Question

我正在尝试更新website_id等于6的客户，并且当website_id为3时，他们的电子邮件不会重复。

我已经能够让website_id为6的所有客户以及website_id为3时他们的电子邮件不重复，SQL语句如下所示。

SELECT
  *
FROM customer_entity
WHERE website_id = 6
AND email NOT IN (SELECT
  email
FROM customer_entity
WHERE website_id = 3);

现在，当我尝试更新所有site_id和store_id等于3到6的客户时，他们的电子邮件在store_id = 3中不重复

UPDATE customer_entity customers
    SET customers.website_id = 3, customers.store_id = 3 
    WHERE customers.website_id = 6 AND customers.email 
      NOT IN (SELECT email FROM customer_entity WHERE website_id = 3);

我收到以下错误

您无法在FROM子句
中为更新指定目标表'customers'

我如何实现我想要做的事情？

编辑：我也试过没有别名，我仍然得到同样的错误。

Answer 1

从代码中删除别名。试试这个：

UPDATE customer_entity 
SET website_id = 3, store_id = 3 
WHERE website_id = 6 AND email 
  NOT IN (SELECT email FROM customer_entity WHERE website_id = 3);

Answer 2

您必须使用MySQL。您可以使用join执行此操作。我会把它写成：

update customer_entity ce join
       (SELECT email
        FROM customer_entity
        GROUP BY email
        HAVING SUM(website_id = 6) > 0 AND
               SUM(website_id = 3) = 0
      ) e
      on ce.email = e.email
    set ce.website_id = 3,
        ce.store_id = 3 ;

Answer 3

这是为了更正或增强您的更新声明

更新FROM而不是EXISTS否则基本更新而不是

UPDATE customers
    SET website_id = 3
       ,store_id = 3 
    from customer_entity customers
    WHERE website_id = 6 
          AND NOT EXISTS 
            (SELECT 1 
               FROM customer_entity c 
               WHERE c.website_id = 3
               c.email = customers.email
            );

更新而不是

 UPDATE customer_entity 
        SET website_id = 3
           ,store_id = 3 
        WHERE website_id = 6 
              AND email not in
                (SELECT email
                   FROM customer_entity 
                   WHERE website_id = 3
                );

然而从你的问题看这个陈述

现在，当我尝试更新所有拥有website_id和的客户时 store_id等于3到6，其中的电子邮件不重复 store_id = 3

这应该是你的解决方案：更新和不存在

UPDATE customers

        SET website_id = 6      -- Now when I try to update all customers that have website_id and store_id equal to 3 to 6
           ,store_id = 6 

        from customer_entity customers
        WHERE website_id = 3 and store_id = 3     -- Now when I try to update all customers that have website_id and store_id equal to 3 to 6

              AND NOT EXISTS 
                (SELECT 1 
                   FROM customer_entity c 
                   WHERE c.store_id = customers.store_id        -- where their email is not duplicate in store_id = 3 
                   and c.email = customers.email
                );

更新而不是

UPDATE customer_entity
        SET website_id = 6      -- Now when I try to update all customers that have website_id and store_id equal to 3 to 6
           ,store_id = 6 

        WHERE website_id = 3 and store_id = 3     -- Now when I try to update all customers that have website_id and store_id equal to 3 to 6

              AND email NOT IN
                (SELECT email 
                   FROM customer_entity c 
                   WHERE store_id = 3       -- where their email is not duplicate in store_id = 3 

                );

Answer 4

以下是我自己的问题的答案。这里的许多答案都使用JOINS到一个表，所以对于MYSQL，这不起作用，因为你无法在同一个SELECT JOIN上更新。

我的解决方案是创建一个临时表并存储所有具有store_id和website_id = 6的值以及WHERE电子邮件不重复WHERE store_id和website_id = 3.

CREATE TEMPORARY TABLE customer_temp_table(
   SELECT email FROM customer_entity 
      WHERE 
         website_id = 6 
      AND  
         email NOT IN (SELECT email FROM customer_entity WHERE website_id = 3));

然后我使用临时表的结果更新客户实体表。

UPDATE customer_entity AS customers SET customers.website_id = 3, customers.store_id = 3
    WHERE customers.email IN (SELECT email FROM customer_temp_table) AND customers.website_id = 6;

Answer 5

如果您在表格上执行 UPDATE / INSERT / DELETE ，则无法在内部查询中引用该表格（但您可以从该外部查询中引用该字段表......）

解决方案是用（SELECT * FROM customer_entity）替换子查询中的myTable实例，就像这样

SELECT * FROM customer_entity WHERE website_id = 3 as something

您可以在此处找到更多信息：You can't specify target table for update in FROM clause

更新列不在选择中

5 个答案: