用MySQL中的其他5個表填充數據表

Question

我只想使用（My）SQL語句在MySQL中填充關係表。問題是，爲什麼我現在要做的是，我必須用條件填充依賴於其他表格的數據。我從來沒有做過很多有關條件編程和SQL內循環，我現在試過的每一個部分，使它以某種方式工作失敗悲慘。我可能想稍後將結果存儲爲存儲過程，但現在工作語句不會令人滿意。

表設置：

我有5代表具有數據和我需要填寫第六一個與關係的數據。表結構是這樣的：

現在我要填寫僅使用SQL（MySQL的）以下規則CategoryProducts（僞碼）

foreach category-id: 
    if allTags is true then 
    fetch each product-id that has all tags that the category has 
    else 
    fetch each product-id that has at least one tag that the category has 
    fi 

    insert each product-id matching into CategoryProducts 
end foreach 

說實話，我不知道如何做到這一點，只使用普通的SQL。我通常會通過代碼實現這一點，但這次我不能這樣做。我想用它作爲我們某個外部服務器的安裝腳本的一部分，但在安裝例程的這一部分中，我只能執行SQL語句。

編輯：刪除表格設置圖片中的醜陋的鼠標光標。

Answer 1

我想你可以用「簡單」 INSERT..SELECT做到這一點：

insert into CategoryProducts(cID, pID) 
select 
    -- The field list in `select` should match the list in the `insert into` clause. 
    x.cID, 
    x.pID 
from 
    -- Inner select returns all possible combinations, yet unfiltered by the 
    -- conditions you specified, and returns the number of tags 
    -- for each product, each catagory, and the number of tags they share. 
    -- If SharedTagCount = cTagCount, it means that the product has all the tags 
    -- of the category (not necessarily the other way around). 
    (select 
    p.ID as pID, 
    c.ID as cID, 
    c.allTags, 
    (select count(*) from ProductTags pt where pt.pID = p.ID) as pTagCount, 
    (select count(*) from CategoryTags ct where ct.cID = c.ID) as cTagCount, 
    (select 
     count(*) 
    from 
     ProductTags pt 
     inner join CategoryTags ct on ct.tCode = pt.tCode 
    where 
     pt.pID = p.ID and 
     ct.cID = c.ID) as SharedTagCount 
    from 
    Product p 
    cross join Categories c) x 
where 
    -- Outer select filters. 
    -- if SharedTagCount = cTagCount, it means that the product has all 
    -- the tags of the category. 
    (x.AllTags and 
    x.cTagCount = x.SharedTagCount) or 
    -- if SharedTagCount > 0, it means the product has at least one tag of the category. 
    ((not x.AllTags) and 
    x.SharedTagCount > 0)