Query that Counts records with a WHERE clause

问题

                SELECT 
                EmailOfConsumer, 
                COUNT(EmailOfConsumer) as 'NumberOfOrders',
                SUM(CAST(Total as money)) as 'TotalValue',
                (SUM(CAST(Total as money))/COUNT(EmailOfConsumer)) as 'AverageValue'
                FROM webshop
                GROUP BY EmailOfConsumer 
                ORDER BY TotalValue DESC

This brings back:

EmailOfConsumer NumberOfOrders  TotalValue                AverageValue
test                   1              2000000000.10           2000000000.10 

I would like to add a search on WHERE NumberOfOrders = '1'

I have tried adding WHERE COUNT(EmailOfConsumer) = '1'

but I get this error:

    An aggregate may not appear in the WHERE clause unless it is in a subquery contained in a 
HAVING clause or a select list, and the column being aggregated is an outer reference.

回答1:

use

HAVING COUNT(EmailOfConsumer) = 1

The having clause restricts a aggregate whereas the where clause only put restrictions on individual column data

回答2:

Using group by and then having clause. Refer this

SELECT 
            EmailOfConsumer, 
            COUNT(EmailOfConsumer) as 'NumberOfOrders',
            SUM(CAST(Total as money)) as 'TotalValue',
            (SUM(CAST(Total as money))/COUNT(EmailOfConsumer)) as 'AverageValue'
            FROM webshop
            GROUP BY EmailOfConsumer 
            HAVING COUNT(EmailOfConsumer) = '1'
            ORDER BY TotalValue DESC

回答3:

SELECT 
                EmailOfConsumer, 
                COUNT(EmailOfConsumer) as 'NumberOfOrders',
                SUM(CAST(Total as money)) as 'TotalValue',
                (SUM(CAST(Total as money))/COUNT(EmailOfConsumer)) as 'AverageValue'
                FROM webshop
                GROUP BY EmailOfConsumer 

HAVING COUNT(EmailOfConsumer) = '1'
                ORDER BY TotalValue DESC

回答4:

You can try

SELECT 
            EmailOfConsumer, 
            COUNT(EmailOfConsumer) as 'NumberOfOrders',
            SUM(CAST(Total as money)) as 'TotalValue',
            (SUM(CAST(Total as money))/COUNT(EmailOfConsumer)) as 'AverageValue'
            FROM webshop
            GROUP BY EmailOfConsumer 
            HAVING COUNT(EmailOfConsumer) = 1
            ORDER BY TotalValue DESC

回答5:

Use this query:

SELECT 
    EmailOfConsumer, 
    COUNT(EmailOfConsumer) as 'NumberOfOrders',
    SUM(CAST(Total as money)) as 'TotalValue',
    (SUM(CAST(Total as money))/COUNT(EmailOfConsumer)) as 'AverageValue'
FROM webshop
GROUP BY EmailOfConsumer 
HAVING COUNT(*) = 1
ORDER BY TotalValue DESC

Notes:

• WHERE clause precedes GROUP BY, while HAVING clause follows GROUP BY
• WHERE clause filters rows before aggregation; HAVING clause filters aggregated rows
• COUNT(...) returns a number, so the constant 1 should not be enclosed in quotes.

回答6:

You need to add filter condition in HAVING clause. Also COUNT returns numeric value so no need to add quote to check it :

SELECT     EmailOfConsumer, 
           COUNT(EmailOfConsumer) as 'NumberOfOrders',
           SUM(CAST(Total as money)) as 'TotalValue',
           (SUM(CAST(Total as money))/COUNT(EmailOfConsumer)) as 'AverageValue'
           FROM webshop
           GROUP BY EmailOfConsumer 
           HAVING COUNT(EmailOfConsumer) = 1
           ORDER BY TotalValue DESC

You can't check condition with COUNT in WHERE clause because it will execute before aggregation, so you need to check it after aggregation in HAVING clause.

来源：https://stackoverflow.com/questions/20098378/query-that-counts-records-with-a-where-clause