[英]Show Results not in joining table
我有三张表:
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 JOIN
和NULL
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`);
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