MySQL 连接和来自另一个表的 COUNT(*)

发布于 2024-10-15 10:10:48 字数 450 浏览 2 评论 0原文

我有两个表:groupsgroup_members

groups 表包含每个群组的所有信息,例如 ID、标题、描述等。

group_members 表中,它列出了属于该群组的所有成员。每个组都是这样的:

group_id | user_id
1 | 100
2 | 23
2 | 100
9 | 601

基本上,我想在页面上列出三个组,并且我只想列出拥有四个以上成员的组。在 循环内,我想要该组中的四个成员。我在列出组和在另一个内部循环中列出成员时没有遇到任何问题,我只是无法细化组,以便只有那些拥有超过 4 个成员的组才会显示。

有人知道该怎么做吗?我确信它与 MySQL 连接有关。

I have two tables: groups and group_members.

The groups table contains all the information for each group, such as its ID, title, description, etc.

In the group_members table, it lists all the members who are apart of each group like this:

group_id | user_id
1 | 100
2 | 23
2 | 100
9 | 601

Basically, I want to list THREE groups on a page, and I only want to list groups which have MORE than four members. Inside the <?php while ?> loop, I then want to four members who are apart of that group. I'm having no trouble listing the groups, and listing the members in another internal loop, I just cannot refine the groups so that ONLY those with more than 4 members show.

Does anybody know how to do this? I'm sure it's with MySQL joins.

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嗫嚅 2024-10-22 10:10:48

MySQL 使用 HAVING 语句来执行此任务。

您的查询将如下所示:

SELECT g.group_id, COUNT(m.member_id) AS members
FROM groups AS g
LEFT JOIN group_members AS m USING(group_id)
GROUP BY g.group_id
HAVING members > 4

引用具有不同名称时的示例

SELECT g.id, COUNT(m.member_id) AS members
FROM groups AS g
LEFT JOIN group_members AS m ON g.id = m.group_id
GROUP BY g.id
HAVING members > 4

另外,请确保在数据库架构内为 JOINS 中使用的键设置索引,因为它会影响您的站点性能。

MySQL use HAVING statement for this tasks.

Your query would look like this:

SELECT g.group_id, COUNT(m.member_id) AS members
FROM groups AS g
LEFT JOIN group_members AS m USING(group_id)
GROUP BY g.group_id
HAVING members > 4

example when references have different names

SELECT g.id, COUNT(m.member_id) AS members
FROM groups AS g
LEFT JOIN group_members AS m ON g.id = m.group_id
GROUP BY g.id
HAVING members > 4

Also, make sure that you set indexes inside your database schema for keys you are using in JOINS as it can affect your site performance.

绝影如岚 2024-10-22 10:10:48
SELECT DISTINCT groups.id, 
       (SELECT COUNT(*) FROM group_members
        WHERE member_id = groups.id) AS memberCount
FROM groups
SELECT DISTINCT groups.id, 
       (SELECT COUNT(*) FROM group_members
        WHERE member_id = groups.id) AS memberCount
FROM groups
薄暮涼年 2024-10-22 10:10:48

也许我在这里偏离了主题,不理解OP,但你为什么要加入表格?

如果您有一个包含成员的表,并且该表有一个名为“group_id”的列,则您只需对members 表运行查询即可获取按group_id 分组的成员计数。

SELECT group_id, COUNT(*) as membercount 
FROM members 
GROUP BY group_id 
HAVING membercount > 4

这应该具有最少的开销,因为您避免了连接,但仍然应该给您您想要的东西。

如果您想要组详细信息和描述等,则将成员表中的联接添加回组表以检索名称将为您提供最快的结果。

Maybe I am off the mark here and not understanding the OP but why are you joining tables?

If you have a table with members and this table has a column named "group_id", you can just run a query on the members table to get a count of the members grouped by the group_id.

SELECT group_id, COUNT(*) as membercount 
FROM members 
GROUP BY group_id 
HAVING membercount > 4

This should have the least overhead simply because you are avoiding a join but should still give you what you wanted.

If you want the group details and description etc, then add a join from the members table back to the groups table to retrieve the name would give you the quickest result.

逆蝶 2024-10-22 10:10:48

您的 groups_main 表有一个名为 id 的键列。我相信,如果 groups_fans 表具有同名的键列,则只能使用 USING 语法进行连接,但它可能没有。因此,请尝试以下操作:

LEFT JOIN groups_fans AS m ON m.group_id = g.id

或将 group_id 替换为 groups_fans< 中适当的列名称/代码>表。

Your groups_main table has a key column named id. I believe you can only use the USING syntax for the join if the groups_fans table has a key column with the same name, which it probably does not. So instead, try this:

LEFT JOIN groups_fans AS m ON m.group_id = g.id

Or replace group_id with whatever the appropriate column name is in the groups_fans table.

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