Question

我有多个表遵循从其成本计算阶段到开票阶段的报销请求，在请求的每个阶段记录订单项及其值。

为了获得每个阶段的每个费用类别的总数，我使用选择对它们进行求和 - 我无法弄清楚的是一种有效的方法来总结整个阶段的总计，而无需另外求和请求。由于这会查询数十万行 - 响应时间正在受到影响。当前的方法需要不到4.59秒 - 而当我删除总计的总和需要4.02秒时 - 任何关于如何使用原始总和来获得总计的建议都是非常受欢迎的：

SELECT p.program_name, p.id as program_id, p.entity_id, a.id as account_id,


(SELECT COALESCE(SUM(amount),0) FROM cost_items c WHERE expense_category_id = 9 and c.account_id = a.id) as NIPLoanCost,
(SELECT COALESCE(SUM(amount),0) FROM cost_items c WHERE expense_category_id = 2 and c.account_id = a.id) as AcquisitionCost,
(SELECT COALESCE(SUM(amount),0) FROM cost_items c WHERE expense_category_id = 3 and c.account_id = a.id) as PreDemoCost,
(SELECT COALESCE(SUM(amount),0) FROM cost_items c WHERE expense_category_id = 4 and c.account_id = a.id) as DemolitionCost,
(SELECT COALESCE(SUM(amount),0) FROM cost_items c WHERE expense_category_id = 5 and c.account_id = a.id) as GreeningCost,
(SELECT COALESCE(SUM(amount),0) FROM cost_items c WHERE expense_category_id = 6 and c.account_id = a.id) as MaintenanceCost,
(SELECT COALESCE(SUM(amount),0) FROM cost_items c WHERE expense_category_id = 7 and c.account_id = a.id) as AdministrationCost,
(SELECT COALESCE(SUM(amount),0) FROM cost_items c WHERE expense_category_id = 8 and c.account_id = a.id) as OtherCost,

(SELECT COALESCE(SUM(amount),0) FROM cost_items c WHERE c.account_id = a.id) 
as TotalCost,

/* I have removed the other 4 table's requests that follow this same methodology */

FROM programs p, accounts a  WHERE p.entity_id = a.entity_id;

Answer 1

您可以使用带有SELECT p.program_name, p.id as program_id, p.entity_id, t.* FROM programs p INNER JOIN accounts a ON p.entity_id = a.entity_id INNER JOIN ( SELECT a.id AS account_id, SUM(CASE WHEN c.expense_category_id = 9 THEN c.amount ELSE 0 END) AS NIPLoanCost, SUM(CASE WHEN c.expense_category_id = 2 THEN c.amount ELSE 0 END) AS AcquisitionCost, SUM(CASE WHEN c.expense_category_id = 3 THEN c.amount ELSE 0 END) AS PreDemoCost, SUM(CASE WHEN c.expense_category_id = 4 THEN c.amount ELSE 0 END) AS DemolitionCost, SUM(CASE WHEN c.expense_category_id = 5 THEN c.amount ELSE 0 END) AS GreeningCost, SUM(CASE WHEN c.expense_category_id = 6 THEN c.amount ELSE 0 END) AS MaintenanceCost, SUM(CASE WHEN c.expense_category_id = 7 THEN c.amount ELSE 0 END) AS AdministrationCost, SUM(CASE WHEN c.expense_category_id = 8 THEN c.amount ELSE 0 END) AS OtherCost, COALESECE(SUM(c.amount), 0) AS TotalCost FROM accounts a LEFT JOIN cost_items c ON a.id = c.account_id GROUP BY a.id ) t ON a.id = t.account_id表连接的条件聚合来替换所有子查询。这样的事情应该有效：

(SELECT COALESCE(SUM(amount),0) FROM cost_items c WHERE expense_category_id = 9 and c.account_id = a.id)

请注意，原始方法的缺陷是相关的子查询，如下所示：

programs

对于accounts和cost_items表的连接产品中的每条记录，您发出了一个单独的查询（实际上是其中许多），它正在扫描整个accounts表以确定和。在我的方法中，我们只通过cost_items和{{1}}的联接，这就是节省时间的地方。

简化mysql中求和的总和

1 个答案: