[英]Left join the result of subquery in mysql
我想做以下事情,但它不起作用。 我想離開加入子查詢的結果。
select result1.id, result.name from (select cust.id as id, cust.name as name, ss.sold_date as soldDate, pp.product_name as productName from customers cust
left join sales ss on ss.customer_id = cust.id
left join products pp on pp.id = ss.product_id) as result
left join result as result1 on result.id = result1.id
當我這樣做時,它說表“結果”不存在。 我如何離開加入結果別名?
根據您的評論,您正在嘗試將子查詢結果與其自身結合起來。
在這種特殊情況下,它沒有任何意義,因為您只會獲得兩次相同的數據。 因此,使用子查詢一次將起作用
select result1.id, result.name
from (select cust.id as id, cust.name as name, ss.sold_date as soldDate, pp.product_name as productName
from customers cust
left join sales ss on ss.customer_id = cust.id
left join products pp on pp.id = ss.product_id) as result
left join result as result1 on result.id = result1.id
一般而言,如果您需要使用相同的子查詢兩次,您可以使用 CTE(公用表表達式):
with sub_q as (select cust.id as id, cust.name as name, ss.sold_date as soldDate, pp.product_name as productName
from customers cust
left join sales ss on ss.customer_id = cust.id
left join products pp on pp.id = ss.product_id)
select *
from sub_q res
left join sub_q res1
on res.id = res1.id
CTE(上面查詢的“with”部分)就像一個變量。 在“通常的”編程語言中,變量用於存儲值,而在查詢語言中,它的工作是存儲查詢
更新。 OP 似乎在 8.0 之前的 mysql 版本上,並且 db OP 不支持 CTE
因此,在這里您可能最終會使用視圖,例如
首先,一個創建視圖的腳本
create view sub_q as select cust.id as id, cust.name as name, ss.sold_date as soldDate, pp.product_name as productName
from customers cust
left join sales ss on ss.customer_id = cust.id
left join products pp on pp.id = ss.product_id;
二、運行查詢
select *
from sub_q res
left join sub_q res1
on res.id = res1.id;
或者,您可以在 select 語句中重復子查詢兩次
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