SQL Server GROUP BY multiple columns - sql-server

Question

Let's assume I have in SQL Server the following table with only seven days available (SUN - SAT):
Orders
| Day | ProductType | Price |
| SUN | 1 | 10 |
| MON | 1 | 15 |
| MON | 2 | 20 |
| MON | 3 | 10 |
| TUE | 1 | 5 |
| TUE | 3 | 5 |
...
I need to group the data in a way so that to see the Total sum of Prices by each distinct Day and two groups of ProductType (= 1 and > 1):
| Day | FirstProductTypeTotal | RestProductsTypesTotal | GrandTotal |
| SUN | 10 | 0 | 10 |
| MON | 15 | 30 | 45 |
| TUE | 5 | 5 | 10 |
...
where FirstProductTypeTotal is ProductType = 1 and RestProductTypesTotal is ProductType > 1.
Is it possible to select this in one select instead of writing two different selects:
Select Day, SUM(Price) as FirstTotal from Orders where ProductType = 1 group by Day
and
Select Day, SUM(Price) as SecondTotal from Orders where ProductType > 1 group by Day
And then add FirstTotal and SecondTotal manually in the code to get the Grand total for each day of the week?

Answer 1

Use CASE Expression
Select Day, SUM(CASE WHEN ProductType = 1 THE Price ELSE 0 END) AS FirstTotal,
SUM(CASE WHEN ProductType > 1 THE Price ELSE 0 END) AS SecondTotal,
SUM(Price) AS GrandTotal
FROM Orders
group by Day

Answer 2

Try conditional aggregation;
Sample data;
CREATE TABLE #Orders ([Day] varchar(10), ProductType int, Price int)
INSERT INTO #Orders ([Day],ProductType, Price)
VALUES
('SUN',1,10)
,('MON',1,15)
,('MON',2,20)
,('MON',3,10)
,('TUE',1,5)
,('TUE',3,5)
Query;
SELECT
o.[Day]
,SUM(CASE WHEN o.ProductType = 1 THEN o.Price ELSE 0 END) FirstTotal
,SUM(CASE WHEN o.ProductType > 1 THEN o.Price ELSE 0 END) SecondTotal
,SUM(o.Price) GrandTotal
FROM #Orders o
GROUP BY o.[Day]
Result
Day FirstTotal SecondTotal GrandTotal
MON 15 30 45
SUN 10 0 10
TUE 5 5 10
You'd just need to sort out the ordering of the days because SQL Server by definition doesn't store the data in any particular order.