需要根据另一列否定一列，作为总计的一部分

Question

我在会计系统中生成损益报告时遇到问题。

每个普通日记帐分录都有一个金额、一个来源帐户和一个目标帐户。这是一个简化的架构和一些示例数据：

CREATE TABLE `sa_general_journal` (
  `ID` int(10) unsigned NOT NULL AUTO_INCREMENT,
  `Date` timestamp NOT NULL DEFAULT current_timestamp(),
  `Item` varchar(1024) NOT NULL DEFAULT '',
  `Amount` decimal(9,2) NOT NULL DEFAULT 0.00,
  `Source` int(10) unsigned NOT NULL,
  `Destination` int(10) unsigned NOT NULL,
  PRIMARY KEY (`ID`),
  KEY `Date` (`Date`),
  KEY `Source` (`Source`),
  KEY `Destination` (`Destination`),
  CONSTRAINT `sa_credit-account` FOREIGN KEY (`Destination`) REFERENCES `sa_accounts` (`ID`),
  CONSTRAINT `sa_debit-account` FOREIGN KEY (`Source`) REFERENCES `sa_accounts` (`ID`)
) ENGINE=InnoDB AUTO_INCREMENT=21561 DEFAULT CHARSET=utf8;

CREATE TABLE `sa_accounts` (
  `ID` int(10) unsigned NOT NULL,
  `Name` varchar(255) NOT NULL,
  `Type` enum('Asset','Liability','Income','Expense'),
  `Report` enum('BS','PL'), -- for "Balance Sheet" or "Profit & Loss"
  PRIMARY KEY (`ID`),
) ENGINE=InnoDB DEFAULT CHARSET=utf8;


INSERT INTO sa_account (`ID`, `Name`, `Type`, `Report`)
  VALUES (1009999, "Test chequing account", "Asset", "BS"),
         (4059999, "Test Income account", "Income", "PL"),
         (5059999, "Test Expense account", "Expense", "PL");
INSERT INTO sa_general_journal (`ID`, `Date`, `Item`, `Amount`, `Source`, `Destination`)
  VALUES (NULL, "2020-05-03", "Test income transaction", 10.10, 4059999, 1009999),
 (NULL, "2020-05-03", "Test expense transaction", 1.01, 1009999, 5059999);

这是存入支票账户的 10.10 美元收入和从支票账户中支出的 1.01 美元。

要获得损益表的余额，需要对 Source 列中每个出现的帐户的所有 Amount 条目求和，然后减去所有 Amount 个条目，其中该帐户位于 Destination 列。

预期结果为：

<table>
<th>ID</th><th>Name</th><th>Debits</th><th>Credits</th><th>Net</th></tr>
<tr><td>1009999</td><td>Test chequing account</td><td>-1.01</td><td>10.10</td><td>9.09</td></tr>
<tr><td>4059999</td><td>Test income transaction</td><td>-10.10</td><td><i>NULL</i></td><td>-10.10</td></tr>
<tr><td>5059999</td><td>Test expense transaction</td><td><i>NULL</i></td><td>1.01</td><td>1.01</td></tr>
</table>

我的第一个方法有点天真，查询 sa_general_journal table 与 sa_accounts table 连接，由 Source 和 Destination 列，如果 Destination 包含感兴趣的帐户，则使用 IF 函数取反 Amount。我在使用准备好的语句查询个人帐户时成功地做到了这一点：

SELECT
  DATE_FORMAT(exp.Date, '%Y') AS `Year`,
  AVG(exp.Amount) AS `Avg`,
  SUM(IF(Source = ?, 0 - exp.Amount, NULL)) AS `Debits`,
  SUM(IF(Destination = ?, exp.Amount, NULL)) AS `Credits`,
  SUM(IF(Source = ?, 0 - exp.Amount, exp.Amount)) AS `Net`
FROM sa_general_journal exp
  LEFT JOIN sa_accounts Destination ON exp.Destination = Destination.ID
WHERE Destination = ?
  OR Source = ?
GROUP BY `Year`

所有四个占位符都拥有相同的帐户 ID。

但是，当在没有 WHERE 子句的情况下按帐户 ID 进行分组以获取所有帐户余额的列表时，这会失败——用动态帐户 ID 代替占位符中的静态帐户 ID 似乎永远不会达到“0 - exp.Amount" 代码……呃！查询被选择为一个集合，而不是过程中的步骤。我明白了。

使 SUM 语句子查询可行，但速度慢得可怕，显然要对数万条记录中的每条记录执行三个子查询！

所以，我想我可以用几个常见的 Table 表达式分解子查询，但这似乎也不能正常工作，因为它似乎仍然只是简单地返回 SUM(Amount), 不减去 Amounts 即 Destinations 而不是 Sources.

WITH
 source1 AS (SELECT
  src.ID,
  src.Name,
  SUM(Amount) Amount
FROM sa_general_journal gj
  LEFT JOIN sa_accounts src ON gj.`Source` = src.ID
WHERE src.Report = "PL" -- AND YEAR(`Date`) = 2019
GROUP BY src.ID),
 destination1 AS (SELECT
  dst.ID,
  dst.Name,
  SUM(0-Amount) Amount
FROM sa_general_journal gj
  LEFT JOIN sa_accounts dst ON gj.`Destination` = dst.ID
WHERE dst.Report = "PL" --  AND YEAR(`Date`) = 2019
GROUP BY dst.ID)
SELECT ID, Name, sum(Amount)
FROM source1
UNION ALL
SELECT ID, Name, sum(Amount)
FROM destination1
GROUP BY ID

我猜我在做一些愚蠢的假设，或者做一些愚蠢的事情，所以任何建议都将不胜感激！

Answer 1

没有样本数据很难 100% 确定，但此查询应该会给出您想要的结果。它使用 UNION 查询将交易拆分为 Source 和 Destination 账户，然后使用条件聚合 SUM 每个账户的交易。

WITH CTE AS (
  SELECT Source AS account, 'debits' AS type, -Amount AS Amount
  FROM sa_general_journal
  UNION ALL 
  SELECT Destination, 'credits', Amount
  FROM sa_general_journal
)
SELECT acc.ID, acc.Name, 
       SUM(CASE WHEN CTE.type = 'debits'  THEN Amount END) AS Debits,
       SUM(CASE WHEN CTE.type = 'credits' THEN Amount END) AS Credits,
       SUM(Amount) AS Net
FROM CTE
JOIN sa_accounts acc ON CTE.account = acc.ID
GROUP BY acc.ID, acc.Name

输出（对于您的样本数据）：

ID          Name                    Debits  Credits Net
4059999     Test Income account     -10.1   null    -10.1
1009999     Test chequing account   -1.01   10.1    9.09
5059999     Test Income account     null    1.01    1.01

Demo on dbfiddle

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