0

我有一张包含各种成本明细的发票表。我也有一个包含杂项费用的链接表。关系是一张发票到许多杂项费用

当我执行 SUM() 并且杂项成本表 (RIGHT) 上存在多行时，发票表 (LEFT) 中的值乘以找到的行数。

DB小提琴：https ://www.db-fiddle.com/f/9CKGRShXPoT3k4Wcm83fVD/5

LABOUR、PARTS、POLS 和 SUBLET 的原始查询结果不正确：

SELECT  SUM(ad.labour_cost) AS LABOUR,
    SUM(ad.part_cost) AS PARTS,
    SUM(ad.pol_cost) AS POLS,
    SUM(ad.sublet_cost) AS SUBLET,
    SUM(am.misc_sales_amt)  AS MISC
FROM AdvisorSalesData ad 
LEFT JOIN AdvisorMiscSalesData am 
  ON (ad.customer_id=am.customer_id AND ad.invoice_no=am.invoice_no)
WHERE ad.customer_id IN (3)

调整后的查询 - 删除了 INNER JOIN 并在 misc 表上尝试了一个 SELECT，但 MISC 的结果不正确（NULL）：

SELECT  SUM(ad.labour_cost) AS LABOUR,
    SUM(ad.part_cost) AS PARTS,
    SUM(ad.pol_cost) AS POLS,
    SUM(ad.sublet_cost) AS SUBLET,
    (SELECT SUM(misc_sales_amt) FROM AdvisorMiscSalesData WHERE customer_id IN (3) AND invoice_no=ad.invoice_no ) AS MISC
FROM AdvisorSalesData ad 
WHERE ad.customer_id IN (3)

如何调整以在一个查询中获得所有 5 列的正确结果？

4

1 回答 1

0

加入前聚合：

SELECT SUM(ad.labour) AS LABOUR,
       SUM(ad.parts) AS PARTS,
       SUM(ad.pols) AS POLS,
       SUM(ad.sublet) AS SUBLET,
       SUM(am.misc)  AS MISC
FROM (SELECT customer_id, invoice_no,
             SUM(ad.labour_cost) AS LABOUR,
             SUM(ad.part_cost) AS PARTS,
             SUM(ad.pol_cost) AS POLS,
             SUM(ad.sublet_cost) as sublet
      FROM AdvisorSalesData ad 
      GROUP BY customer_id, invoice_no
     ) ad LEFT JOIN
     (SELECT customer_id, invoice_no,
             SUM(am.misc_sales_amt) AS MISC
      FROM AdvisorMiscSalesData am 
      GROUP BY customer_id, invoice_no
     ) am
     ON ad.customer_id = am.customer_id AND    
        ad.invoice_no = am.invoice_no
WHERE ad.customer_id IN (3);

这是一个 db<>fiddle。

您可以在子查询中重复过滤器以获得更好的性能：

SELECT SUM(ad.labour) AS LABOUR,
       SUM(ad.parts) AS PARTS,
       SUM(ad.pols) AS POLS,
       SUM(ad.sublet) AS SUBLET,
       SUM(am.misc)  AS MISC
FROM (SELECT customer_id, invoice_no,
             SUM(ad.labour_cost) AS LABOUR,
             SUM(ad.part_cost) AS PARTS,
             SUM(ad.pol_cost) AS POLS,
             SUM(ad.sublet_cost) as sublet
      FROM AdvisorSalesData ad 
      WHERE ad.customer_id IN (3)
      GROUP BY customer_id, invoice_no
     ) ad LEFT JOIN
     (SELECT customer_id, invoice_no,
             SUM(am.misc_sales_amt) AS MISC
      FROM AdvisorMiscSalesData am 
      WHERE ad.customer_id IN (3)
      GROUP BY customer_id, invoice_no
     ) am
     ON ad.customer_id = am.customer_id AND    
        ad.invoice_no = am.invoice_no
WHERE ad.customer_id IN (3);
于 2020-02-26T22:42:36.447 回答