以下SQL查询有什么问题？

Question

I have the following SQL query which finds the name and age of the oldest snowboarders.

SELECT c.cname, MAX(c.age)  
FROM Customers c  
WHERE c.type = 'snowboard';

But the guide I am reading says that query is wrong because the non-aggregate columns in the SELECT clause must come from the attributes in the GROUP BY clause.

I think this query serves its purpose because the aggregate MAX(c.age) corresponds to a single value. It does find the age of the oldest snowboarder.

Answer 1

You need to group by c.cname column. Whenever you do any aggregation on some column like SUM, COUNT, etc. you need to provide another column(s) by which you want to aggregate. You generally provide these columns in your SELECT clause(here c.cname ). These same columns should be mentioned in the GROUP BY clause else you will get a syntax error.

The form should be

SELECT A, B, C, SUM(D)
FROM TABLE_NAME
GROUP BY A, B, C;

Your query should be like below

SELECT  c.cname,    MAX(c.age)  
FROM    Customers   c  
WHERE   c.type=‘snowboard’
GROUP BY c.cname;

Answer 2

If you want to display the name and age of the oldest snowboarder(s), you have to do this:

SELECT  c.cname, c.age
FROM    Customers   c  
WHERE   c.type = 'snowboard'
AND c.age = (SELECT MAX(age)
             FROM Customers
             WHERE type = 'snowboard')

Answer 3

Whenever some other table attributes are selected along with the aggregate function then it has to be accompanied by group by clause, otherwise a situation may occur where database may not know which row has to be selected. Let us say in your example there are two customers who have same maximum age, now database will try to pull out the age from both the rows but it will get confused, which name to pick. Here your group by clause comes into picture, which will instruct to display two different rows with different customer names but same maximum age. Thus, your query should look like this:

SELECT  c.cname,    MAX(c.age)  
FROM    Customers   c  
WHERE   c.type=‘snowboard’
GROUP BY c.cname;