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How to Use Group By clause when we use Aggregate function in the Joins?

I want to join three tables and to calculate the Sum(Quantity) of the Table A. I tried something and I get the desired output. But still I have confusion based on aggregate function and Group By clause.

While calculating the sum value by joining two or more tables, what are the columns we need to mention in the Group By clause and why do we need to give those columns?

For Example: Here is my table and the desired query.

TableA: ItemID, JobOrderID, CustomerID, DivisionID, Quantity
TableB: ItemID, ItemName, SpecificationID
TableC: SpecificationID, SpecificationName
TableD: DivisionID, DivisionName
TableE: JobOrderID, JobOrderNo.
TableF: CustomerID, CustomerName

I want to get the Sum(Quantity) based on ItemID, CustomerID, JobOrderID and DivisionID.

I wrote the following query and it's working fine. But if I remove any column in the Group By clause, it doesn't give the desired result. Why? What does the Group By clause do here? How to specify the Group By clause when using Aggregate function? Here is my Query.

    SELECT 
            B.ItemName + ' - ' + C.SpecificationName AS 'ItemName',
            SUM(A.Quantity) AS 'Quantity',
            A.ItemID,
            D.DivisionName,
            F.CustomerName,
            E.JobOrderNo,
            A.DivisionID,
            A.JobOrderID,
            A.CustomerID

    FROM
            TableA A  
            INNER JOIN TableB B ON B.ItemID = A.ItemID 
            INNER JOIN TableC C ON C.SpecificationID = B.SpecificationID
            INNER JOIN TableD D ON D.DivisionID = A.DivisionID
            LEFT JOIN TableE E ON E.JobOrderID = A.JobOrderID
            LEFT JOIN TableF F ON F.CustomerID = A.CustomerID
    WHERE
            A.ItemID = @ItemID
    GROUP BY
            A.ItemID,
            A.JobOrderID,
            A.DivisionID,
            A.CustomerID,
            D.DivisionName,
            F.CustomerName,
            E.JobOrderNo,
            B.ItemName,
            C.SpecificationName

Any one please give suggestion about the Group By Clause by considering this as an example.

like image 935
thevan Avatar asked Jun 21 '11 10:06

thevan


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2 Answers

GROUP BY for any unique combination of the specified columns does aggregation (like sum, min etc). If you don't specify some column name in the GROUP BY clause or in the aggregate function its unknown to the SQL engine which value it should return for that kind of column.

like image 162
Piotr Auguscik Avatar answered Sep 28 '22 01:09

Piotr Auguscik


GROUP BY (Transact-SQL) groups a selected set of rows into a set of summary rows by the values of one or more columns or expressions in SQL Server 2008 R2. One row is returned for each group. Aggregate functions in the SELECT clause list provide information about each group instead of individual rows.

SELECT a.City, COUNT(bea.AddressID) AS EmployeeCount
FROM Person.BusinessEntityAddress AS bea 
    INNER JOIN Person.Address AS a
        ON bea.AddressID = a.AddressID
GROUP BY a.City

The GROUP BY clause has an ISO-compliant syntax and a non-ISO-compliant syntax. Only one syntax style can be used in a single SELECT statement. Use the ISO compliant syntax for all new work. The non-ISO compliant syntax is provided for backward compatibility.

In ISO-compliant syntax each table or view column in any nonaggregate expression in the list must be included in the GROUP BY list.

select pub_id, type, avg(price), sum(total_sales)
from titles
group by pub_id, type

Refering to Organizing query results into groups: the group by clause

Sybase or non-ISO-compliant syntax lifts restrictions on what you can include or omit in the select list of a query that includes group by:

  • The columns in the select list are not limited to the grouping columns and columns used with the vector aggregates.

  • The columns specified by group by are not limited to those non-aggregate columns in the select list.

Example:

select type, title_id, avg(price), avg(advance) 
from titles 
group by type 
like image 37
Akram Shahda Avatar answered Sep 28 '22 01:09

Akram Shahda