mysql - 如何在sql中对2个不同的列进行排序

Question

你们能告诉我如何在 sql 中对用户定义列和固定列名进行排序吗？我需要显示最高交易和outletid，而我只得到最高交易但outlet id不在分组中。

对不起，我的英语很差

问题来了

outlet id | revenue code | total transaction | total amount

6837      |     014  |        326        |   39158.94
6821      |     408  |        291        |   48786.50
6814      |     014  |        285        |   74159.76
6837      |     452  |        282        |   8846.80

这是我的sql

SELECT 
                outletid, 
                revcode,
                count(receiptnumbe) as Transactions,
                sum(amount) as total
            FROM 
                user_payment
            WHERE
                date = (SELECT MAX(date) FROM user_payment GROUP BY date desc LIMIT 0, 1)
            GROUP BY 
                outletid, revcode
            ORDER BY Transactions desc

我需要它是这样的。按网点 ID 和最高交易排序。

outlet id | revenue code | total transaction | total amount

    6837      |     014  |        326        |   39158.94
    6837      |     452  |        282        |   8846.80
    6821      |     408  |        291        |   48786.50
    6814      |     014  |        285        |   74159.76

Answer 1

这是你想要的吗？

   ORDER BY OutletId, Transactions desc

编辑:

如果我没理解错的话，您希望它按总交易量最多的商店排序。然后通过该组内的交易。为此，您需要在 outlet 级别再次汇总并返回结果:

select outor.*
from (SELECT up.outletid, up.revcode, count(up.receiptnumbe) as Transactions,
             sum(up.amount) as total
      FROM user_payment up
      WHERE date = (SELECT MAX(date) FROM user_payment)
      GROUP BY outletid, revcode
     ) outor join
     (SELECT up.outletid, count(up.receiptnumbe) as Transactions,
             sum(up.amount) as total
      FROM user_payment up
      WHERE date = (SELECT MAX(date) FROM user_payment)
      GROUP BY outletid
     ) o
     on outor.outletid = o.outletid
order by o.Transactions desc, outor.outletid, outor.Transactions desc;