Php 选择3个不同的表

我有3个表名为

rsales

，

rreturn

，

productlist

。表

rsales

和

rreturn

有

total

列。我的目标是，我要将

rsales

中的所有

total

值和

rreturn

表中的

total

值相加，并仅从

productlist

表中选择特定值

比如说，我从

productlist

表中获得了以下数据：

id | pcode | pname | pdesc | 
1  |  222  | 33uf  |  10v  |

id | total | pcode |
1  |  200  | 222   |
2  |  200  | 222   |

id | total | pcode |
1  |  50   | 222   |
2  |  20   | 222   |

来自

rsales

表的数据：

id | pcode | pname | pdesc | 
1  |  222  | 33uf  |  10v  |

id | total | pcode |
1  |  200  | 222   |
2  |  200  | 222   |

id | total | pcode |
1  |  50   | 222   |
2  |  20   | 222   |

来自

rreturn

表的数据：

id | pcode | pname | pdesc | 
1  |  222  | 33uf  |  10v  |

id | total | pcode |
1  |  200  | 222   |
2  |  200  | 222   |

id | total | pcode |
1  |  50   | 222   |
2  |  20   | 222   |

输出必须如下所示：

   id | pcode | pname | pdesc | total 
   1  |  222  | 33uf  |  10v  |  470

我的问题是：我想对

rsales

和

rreturn

表中的所有

total

值求和，然后从

productlist

中选择所有值。我有下面的代码，运行得很好。但它只能对

rsales

表中的总值求和，或者更确切地说，它只能对单个表中的值求和

$result = mysql_query("SELECT 
    productlist.*, 
    SUM(rsales.total) as total, 
        SUM(rsales.vatable_sales) as vatable_sales, 
    SUM(rsales.vats) as vats,
     SUM(rsales.discount) as discount

FROM productlist
LEFT JOIN rsales ON rsales.pcode = productlist.pcode
GROUP BY pcode
ORDER BY total ASC");

---

你不太清楚什么不适合你。你是否得到了结果，但数字并不是你所期望的？好吧，假设你有三次销售和两次退货，那么通过合并所有表，你会得到六条记录。因此，你的销售额翻了一番，回报翻了三倍

对于一个解决方案，您可以加入聚合：

$result = mysql_query("SELECT 
    productlist.*, 
    sumsales.sum_total + sumreturns.sum_total as total, 
    sumsales.sum_vatable_sales as vatable_sales, 
    sumsales.sum_vats as vats,
    sumsales.sum_discount as discount
    FROM productlist
    LEFT JOIN
    (
      SELECT pcode, SUM(total) as sum_total, SUM(vatable_sales) as sum_vatable_sales, SUM(vats) as sum_vats, SUM(discount) as sum_discount
      FROM rsales
      GROUP BY pcode
    ) AS sumsales ON sumsales.pcode = productlist.pcode
    LEFT JOIN
    (
      SELECT pcode, SUM(total) as sum_total 
      FROM rreturn 
      GROUP BY pcode
    ) AS sumreturns ON sumreturns.pcode = productlist.pcode
    ORDER BY total ASC");

---

你不太清楚什么不适合你。你是否得到了结果，但数字并不是你所期望的？好吧，假设你有三次销售和两次退货，那么通过合并所有表，你会得到六条记录。因此，你的销售额翻了一番，回报翻了三倍

对于一个解决方案，您可以加入聚合：

$result = mysql_query("SELECT 
    productlist.*, 
    sumsales.sum_total + sumreturns.sum_total as total, 
    sumsales.sum_vatable_sales as vatable_sales, 
    sumsales.sum_vats as vats,
    sumsales.sum_discount as discount
    FROM productlist
    LEFT JOIN
    (
      SELECT pcode, SUM(total) as sum_total, SUM(vatable_sales) as sum_vatable_sales, SUM(vats) as sum_vats, SUM(discount) as sum_discount
      FROM rsales
      GROUP BY pcode
    ) AS sumsales ON sumsales.pcode = productlist.pcode
    LEFT JOIN
    (
      SELECT pcode, SUM(total) as sum_total 
      FROM rreturn 
      GROUP BY pcode
    ) AS sumreturns ON sumreturns.pcode = productlist.pcode
    ORDER BY total ASC");

---

请向我们展示样本数据。以及该数据的预期输出。我已经提供了一个示例。请向我们展示样本数据。以及该数据的预期输出。我已经提供了一个示例