Php 如何在mysql中连接查询和计算？

如何在mysql中连接查询和计算

表中的fab_

制表

我的问题如下

  SELECT fabric_in.fabric_id, fabric_in.fabric_date, fabric_in.fabric_order_qty, fabric_in.fabric_rejected_qty, fabric_in.fabric_rejected_reason, fabric_in.fabric_total_qty, fabric_in.fabric_company_name, fabric_out.fabric_out_qty 
    FROM fabric_out 
    RIGHT JOIN fabric_in 
ON fabric_out.fabric_id=fabric_in.fabric_id

结果查询

但是我想要和面料的关系，你能帮忙吗

---

作为新版本，您需要添加所有列作为分组依据或加重功能。因此，您可以这样使用查询

    SELECT fabric_in.fabric_id,fabric_in.fabric_date, 
    fabric_in.fabric_order_qty, fabric_in.fabric_rejected_qty,
    fabric_in.fabric_rejected_reason, fabric_in.fabric_total_qty,
    fabric_in.fabric_company_name,a.fabric_out_qty 
        FROM fabric_in 
        RIGHT JOIN  (select fabric_id,sum(fabric_out_qty) as total from 
    fabric_out group by fabric_id ) as a
    ON a.fabric_id=fabric_in.fabric_id

---

作为新版本，您需要将所有列添加为group by或EARGER函数。因此，您可以这样使用查询

    SELECT fabric_in.fabric_id,fabric_in.fabric_date, 
    fabric_in.fabric_order_qty, fabric_in.fabric_rejected_qty,
    fabric_in.fabric_rejected_reason, fabric_in.fabric_total_qty,
    fabric_in.fabric_company_name,a.fabric_out_qty 
        FROM fabric_in 
        RIGHT JOIN  (select fabric_id,sum(fabric_out_qty) as total from 
    fabric_out group by fabric_id ) as a
    ON a.fabric_id=fabric_in.fabric_id

---

您必须使用MySQL的聚合函数：

SELECT MAX(fabric_in.fabric_id) as fabric_id, 
       MAX(fabric_in.fabric_date) as fabric_date, 
       MAX(fabric_in.fabric_order_qty) as fabric_order_qty, 
       MAX(fabric_in.fabric_rejected_qty) as fabric_rejected_qty, 
       MAX(fabric_in.fabric_rejected_reason) as fabric_rejected_reason, 
       MAX(fabric_in.fabric_total_qty) as fabric_total_qty, 
       MAX(fabric_in.fabric_company_name) as fabric_company_name, 
       SUM(fabric_out.fabric_out_qty) AS fabric_out_qty
    FROM fabric_out 
    RIGHT JOIN fabric_in 
    ON fabric_out.fabric_id=fabric_in.fabric_id
    GROUP BY fabric_in.fabric_id

---

这本书适合您，未经测试。

您必须使用MySQL的聚合函数：

SELECT MAX(fabric_in.fabric_id) as fabric_id, 
       MAX(fabric_in.fabric_date) as fabric_date, 
       MAX(fabric_in.fabric_order_qty) as fabric_order_qty, 
       MAX(fabric_in.fabric_rejected_qty) as fabric_rejected_qty, 
       MAX(fabric_in.fabric_rejected_reason) as fabric_rejected_reason, 
       MAX(fabric_in.fabric_total_qty) as fabric_total_qty, 
       MAX(fabric_in.fabric_company_name) as fabric_company_name, 
       SUM(fabric_out.fabric_out_qty) AS fabric_out_qty
    FROM fabric_out 
    RIGHT JOIN fabric_in 
    ON fabric_out.fabric_id=fabric_in.fabric_id
    GROUP BY fabric_in.fabric_id

这本书适合你，没有经过测试。

试试这个

SELECT 
    fabric_in.fabric_id,
    fabric_in.fabric_date, 
    fabric_in.fabric_order_qty, 
    fabric_in.fabric_rejected_qty,
    fabric_in.fabric_total_qty,
    fabric_in.fabric_company_name,
    a.fabric_out_qty 
FROM fabric_in RIGHT JOIN  
    (select sum(fabric_out_qty) as out_total FROM fabric_out group by fabric_id ) as a ON a.fabric_id=fabric_in.fabric_id;

试试这个

SELECT 
    fabric_in.fabric_id,
    fabric_in.fabric_date, 
    fabric_in.fabric_order_qty, 
    fabric_in.fabric_rejected_qty,
    fabric_in.fabric_total_qty,
    fabric_in.fabric_company_name,
    a.fabric_out_qty 
FROM fabric_in RIGHT JOIN  
    (select sum(fabric_out_qty) as out_total FROM fabric_out group by fabric_id ) as a ON a.fabric_id=fabric_in.fabric_id;

---

按结构使用MySQL组的SUM（fabric_out_qty）函数按结构使用MySQL组的SUM（fabric_out_qty）函数谢谢jazdanowThank you jazdanowThank you monaris谢谢你