Question

我有三张桌子：

儿童
活动
signupActivity

表'儿童'包含儿童的详细信息。

表'活动'包含活动的详细信息。

当孩子注册活动时，会在表中添加一个带有activityID和childID的条目。我正在尝试获取尚未注册活动的孩子列表。我尝试了以下查询，但收到错误：

“不唯一的表/别名：'children'”

SELECT children.childrenEmailAddress
FROM children 
    INNER JOIN activities ON signupActivity.SignupActivityID = activities.activityID
    INNER JOIN children ON signupActivity.signupActivitychildID = children.childrenID
LEFT JOIN signupActivity ON children.childrenID = signupActivity.signupActivitychildID
WHERE activities.activityID = 8

记录集还需要包含基于加入表Children和Activity中的'section'的结果。 children.childrenSection = activities.activitySection以及通过activities.activityID过滤记录集

这是我改编自Arth的决赛：

 SELECT c.childrenEmailAddress
     FROM children c
     JOIN activities a
       ON a.activitySection = c.childrenSection
        AND a.activityID = 8
LEFT JOIN signupActivity sa 
       ON sa.signupActivitychildID = c.childrenID
    WHERE sa.signupActivitychildID IS NULL

Answer 1

SELECT
    *
FROM
    Childeren C1
WHERE
    NOT EXISTS (SELECT *
        FROM
           signupActivity S1
        WHERE
           S1.activityID= 8 AND
           C1.ChildID= S1.ChildID)

Answer 2

我使用LEFT JOIN和NULL支票

   SELECT c.childrenEmailAddress
     FROM children c
LEFT JOIN signupActivity sa 
       ON sa.signupActivitychildID = c.childrenID
      AND sa.SignupActivityID = 8
    WHERE sa.signupActivitychildID IS NULL

就我个人而言，我不喜欢在每个列名中重复表名。我发现使用它是不必要和令人沮丧的。您的查询可以简单如下：

   SELECT c.emailAddress
     FROM child c
LEFT JOIN signupActivity sa 
       ON sa.childId = c.id 
      AND sa.activityID = 8
    WHERE sa.childId IS NULL

<强>更新

   SELECT c.childrenEmailAddress
     FROM children c
     JOIN activities a
       ON a.activitySection = c.childrenSection
      AND a.activityID = 8
LEFT JOIN signupActivity sa 
       ON sa.signupActivitychildID = c.childrenID
      AND sa.signupActivityactivityID = a.activityID
    WHERE sa.signupActivitychildID IS NULL

Answer 3

SELECT *
FROM `children` c
WHERE c.childID NOT IN(
SELECT `childID`
FROM `activities`);

显示结果不在连接表中

3 个答案: