MySQL总结了两个相反的货币对

Question

我有以下购买和出售货币对和它们的价值在一张桌子上，我试图写一个小的选择语句，获得每对的总数，但是当对有对立面时，他们可以互相扣除，例如，卖出任何欧元都没有意义，因为它可以用来抵消部分买入欧元。所以，这两条线英镑/欧元，欧元/英镑将成为： 注意对可能不仅EUR/GBP匹配所以不能在

EUR 852.07 GBP -636.1 

被硬编码（也许侧重于减少量买就是要走的路？）

我很想给你举个例子，但我完全卡住了！任何想法，不胜感激。

Buy    Sell 
ccy amt  ccy amt 
EUR, 1409.7600, AUD, -1965.4900 
GBP, 2801.8500, AUD, -5446.5100 
EUR, 1116.2800, CAD, -1472.1500 
EUR, 4862.2000, CHF, -5013.9000 
GBP, 3937.6700, CHF, -5661.5700 
EUR, 2066.8900, DKK, -15505.0000 
GBP, 1688.7200, DKK, -17640.6400 
** 
GBP, 1986.4400, EUR, -2778.7500 
EUR, 3630.8200, GBP, -2622.5400 
** 
EUR, 358.4100, NOK, -3046.6600 
GBP, 2865.9800, NOK, -33965.5900 
EUR, 1574.9800, SEK, -14746.0600 
GBP, 1511.3100, SEK, -19727.5900 
EUR, 724.3600, USD, -777.8900 
GBP, 34.7400, USD, -52.0200 


ccy = currency char(3) 
amt = amount decimal 

最终结果应该如下所示。

Buy    Sell 
ccy amt  ccy amt 
EUR, 1409.7600, AUD, -1965.4900 
GBP, 2801.8500, AUD, -5446.5100 
EUR, 1116.2800, CAD, -1472.1500 
EUR, 4862.2000, CHF, -5013.9000 
GBP, 3937.6700, CHF, -5661.5700 
EUR, 2066.8900, DKK, -15505.0000 
GBP, 1688.7200, DKK, -17640.6400 
** 
EUR, 852.07  GBP, -636.1 
** 
EUR, 358.4100, NOK, -3046.6600 
GBP, 2865.9800, NOK, -33965.5900 
EUR, 1574.9800, SEK, -14746.0600 
GBP, 1511.3100, SEK, -19727.5900 
EUR, 724.3600, USD, -777.8900 
GBP, 34.7400, USD, -52.0200 

Answer 1

尝试这样的事情，并假设你使用PHP ...

$sql = " 
select 
    sum( 
    case 
     when buy_currency = '$currency_one' then buy_amount 
     when sell_currency = '$currency_one' then sell_amount 
    end 
) as first_currency, 
    sum( 
    case 
     when buy_currency = '$currency_twp' then buy_amount 
     when sell_currency = '$currency_two' then sell_amount 
    end 
) as second_currency 
from test"; 

假设模式一样

CREATE TABLE `test` (
    `id` INT(11) NOT NULL AUTO_INCREMENT, 
    `buy_currency` VARCHAR(50) NOT NULL DEFAULT '0', 
    `buy_amount` VARCHAR(50) NOT NULL DEFAULT '0', 
    `sell_currency` VARCHAR(50) NOT NULL DEFAULT '0', 
    `sell_amount` VARCHAR(50) NOT NULL DEFAULT '0', 
    PRIMARY KEY (`id`) 
) 
COLLATE='latin1_swedish_ci' 
ENGINE=InnoDB 
AUTO_INCREMENT=6 
; 

Answer 2

试试这个：

SELECT SUM(CASE 
      WHEN `ccy_buy` = 'EUR' AND `ccy_sell` = 'GBP' THEN `amt_buy` 
      WHEN `ccy_sell` = 'EUR' AND `ccy_buy` = 'GBP' THEN `amt_sell` 
     END 
    ) AS `ccy1`, 
     SUM(CASE 
      WHEN `ccy_buy` = 'GBP' and `ccy_sell` = 'EUR' THEN `amt_buy` 
      WHEN `ccy_sell` = 'GBP' and `ccy_buy` = 'EUR' THEN `amt_sell` 
     END 
    ) AS `ccy2` 
    FROM `test` 

您可能想要将其包装到一个存储过程中并通过这两个货币作为参数，像这样：

DELIMITER ;; 
CREATE PROCEDURE `sp_compareCurrencyPairs`(IN pCur1 CHAR(3), IN pCur2 CHAR(3)) 
    READS SQL DATA 
BEGIN 
    SELECT SUM(CASE 
      WHEN `ccy_buy` = pCur1 AND `ccy_sell` = pCur2 THEN `amt_buy` 
      WHEN `ccy_sell` = pCur1 AND `ccy_buy` = pCur2 THEN `amt_sell` 
     END 
    ) AS `ccy1`, 
     SUM(CASE 
      WHEN `ccy_buy` = pCur2 AND `ccy_sell` = pCur1 THEN `amt_buy` 
      WHEN `ccy_sell` = pCur2 AND `ccy_buy` = pCur1 THEN `amt_sell` 
     END 
    ) AS `ccy2` 
    FROM `test`; 
END;; 
DELIMITER ; 

然后使用它是这样的：

CALL `sp_compareCurrencyPairs`('GBP', 'EUR'); 

Answer 3

我不得不稍微改变您的列名，以反映有在同为多CCY和AMT的事实记录。

SELECT 
a.buy_ccy, 
SUM(a.buy_amt) + COALESCE(SUM(b.sell_amt),0) as buy_amt, 
a.sell_ccy, 
SUM(a.sell_amt) + COALESCE(SUM(b.buy_amt),0) as sell_amt 
FROM 
transactions as a 
LEFT OUTER JOIN transactions as b ON (a.buy_ccy = b.sell_ccy AND a.sell_ccy = b.buy_ccy) 
GROUP BY 
a.buy_ccy, 
b.buy_ccy, 
a.sell_ccy, 
b.sell_ccy 
HAVING buy_amt > 0; 

例子：http://sqlfiddle.com/#!9/f2006/3

Answer 4

http://sqlfiddle.com/#!9/d5198/1

SELECT 
    IF(buy_ccy>sell_ccy,buy_ccy,sell_ccy) as buy_ccy, 
    SUM(IF(buy_ccy>sell_ccy,buy_amt,sell_amt)) as buy_amt, 
    IF(buy_ccy>sell_ccy,sell_ccy,buy_ccy) as sell_ccy, 
    SUM(IF(buy_ccy>sell_ccy,sell_amt,buy_amt)) as sell_amt 
FROM EXCHANGE 
GROUP BY IF(buy_ccy>sell_ccy,CONCAT(buy_ccy,sell_ccy),CONCAT(sell_ccy,buy_ccy))