Php MySQL通过关系表连接_Php_Mysql

Php MySQL通过关系表连接

php mysql

Php MySQL通过关系表连接,php,mysql,Php,Mysql,嗨，我似乎找不到写这个查询的正确方法。我有两个实体网站和客户端，以及一个通过id字段将它们关联起来的表这是一种多对多的关系。i、一个网站可以有多个客户端，一个客户端可以有多个网站我试图写一个查询，返回所有的网站与客户属于他们。我想返回所有的网站，即使他们没有与他们相关联的客户端。以下是我目前正在处理的问题：这三个表是ost_站点=网站，ost_站点_授权=关系表，ost_客户端=客户端 SELECT ost_sites.site_id, ost_sites.name,

嗨，我似乎找不到写这个查询的正确方法。我有两个实体网站和客户端，以及一个通过id字段将它们关联起来的表

这是一种多对多的关系。i、一个网站可以有多个客户端，一个客户端可以有多个网站

我试图写一个查询，返回所有的网站与客户属于他们。我想返回所有的网站，即使他们没有与他们相关联的客户端。以下是我目前正在处理的问题：

这三个表是ost_站点=网站，ost_站点_授权=关系表，ost_客户端=客户端

SELECT 
    ost_sites.site_id, 
    ost_sites.name,
    ost_sites.site_url,
    ost_site_auth.site_id,
    ost_site_auth.client_id 
    ost_clients.client_id,
    CONCAT_WS(" ", ost_clients.lastname, ost_clients.firstname) as name,
FROM ost_sites
LEFT JOIN (ost_site_auth, ost_clients) 
    ON (ost_sites.site_id=ost_site_auth.site_id 
        AND ost_site_auth.client_id=ost_clients.client_id)
GROUP BY ost_sites.name

我得到一个结果集，但它不会返回所有站点，而且所有行都没有与它们关联的客户端

非常感谢你的帮助

编辑：以下是表的列：

ost_站点

site_id |    name    | site_url
1          facebook    facebook.com
2          twitter     twitter.com
3          tubmblr     tumblr.com
4          google      google.com

ost_站点_授权

（注意认证列表中没有站点id=3）

ost_客户端

client_id  |  firstname  |  lastname
1              wilma         flintstone
2              bam           bam
3              fred          flintstone
4              barney        rubble

预期产出：

site_id |    name    |    site_url    |      client_name     |
1          facebook    facebook.com        wilma flintstone
1          facebook    facebook.com        bam bam
2          twitter     twitter.com         wilma flintstone
2          twitter     twitter.com         bam bam
4          google      google.com          wilma flintstone
4          google      google.com          barney rubble
3          tumblr      tumlr.com           NULL

你的加入看起来有点不对劲。。。试试这个

SELECT 
    ost_sites.site_id, 
    ost_sites.name,
    ost_sites.site_url,
    ost_site_auth.site_id,
    ost_site_auth.client_id 
    ost_clients.client_id,
    CONCAT_WS(" ", ost_clients.lastname, ost_clients.firstname) as name

FROM ost_sites

LEFT OUTER JOIN ost_site_auth 
    ON ost_sites.site_id=ost_site_auth.site_id 

LEFT OUTER JOIN ost_clients
    ON ost_site_auth.client_id=ost_clients.client_id

ORDER BY ost_sites.name

让我试着给你解释一下

我们从
```
ost_sites
```
表开始，不管其他表中是否有匹配的内容，我们都希望从中获得所有结果

然后，我们对表

ost\u site\u auth

执行

左外部联接。这意味着，如果来自ost_site_auth
的内容与ost_sites
中的内容不匹配，则不会返回该内容。但是，ost_站点
中与ost_站点_auth
中的内容不匹配的内容将被返回，因为位于左侧外部
部分
接下来，我们为ost_客户机
重复左外连接




不知道你想要什么。。。让我们假设这些数据在表中表示：

站点1没有客户端
站点2有一个客户端：A
站点3有两个客户端：B、C
站点4有三个客户端：D、E、F
站点#5没有客户端
客户端G和H没有关联的站点

查询一个
      SELECT 
          ost_sites.site_id as SITE,   
          ost_clients.client_id as CLIENT

      FROM ost_sites

      LEFT OUTER JOIN ost_site_auth 
          ON ost_sites.site_id=ost_site_auth.site_id 

      LEFT OUTER JOIN ost_clients
          ON ost_site_auth.client_id=ost_clients.client_id

      ORDER BY ost_sites.site_id, ost_clients.client_id

这将返回（基本上）
查询两个
      SELECT 
          ost_sites.site_id as SITE,   
          ost_clients.client_id as CLIENT

      FROM ost_sites

      JOIN ost_site_auth 
          ON ost_sites.site_id=ost_site_auth.site_id 

      JOIN ost_clients
          ON ost_site_auth.client_id=ost_clients.client_id

      ORDER BY ost_sites.site_id, ost_clients.client_id

      SELECT 
          ost_sites.site_id as SITE,   
          ost_clients.client_id as CLIENT

      FROM ost_sites

      FULL OUTER JOIN ost_site_auth 
          ON ost_sites.site_id=ost_site_auth.site_id 

      FULL OUTER JOIN ost_clients
          ON ost_site_auth.client_id=ost_clients.client_id

      ORDER BY ost_sites.site_id, ost_clients.client_id

      SELECT DISTINCT ost_sites.site_id as SITE 

      FROM ost_sites

      LEFT OUTER JOIN ost_site_auth 
          ON ost_sites.site_id=ost_site_auth.site_id 

      LEFT OUTER JOIN ost_clients
          ON ost_site_auth.client_id=ost_clients.client_id

      ORDER BY ost_sites.site_id 
      ORDER BY ost_sites.site_id 

      SELECT 
          ost_sites.site_id as SITE,
          count(ost_clients.client_id) as CLIENT_COUNT

      FROM ost_sites

      JOIN ost_site_auth 
          ON ost_sites.site_id=ost_site_auth.site_id 

      JOIN ost_clients
          ON ost_site_auth.client_id=ost_clients.client_id

      GROUP BY ost_sites.site_id 
      ORDER BY ost_sites.site_id 

      SELECT 
          ost_sites.site_id as SITE,
          count(ost_clients.client_id) as CLIENT_COUNT

      FROM ost_sites

      LEFT OUTER JOIN ost_site_auth 
          ON ost_sites.site_id=ost_site_auth.site_id 

      LEFT OUTER JOIN ost_clients
          ON ost_site_auth.client_id=ost_clients.client_id

      GROUP BY ost_sites.site_id 
      ORDER BY ost_sites.site_id 

这将返回（基本上）
查询三个
      SELECT 
          ost_sites.site_id as SITE,   
          ost_clients.client_id as CLIENT

      FROM ost_sites

      JOIN ost_site_auth 
          ON ost_sites.site_id=ost_site_auth.site_id 

      JOIN ost_clients
          ON ost_site_auth.client_id=ost_clients.client_id

      ORDER BY ost_sites.site_id, ost_clients.client_id

      SELECT 
          ost_sites.site_id as SITE,   
          ost_clients.client_id as CLIENT

      FROM ost_sites

      FULL OUTER JOIN ost_site_auth 
          ON ost_sites.site_id=ost_site_auth.site_id 

      FULL OUTER JOIN ost_clients
          ON ost_site_auth.client_id=ost_clients.client_id

      ORDER BY ost_sites.site_id, ost_clients.client_id

      SELECT DISTINCT ost_sites.site_id as SITE 

      FROM ost_sites

      LEFT OUTER JOIN ost_site_auth 
          ON ost_sites.site_id=ost_site_auth.site_id 

      LEFT OUTER JOIN ost_clients
          ON ost_site_auth.client_id=ost_clients.client_id

      ORDER BY ost_sites.site_id 
      ORDER BY ost_sites.site_id 

      SELECT 
          ost_sites.site_id as SITE,
          count(ost_clients.client_id) as CLIENT_COUNT

      FROM ost_sites

      JOIN ost_site_auth 
          ON ost_sites.site_id=ost_site_auth.site_id 

      JOIN ost_clients
          ON ost_site_auth.client_id=ost_clients.client_id

      GROUP BY ost_sites.site_id 
      ORDER BY ost_sites.site_id 

      SELECT 
          ost_sites.site_id as SITE,
          count(ost_clients.client_id) as CLIENT_COUNT

      FROM ost_sites

      LEFT OUTER JOIN ost_site_auth 
          ON ost_sites.site_id=ost_site_auth.site_id 

      LEFT OUTER JOIN ost_clients
          ON ost_site_auth.client_id=ost_clients.client_id

      GROUP BY ost_sites.site_id 
      ORDER BY ost_sites.site_id 

这将返回（基本上）
查询四个
      SELECT 
          ost_sites.site_id as SITE,   
          ost_clients.client_id as CLIENT

      FROM ost_sites

      JOIN ost_site_auth 
          ON ost_sites.site_id=ost_site_auth.site_id 

      JOIN ost_clients
          ON ost_site_auth.client_id=ost_clients.client_id

      ORDER BY ost_sites.site_id, ost_clients.client_id

      SELECT 
          ost_sites.site_id as SITE,   
          ost_clients.client_id as CLIENT

      FROM ost_sites

      FULL OUTER JOIN ost_site_auth 
          ON ost_sites.site_id=ost_site_auth.site_id 

      FULL OUTER JOIN ost_clients
          ON ost_site_auth.client_id=ost_clients.client_id

      ORDER BY ost_sites.site_id, ost_clients.client_id

      SELECT DISTINCT ost_sites.site_id as SITE 

      FROM ost_sites

      LEFT OUTER JOIN ost_site_auth 
          ON ost_sites.site_id=ost_site_auth.site_id 

      LEFT OUTER JOIN ost_clients
          ON ost_site_auth.client_id=ost_clients.client_id

      ORDER BY ost_sites.site_id 
      ORDER BY ost_sites.site_id 

      SELECT 
          ost_sites.site_id as SITE,
          count(ost_clients.client_id) as CLIENT_COUNT

      FROM ost_sites

      JOIN ost_site_auth 
          ON ost_sites.site_id=ost_site_auth.site_id 

      JOIN ost_clients
          ON ost_site_auth.client_id=ost_clients.client_id

      GROUP BY ost_sites.site_id 
      ORDER BY ost_sites.site_id 

      SELECT 
          ost_sites.site_id as SITE,
          count(ost_clients.client_id) as CLIENT_COUNT

      FROM ost_sites

      LEFT OUTER JOIN ost_site_auth 
          ON ost_sites.site_id=ost_site_auth.site_id 

      LEFT OUTER JOIN ost_clients
          ON ost_site_auth.client_id=ost_clients.client_id

      GROUP BY ost_sites.site_id 
      ORDER BY ost_sites.site_id 

这将返回（基本上）
查询五个
      SELECT 
          ost_sites.site_id as SITE,   
          ost_clients.client_id as CLIENT

      FROM ost_sites

      JOIN ost_site_auth 
          ON ost_sites.site_id=ost_site_auth.site_id 

      JOIN ost_clients
          ON ost_site_auth.client_id=ost_clients.client_id

      ORDER BY ost_sites.site_id, ost_clients.client_id

      SELECT 
          ost_sites.site_id as SITE,   
          ost_clients.client_id as CLIENT

      FROM ost_sites

      FULL OUTER JOIN ost_site_auth 
          ON ost_sites.site_id=ost_site_auth.site_id 

      FULL OUTER JOIN ost_clients
          ON ost_site_auth.client_id=ost_clients.client_id

      ORDER BY ost_sites.site_id, ost_clients.client_id

      SELECT DISTINCT ost_sites.site_id as SITE 

      FROM ost_sites

      LEFT OUTER JOIN ost_site_auth 
          ON ost_sites.site_id=ost_site_auth.site_id 

      LEFT OUTER JOIN ost_clients
          ON ost_site_auth.client_id=ost_clients.client_id

      ORDER BY ost_sites.site_id 
      ORDER BY ost_sites.site_id 

      SELECT 
          ost_sites.site_id as SITE,
          count(ost_clients.client_id) as CLIENT_COUNT

      FROM ost_sites

      JOIN ost_site_auth 
          ON ost_sites.site_id=ost_site_auth.site_id 

      JOIN ost_clients
          ON ost_site_auth.client_id=ost_clients.client_id

      GROUP BY ost_sites.site_id 
      ORDER BY ost_sites.site_id 

      SELECT 
          ost_sites.site_id as SITE,
          count(ost_clients.client_id) as CLIENT_COUNT

      FROM ost_sites

      LEFT OUTER JOIN ost_site_auth 
          ON ost_sites.site_id=ost_site_auth.site_id 

      LEFT OUTER JOIN ost_clients
          ON ost_site_auth.client_id=ost_clients.client_id

      GROUP BY ost_sites.site_id 
      ORDER BY ost_sites.site_id 

这将返回（基本上）
查询五个
      SELECT 
          ost_sites.site_id as SITE,   
          ost_clients.client_id as CLIENT

      FROM ost_sites

      JOIN ost_site_auth 
          ON ost_sites.site_id=ost_site_auth.site_id 

      JOIN ost_clients
          ON ost_site_auth.client_id=ost_clients.client_id

      ORDER BY ost_sites.site_id, ost_clients.client_id

      SELECT 
          ost_sites.site_id as SITE,   
          ost_clients.client_id as CLIENT

      FROM ost_sites

      FULL OUTER JOIN ost_site_auth 
          ON ost_sites.site_id=ost_site_auth.site_id 

      FULL OUTER JOIN ost_clients
          ON ost_site_auth.client_id=ost_clients.client_id

      ORDER BY ost_sites.site_id, ost_clients.client_id

      SELECT DISTINCT ost_sites.site_id as SITE 

      FROM ost_sites

      LEFT OUTER JOIN ost_site_auth 
          ON ost_sites.site_id=ost_site_auth.site_id 

      LEFT OUTER JOIN ost_clients
          ON ost_site_auth.client_id=ost_clients.client_id

      ORDER BY ost_sites.site_id 
      ORDER BY ost_sites.site_id 

      SELECT 
          ost_sites.site_id as SITE,
          count(ost_clients.client_id) as CLIENT_COUNT

      FROM ost_sites

      JOIN ost_site_auth 
          ON ost_sites.site_id=ost_site_auth.site_id 

      JOIN ost_clients
          ON ost_site_auth.client_id=ost_clients.client_id

      GROUP BY ost_sites.site_id 
      ORDER BY ost_sites.site_id 

      SELECT 
          ost_sites.site_id as SITE,
          count(ost_clients.client_id) as CLIENT_COUNT

      FROM ost_sites

      LEFT OUTER JOIN ost_site_auth 
          ON ost_sites.site_id=ost_site_auth.site_id 

      LEFT OUTER JOIN ost_clients
          ON ost_site_auth.client_id=ost_clients.client_id

      GROUP BY ost_sites.site_id 
      ORDER BY ost_sites.site_id 

这将返回（基本上）
退房

我认为您真正需要做的就是执行与右外部联接相同的查询，并将它们联合起来。
此查询得到什么结果？你能发布一个输出示例吗？或者你的一部分数据？我会用一般数据编辑帖子，但我不能输入具体数据，因为名字和网站与我的工作有关。好的。。。这里有四个问题：我刚刚编辑了这篇文章，让我知道这是否有点清楚，我甚至在这个链接的顶部添加了第五个问题…-->从你的例子来看。。。您只需要使用一个左外部联接
。我不确定我最初建议的第一个查询如何不符合您的要求…？在第二个加入时，我意外地离开了和，而不是将其更改为。。。现在已经更新/编辑了。这可以工作并为我提供所有网站和一个与该网站关联的客户端（如果该网站没有客户端，则为空）。我只需要它返回多行，如果该网站有多个客户端。也许我可以让query返回所有有客户端的网站，然后再执行另一个查询，找出哪些网站没有客户端？这就是你想要的吗？如果直接执行联接
，它将只返回在所有三个表中都有匹配行的内容。如果您要执行一个右外部联接
，它将以反向联接顺序进行，因此您将获得所有客户端，无论它们是否在ost\u site\u auth
中有行，然后是ost\u sites
。（最好使用左外部联接
，因为它更常见，因此更容易可视化/理解。）如果要返回所有匹配的内容，即使不匹配，也应该使用另一个答案所指的完全外部联接
。。。在数据匹配的地方，您将得到一整行。但是你也会得到没有客户端和客户端的站点。没有站点。所以你不想看到没有任何客户端的站点？