子表问题 - MySql 查询

Question

我的以下查询有问题，销售订单项目是销售订单表中的一个子表，具有多行的那些表（销售订单项目）仅显示第一个项目我的意思是第一行并且所有其他行都没有显示。请给一些小伙伴们一些启发。

select 
 `tabSales Order`.`name` as "Sales Order:Link/Sales Order:120",
`tabSales Order`.`customer` as "Customer:Link/Customer:120",
 `tabSales Order Item`.item_code as "Item:Link/Item:120",
`tabSales Order Item`.item_name as "Des:Link/Item:120",
`tabSales Order Item`.qty as "Qty:Int:100",
 `tabSales Order Item`.delivered_qty as "Delivered Qty:Int:100",
 (`tabSales Order Item`.qty - ifnull(`tabSales Order Item`.delivered_qty, 0)) as "Qty to Deliver:Int:140",
`tabSales Order Item`.date as "Sch date:Date:100",
`tabItem`.thumbnail as "Cut wt:Data:60",
(`tabSales Order Item`.qty * `tabItem`.thumbnail) as "Tonnage:Float:80",
`tabProduct master`.grade as "Grade:Data:100",
`tabProduct master`.dieno as "Die no:Data:100",
`tabProduct master`.hammer as "Hammer:Data:50",
`tabProduct master`.rm_idl_cs as "Idl sec:Data:60",
`tabProduct master`.rm_alt_cs as "Alt sec:Data:60",
ifnull(SUM(CASE WHEN `tabSales Order`.`name` = `tabProduction Order`.`sales_order` and `tabSales Order Item`.`item_code` = `tabProduction Order`.`production_item`
         THEN `tabProduction Order`.`qty`
    END),0) as "Issued:Int:80",
SUM(CASE WHEN `tabSales Order`.`name` = `tabProduction Order`.`sales_order` and `tabSales Order Item`.`item_code` = `tabProduction Order`.`production_item`
         THEN `tabProduction Order`.`produced_qty`
    END) as "Cut:Int:80",
(`tabSales Order Item`.qty - ifnull(SUM(CASE WHEN `tabSales Order`.`name` = `tabProduction Order`.`sales_order` and `tabSales Order Item`.`item_code` = `tabProduction Order`.`production_item`
         THEN `tabProduction Order`.`qty`
    END),0)) as "To be issued:Int:80"

    from  `tabSales Order`
    LEFT JOIN  `tabSales Order Item`  ON ( `tabSales Order Item`.`parent` = `tabSales Order`.`name`
              and  `tabSales Order`.docstatus = 1  )
    LEFT JOIN  `tabItem`  ON `tabItem`.`item_code` = `tabSales Order Item`.`item_code`
    LEFT JOIN  `tabProduct master`  ON `tabProduct master`.`part_no` = `tabSales Order Item`.`item_code`
    LEFT JOIN  `tabProduction Order`
               ON (`tabSales Order Item`.`item_code` = `tabProduction Order`.`production_item`
              and  `tabSales Order`.`name` = `tabProduction Order`.`sales_order` )
    where  `tabSales Order`.status not in ("Stopped", "Closed", "Completed" )
      and  ifnull(`tabSales Order Item`.delivered_qty,0) < ifnull(`tabSales Order Item`.qty,  0 )
    GROUP BY  `tabSales Order`.`name`
    HAVING  (`tabSales Order Item`.qty -
                SUM(CASE WHEN `tabSales Order`.`name` = `tabProduction Order`.`sales_order`
                      and  `tabSales Order Item`.`item_code` = `tabProduction Order`.`production_item`
                         THEN `tabProduction Order`.`qty` END)
            ) > 0
      or  (`tabSales Order Item`.qty -
                SUM(CASE WHEN `tabSales Order`.`name` = `tabProduction Order`.`sales_order`
                      and  `tabSales Order Item`.`item_code` = `tabProduction Order`.`production_item`
                         THEN `tabProduction Order`.`qty` END)
            ) is NULL;

Answer 1

首先，将其缩短以使其更具可读性：

    FROM  `tabSales Order` AS so
    LEFT JOIN  `tabSales Order Item` AS soi
               ON ( soi.`parent` = so.`name`
              AND  so.docstatus = 1  )
    LEFT JOIN  `tabItem` AS i  ON i.`item_code` = soi.`item_code`
    LEFT JOIN  `tabProduct master` AS pm
               ON pm.`part_no` = soi.`item_code`
    LEFT JOIN  `tabProduction Order` AS po
               ON (soi.`item_code` = po.`production_item`
              AND  so.`name` = po.`sales_order` )
    WHERE  so.status NOT IN ("Stopped", "Closed", "Completed" )
      AND  IFNULL(soi.delivered_qty,0) < IFNULL(soi.qty,  0 )
    GROUP BY  so.`name`
    HAVING  IFNULL(soi.qty -
                SUM(CASE WHEN so.`name` = po.`sales_order`
                   AND  soi.`item_code` = po.`production_item`
                         THEN po.`qty` END), 1) > 0

然后进行一些一般性的分析，不一定能解决问题：

• 你真的需要LEFT吗？也就是说，“正确”表是可选的吗？您想要可能得到的NULLs？
• LEFT JOIN b ON b.x = 1 和 LEFT JOIN b WHERE b.x = 1 之间存在差异。前者可能包含比后者更多的b 行。
• NULLs 在很多情况下都很好，但如果你能避免它们，SQL 就会变得更简单。

更多

首先在没有“GROUP BY tabSales Order.name”的情况下运行它。你会得到你丢失的线条，但你可能会得到比你想要的更多的东西。如果是这样，那么删除其余的连接，让我们看看我们是否可以在没有项目“内爆”的情况下重新添加它们。

为了帮助理解数据，请说明每个关系是 1:1 还是 1:many 或 many:many。