Question

我正在创建一个以230美元的基本成本开始的查询，然后将参加每个人注册到课堂会话的父母数量乘以55美元（每个父母的成本）参加会议）。然后，它将检查参加者迄今已支付的金额，并将从刚刚计算的总费用中减去该金额。我的问题是我还需要在每一行中包含电子邮件，因此我需要查询GROUP BY GuestEmail部分。使用此GROUP BY行会导致COUNT(Guest.ID)始终返回1，即使在多种情况下也应返回2,3或4。

SELECT Registration.ID, Guest.GuestEmail,
    CASE
        WHEN ((COUNT(Guest.ID) * 55) + 230) - SUM(DISTINCT Transaction.AuthorizationAmount) is not null
            THEN ((COUNT(Guest.ID) * 55) + 230) - SUM(DISTINCT  Transaction.AuthorizationAmount)
        WHEN ((COUNT(Guest.ID) * 55) + 230) is not null AND SUM(DISTINCT Transaction.AuthorizationAmount) is null
            THEN ((COUNT(Guest.ID) * 55) + 230)
        WHEN ((COUNT(Guest.ID) * 55) + 230) is null AND SUM(DISTINCT Transaction.AuthorizationAmount) is not null
            THEN - SUM(DISTINCT Transaction.AuthorizationAmount)
        ELSE '0'
    END AS Balance

    FROM Registration 
    LEFT JOIN Transaction ON Registration.ID = Transaction.ID 
    LEFT JOIN Guest ON Registration.ID = Guest.ID 

    WHERE Registration.IsActive = 'True' AND Registration.ID = 111111 AND ConfirmationEmail = 0 AND Guest.IsActive = 'True' AND Registration.ID = Guest.ID
    GROUP BY Registration.ID, Guest.GuestEmail

注册是保存学生信息的表，而访客持有父母的信息。通过注册和访客中包含的ID将每个父母标识为其子女的父母。 ID与父母和学生的编号相同。 Transaction.AuthorizationAmount是已注册人已支付的金额。每位客人都有一个单独的行。返回的电子邮件是父母的电子邮件，该电子邮件存储在Guest表中。问题：如何更改查询以使COUNT(Guest.ID)返回1以外的数字？

Answer 1

只是盲目地尝试这个，并且不确定这个查询将如何执行，但它可能会给你一些想法：

SELECT r.ID, r.balance, g.GuestEmail
FROM Guest g
JOIN (
    SELECT Registration.ID,
    CASE
        WHEN ((COUNT(Guest.ID) * 55) + 230) - SUM(DISTINCT Transaction.AuthorizationAmount) is not null
            THEN ((COUNT(Guest.ID) * 55) + 230) - SUM(DISTINCT  Transaction.AuthorizationAmount)
        WHEN ((COUNT(Guest.ID) * 55) + 230) is not null AND SUM(DISTINCT Transaction.AuthorizationAmount) is null
            THEN ((COUNT(Guest.ID) * 55) + 230)
        WHEN ((COUNT(Guest.ID) * 55) + 230) is null AND SUM(DISTINCT Transaction.AuthorizationAmount) is not null
            THEN - SUM(DISTINCT Transaction.AuthorizationAmount)
        ELSE '0'
    END AS Balance
    FROM Registration 
    LEFT JOIN Transaction ON Registration.ID = Transaction.ID 
    LEFT JOIN Guest ON Registration.ID = Guest.ID 
    WHERE Registration.IsActive = 'True' AND Registration.ID = 111111 AND ConfirmationEmail = 0 AND Guest.IsActive = 'True' AND Registration.ID = Guest.ID
    GROUP BY Registration.ID
) r ON (g.ID = r.ID)
GROUP BY g.GuestEmail

更改计数功能的输出

1 个答案: