 1:48am

OriginalGriff

Comments

Sascha Manns 5-Jun-19 8:34am

Thanks for all for trying to help. I'll try this solution. Thanks @all.