繁体   English   中英

显示结果不在连接表中

[英]Show Results not in joining table

我有三张表:

  1. 孩子们
  2. 活动
  3. 注册活动

表“儿童”包含儿童的详细信息。

表“活动”包含活动的详细信息。

当一个孩子注册一个活动时,一个带有activityID 和childID 的条目被添加到表中。 我正在尝试获取尚未报名参加活动的儿童名单。 我尝试了以下查询,但出现错误:

“不是唯一的表/别名:‘儿童’”

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“部分”的结果。 children.childrenSection = activity.activitySection 以及通过 activity.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
SELECT
    *
FROM
    Childeren C1
WHERE
    NOT EXISTS (SELECT *
        FROM
           signupActivity S1
        WHERE
           S1.activityID= 8 AND
           C1.ChildID= S1.ChildID)

我会使用LEFT JOINNULL检查

   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。我觉得使用它是不必要的并且令人沮丧。 您的查询可能很简单:

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

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

暂无
暂无

声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.

 
粤ICP备18138465号  © 2020-2024 STACKOOM.COM