需要幫助創建mysql查詢多個選擇的具體條件

Question

我需要一些幫助，使SQL查詢。
我有表'stock_exchange'，它記錄在四個不同的商店/網站之間互換的股票。
它記錄

日期，ITEM_ID，數量，價格，TOTAL_PRICE，from_site，to_site

現在我需要創建一個查詢，爲商店/網站將返回總價的一個運行爲其他三家店之間發送和接收的股票。

例如：有商店1，商店，Store3，Store4
如果我需要檢查商店1，它應該返回：

商店Name____Total Sent____Total收到
商店2 _________ 90.90 _______ 50
Store3_________120_______100。 40
Store4_________400_______200

我嘗試過自己，但只能獲得發送或接收的股票的總和。

這裏是fiddle

編輯：
我改變了表中記錄存儲ID（1,2,3,4 ......），而不是現在的店別名（CRW，CHC，cal_rav ....），並且我接受的代碼（在答案中）已停止工作。我試圖在代碼這項修正案找到店鋪標識匹配的行，但簡化版，返回任何東西：

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' 

這裏是updated fiddle。

Answer 1

下面是一個查詢，顯示的發送和接收的數量從咖啡館紫檀所有其他商店：

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' 

如果你想查看全部店鋪之間的交易，只需刪除HAVING條款。該查詢採用了將GREATEST(from_site, to_site)組合在一起的兩個商店的技巧。

​​

Answer 2

在我看來，雖然有可能有多個店鋪的更廣泛的組合，我更喜歡使用，而更多的查詢是更簡單的（而且通常更快執行）：

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; 

結果：

...所以我會測試它針對大數據量，並在生產中使用比較之前的解決方案。

計數總計：

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;