問題與子表 - MySql查詢

Question

我有一個問題，我的下面的查詢，銷售訂單項是銷售訂單表內的子表，這些表（銷售訂單項）有多個行只顯示第一項我的意思是第一行，所有其他行都沒有得到顯示。請擺脫一些燈光傢伙。

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，縮短它，使之更具可讀性：

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 

然後一些一般性的分析，而不必解決問題：

• 你真的需要LEFT？也就是說，「正確」表是可選的，你想要得到的NULLs？
• LEFT JOIN b ON b.x = 1和LEFT JOIN b WHERE b.x = 1之間的差異。前者可能比後者包含更多的b行。
• 擁有NULLs在許多情況下都很好，但如果可以避免它們，SQL就會變得更簡單。

更

開始運行它沒有 「GROUP BY tabSales Order。name」。你會得到你缺失的線條，但你可能會得到比你想要的更多。如果是這樣，然後刪除其餘的連接，讓我們看看我們是否可以將它們添加回來沒有項目的「內爆」。

爲了幫助理解數據，請說明每個關係是1：1還是1：多個還是多個：許多。