SQLFiddle

Question

我需要一些帮助才能进行SQL查询我有表'stock_exchange'，它记录了四个不同商店/网站之间互换的股票它记录

date，item_id，qty，price，total_price，from_site，to_site

现在我需要创建一个查询来运行其中一个商店/网站，这将返回其他三个商店之间发送和接收的股票的总价。

例如：Store1，Store2，Store3，Store4
如果我需要检查Store1，它应该返回：

商店名称_发送总数_收到的总数
商店2 _________ 90.90 _______ 50商店3 _________ 120 _______ 100.40
Store4_________400_______200

我试过自己，但只能得到已发送或收到的股票的总和。

以下是fiddle

修改
我把表更改为记录存储ID（1,2,3,4 ....）而不是存储别名（crw，chc，cal_rav ....），我接受的代码（回答）已经停止工作。我在代码中尝试了这个修正，按商店ID查找匹配的行，但不返回任何内容：

SELECT GREATEST(t2.name, t3.name) AS from_store,
       LEAST(t2.name, t3.name) AS to_store,
       SUM(CASE WHEN t2.name < t3.name THEN t1.total_price ELSE 0 END) AS total_sent,
       SUM(CASE WHEN t2.name > t3.name THEN t1.total_price ELSE 0 END) AS total_received
FROM stock_exchange t1
INNER JOIN sites t2
    ON t1.from_site = t2.id
INNER JOIN sites t3
    ON t1.to_site = t3.id
GROUP BY GREATEST(t2.name, t3.name),
         LEAST(t2.name, t3.name)
HAVING MAX(GREATEST(t2.id, t3.id)) = '1'

这是updated fiddle。

Answer 1

在我看来，虽然有可能有更多更多的商店组合，但我更喜欢使用更简单的查询（通常甚至更快执行）：

SELECT name, SUM(total_price) as total_sent_to FROM stock_exchange se 
LEFT JOIN sites ON sites.alias = se.to_site
WHERE from_site = 'crw' GROUP BY from_site, to_site;
SELECT name, SUM(total_price) as total_received_from FROM stock_exchange se
LEFT JOIN sites ON sites.alias = se.from_site
WHERE to_site = 'crw' GROUP BY from_site, to_site;

结果：

...所以我会针对更大的数据量进行测试，并在生产中使用前比较解决方案。

计算总数：

SELECT *, (SELECT SUM(total_price)
                  FROM stock_exchange se 
                  WHERE se.from_site = sites.alias 
                  GROUP BY se.from_site) 
                as total_sent_sum,
                (SELECT SUM(total_price)
                  FROM stock_exchange se 
                  WHERE se.to_site = sites.alias 
                  GROUP BY se.to_site) 
                as total_received_sum
                FROM sites;

Answer 2

这是一个查询，显示Cafe Rosewood向所有其他商店发送和接收的金额：

SELECT GREATEST(t2.name, t3.name) AS from_store,
       LEAST(t2.name, t3.name) AS to_store,
       SUM(CASE WHEN t2.name < t3.name THEN t1.total_price ELSE 0 END) AS total_sent,
       SUM(CASE WHEN t2.name > t3.name THEN t1.total_price ELSE 0 END) AS total_received
FROM stock_exchange t1
INNER JOIN sites t2
    ON t1.from_site = t2.alias
INNER JOIN sites t3
    ON t1.to_site = t3.alias
GROUP BY GREATEST(t2.name, t3.name),
         LEAST(t2.name, t3.name)
HAVING MAX(GREATEST(t2.name, t3.name)) = 'Cafe Rosewood'

如果要查看所有商店之间的交易，只需删除HAVING子句即可。此查询采用了将GREATEST(from_site, to_site)组合在一起的两个商店的技巧。

需要帮助为特定条件的多个选择创建mysql查询

2 个答案:

SQLFiddle