SQL加入表格并按乘积值排序

所以，简单解释：我想从表中的所有记录中按最高家具值（pricelist_furnis_tradevalue.value）排序的games_records乘以记录数量（= games_records.amount）。 furni值在pricelist_furnis_tradevalue中。现在... pricelist_furnis_tradevalue对于某些家具有多个条目。我只想获得最新的条目，因此应该按pricelist_furnis_tradevalue.time排序。但是，下面的这个查询返回每个furni /记录的pricelist_furnis_tradevalue的所有条目。SQL加入表格并按乘积值排序

SELECT * 
FROM games_records 
JOIN pricelist_furnis_tradevalue 
    ON games_records.furni = pricelist_furnis_tradevalue.furni 
ORDER BY (pricelist_furnis_tradevalue.value*games_records.amount) DESC

这里有2级所需的表的结构

games_records

id  user  furni  amount  time 
1  2   4   40   1338052926 
2  4   30   90   1338054046

pricelist_furnis_tradevalue

id  furni  value  time 
1  2   20   1334065379 
2  2   50   1334067445 
3  2   100   1334092057 
4  4   50   1334065375 
4  4   20   1334067436

我会很感激任何帮助！谢谢！

来源

2012-07-18 albin

你试过选择'pricelist_furnis_tradevalue.value * games_records.amount'？ – Kermit 2012-07-18 19:49:01

@njk并用AS变量定义它们，然后用ORDER BY变量DESC定义它们？ **编辑**试过了。没有工作。 – albin 2012-07-18 19:52:26

您需要以最长时间获取记录。

这里有一个方法：

SELECT * 
FROM games_records join 
    (select furni, max(time) as maxtime 
     from pricelist_furnis_tradevalue 
     group by furni 
    ) as maxf 
    on games_records.furni = maxf.furni JOIN 
    pricelist_furnis_tradevalue 
    ON games_records.furni = pricelist_furnis_tradevalue.furni and 
     pricelist_furnis_tradevalue.time = maxf.maxtime 
ORDER BY (pricelist_furnis_tradevalue.value*games_records.amount) DESC

来源

2012-07-18 19:56:25

谢谢！这完美的作品！ – albin 2012-07-18 20:31:15

SELECT * FROM 
    (SELECT *, pricelist_furnis_tradevalue.value*games_records.amount AS 'mult' FROM games_records 
    JOIN pricelist_furnis_tradevalue 
    ON games_records.furni = pricelist_furnis_tradevalue.furni) a 
ORDER BY a.mult DESC

来源

2012-07-18 19:55:19 Kermit

SQL加入表格并按乘积值排序

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