Question

我想请求帮助和UPDATE命令，这取决于4个表的关系。（或者甚至可能吗？

这是我的桌子：

Characters
guid, ...., totalItemlevel

template_items
id, ... itemlevel, ...,

instances_item
itemInstanceID, itemID, Owner
(itemID is foreign key template_items.id, Owner is foreign key Character.guid)

inventory_character
guid, slot, iteminstance
(guid is foreign key character.guid, item is foreign key instances_item.itemInstanceID)

现在我需要做一个UPDATE脚本，它将totalItemLevel设置为template_items.itemLevel的总和，如果角色在库存中具有插槽＆gt; 20

的项目的实例

在伪陈述中：

UPDATE characters 
SET    totalItemLevel = SUM(FOR EACH ITEM IN INVENTORY(is slot>20?(GET its template(get itemLevel))));

所以我试过这样的事情

SELECT  SUM(template_items.itemLevel) 
FROM    template_items, 
        inventory_character, 
        instances_item, 
        characters 
WHERE   chararacter.guid = inventory_character.guid 
AND     inventory_character.slot > 19 
AND     inventory_character.iteminstance = instances_item.itemInstanceID 
AND     instances_item.itemID = template_items.id);

这绝对不行，它很疯狂，它可能需要一些子查询，但我真的不知道，如何实现这一点。

此外，我无法以其他方式执行此操作，例如在PHP中执行某些查询，然后对其进行计数...我需要直接在数据库中执行此更新。

那么请，任何想法，如何实现这个计算列？

提前致谢:)

Answer 1

假设你需要插槽＆gt; 20（在你的帖子中你有一个插槽＆gt; 20在一个地方，插槽＆gt; 19在另一个地方）：

UPDATE Characters
SET Characters.totalItemLevel = s FROM
(
    SELECT SUM(template_items.itemlevel) AS s, c.[guid]
    FROM Characters c
    INNER JOIN inventory_character ON inventory_character.[guid] = c.[guid]
    INNER JOIN instances_item ON instances_item.itemInstanceID = inventory_character.iteminstance AND instances_item.[Owner] = c.[guid]
    INNER JOIN template_items ON template_items.id = instances_item.itemID 
    WHERE c.[guid] = instances_item.[Owner] AND slot > 20
    GROUP BY c.[guid]
) A
WHERE Characters.[guid] = A.[guid]

MySQL SUM更新超过4个表

1 个答案: