Question

已经有很多left join个问题，但我仍然不能完全理解这个问题。移动时，WHERE条件看起来不合理。

问题是应该返回4行，但只有1行。

在检查左连接条件时，每个左连接返回1行，这对于表中的记录数是正确的，但是下面的查询返回1条记录而不是4条，但是我看不清楚返回4，但是。

查询如下：（给出1个结果不是4;预期4个）

SELECT 
 list.uid,
 list.business_uid,
 list.creator_name,
 business.company_name,
 list_alias.uid AS list_alias_uid,
 list_alias.alias AS list_alias,
 list_member.uid AS list_member_uid,
 mailbox.full_name AS list_member_name,
 mailbox.email_address AS list_member_email_address
FROM
 mailbox,
 business,
 list
LEFT JOIN
 list_alias ON list_alias.list_uid=list.uid
LEFT JOIN
 list_member ON list_member.list_uid=list.uid
WHERE
 list.business_uid='1'
 AND list.business_uid=business.uid
 AND mailbox.uid=list_member.mailbox_uid
ORDER BY
 list.full_name ASC

数据：

商业UID 1有4个列表

SELECT * FROM list WHERE business_uid=1 - 提供4个结果

SELECT * FROM list_alias WHERE list_uid IN (SELECT uid FROM list WHERE business_uid=1) - 给出1个结果

SELECT * FROM list_member WHERE list_uid IN (SELECT uid FROM list WHERE business_uid=1) - 给出1个结果

欢迎任何有关我可以检查的内容的指示。

表格样本数据：

列表：

uid | business_uid | creator_name | full_name
--------------------------------------------------
 1          1         List Maker   Subscribe to W
 2          1         List Maker   Subscribe to X
 3          1         List Maker   Subscribe to Y
 4          1         List Maker   Subscribe to Z

业务：

uid | company_name
-------------------
 1    List Company

list_alias：

uid | list_uid | alias
----------------------------------------
 1       1       subscriber@list-url.com

LIST_MEMBER：

uid | list_uid | mailbox_uid
------------------------------------
 1       1       1

邮箱：

uid | full_name | email_address
-------------------------------
 1     I am He    me@me.com

Answer 1

尝试使用单一的连接方法，就像这样。

SELECT list.uid,
    list.business_uid,
    list.creator_name,
    b.company_name,
    la.uid AS list_alias_uid,
    la.alias AS list_alias,
    lm.uid AS list_member_uid,
    m.full_name AS list_member_name,
    m.email_address AS list_member_email_address
FROM list LEFT JOIN list_member lm ON lm.list_uid=list.uid
    LEFT JOIN mailbox m ON m.uid=lm.mailbox_uid
    LEFT JOIN business b ON list.business_uid=b.uid
    LEFT JOIN list_alias la ON la.list_uid=list.uid
WHERE list.business_uid=1
ORDER BY list.full_name ASC

Answer 2

问题：'list'表中'uid'的值是多少？因为'uid'与'business_uid'不同。我的意思是......

如果'list'表有这个......

'uid''business_uid' 1 1 2 1 3 1 4 1

那就是问题所在。您将返回相同的'busines_uid'但不同的'uid'，这意味着它只会匹配第一条记录。

mysql left join产生的结果少于预期

2 个答案: