Question

请记住，声明和表格都有更多内容，但我会简短地阅读。

我正试图了解用户是否已经进入商店。表格看起来像这样。

--Shop--
ShopID

--Visit--
ShopID
UserID

以下是我现在已经走了多远。

SELECT shop.ShopID, visit.UserID AS Visited FROM shop
LEFT OUTER JOIN visit ON shop.ShopID = visit.ShopID
WHERE UserID = 1

此代码的问题在于，对于非UserID 1的访问，它不会显示NULL。我已尝试将OR UserID IS NULL添加到语句中，但它只留给我所有尚未存在的商店任何用户访问过。

我可以通过检查Visited是否与UserID相同来过滤它，但如果数据库可以将其写为1和0则会更容易。所以要问具体：

如果在访问表中与ShopID连接的UserID 1，如何使Visited输出0或1？

Answer 1

您加入的问题是您将所有商店ID保留在列表中，因为您只需创建一个扩展所有shop和visit s的列表使用未访问过的shop。如果我找到你想要的是shop所访问的user = 1。只需使用Visit表即可轻松实现这一目标。

Select DISTINCT visit.shopID FROM visit
WHERE UserID = 1

Answer 2

Dennis Klopfer's answer是简化和表现的方法。但也许你想要更像这样的东西，检查某个用户曾经访问过每个商店：

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如果是这样的话，我会尝试这样的事情：

shop |  visited by user 1
 #1  |       false
 #2  |       true

当然，您还可以在查询中添加SELECT s.ShopID, IF(ISNULL(v.UserID), FALSE, TRUE) AS visited FROM shop S LEFT JOIN visit v ON s.ShopID = v.ShopID AND v.UserID = 1子句，并按任意字段过滤商店列表。

Answer 3

SELECT shop.ShopID, if(visit.UserID is null,0,1) AS Visited FROM shop
LEFT OUTER JOIN visit ON (shop.ShopID = visit.ShopID
AND visit.UserID = 1 )

Answer 4

我正在努力实现这一目标，我想：

您需要将ON条件中的条件放在WHERE子句而不是SELECT子句中。

但是，如果用户可以多次访问，那么您将获得多行。所以，我建议在select s.ShopId, (select v.UserId from visit v where s.SHopId = v.ShopId and v.userId = 1 limit 1 ) as UserId from shop s;中使用子查询来获取用户的任意行：

MySQL：选择用户是否在另一个表中

4 个答案: