Mysql将4个不同表的信息显示为一张表

Question

如何将 4 个不同表中的信息显示为一张表？我有 4 个包含相关信息的表。这些表包含包、组织子级和系统用户数据。我想查询一个组织订阅了什么类型的包以及在该组织下注册了多少孩子和用户。表中包含的示例数据如下所示。

包表：

package_id	package_name
12	Basic
21	Pro
33	Premium

组织机构表：

org_id	org_name	package_id
18	Marks of Awesomeness	12
24	John Hopkins Hospital	21
38	Teddy and the Wailers	33
78	Lawrence Movers	12

儿童桌：

id_child	id_org child_name	id_org
14	Mark Walker	18
22	Jane Quinn	24
38	Lily Audrey	24
44	Dona Marie	18

用户表：

idu	org_id	fname	lname
87	18	John	Doe
92	33	Jane	Doe
107	18	Martin	Short
112	18	Jason	Seguel
127	33	Josh	Radnor

我的查询如下图：

SELECT SQL_CALC_FOUND_ROWS `id_org`, `org_name`, `package_name`
        ,COUNT(id_child) AS child_count, COUNT(idu) AS user_count FROM organisations,packages,children,system_users
        WHERE organisations.id_org=children.org_id AND organisations.id_org=system_users.org_id
     AND organisations.org_package_id=packages.id_package

查询的问题是它只显示一个组织，如果它有一个孩子和一个用户 使用其 org_id 列出。我想用它订阅的包名列出所有组织的所有数据， 该特定组织下列出的儿童总数，下列出的用户总数 该特定组织，如果组织没有孩子、用户或两者兼有，则为零。 以下是我想要的结果。

期望的结果：

Organisation	package Name	No of Children	Number of Users
Marks of Awesomeness	Basic	2	3
John Hopkins Hospital	Pro	2	0
Teddy and the Wailers	Premium	0	2
Lawrence Movers	Basic	0	0

Answer 1

这将满足您的需求...您需要将两个不包含所有记录的表连接起来，以便用户和孩子...对这些表进行计数并使用 COALESCE 将其取出以处理空值

注意：

您的预期结果不正确...组织表中没有 org_id = 33。 Teddy and the Wailers 的计数不应为 2，而应为 0 - 或者您要将用户中的 org_id 更改为 38 而不是 33。

查询：

SELECT 
    o.org_name AS 'Organisation', 
    p.package_name as 'Package Name', 
    COALESCE(t.num_children, 0) AS 'No of Children', 
    COALESCE(t1.num_users, 0) AS 'Number of Users'
FROM organisation o
JOIN packages p ON p.package_id = o.package_id
LEFT JOIN 
(   SELECT 
        COUNT(*) as num_children, id_org 
    FROM children 
    GROUP BY id_org
) as t ON t.id_org = o.org_id 
LEFT JOIN 
(   SELECT 
       COUNT(*) as num_users, org_id 
    FROM users 
    GROUP BY org_id
) as t1 ON t1.org_id = o.org_id

DEMO

输出：

+-----------------------+-----------------+-----------------+------------------+
| Organisation          | Package Name    | No of Children  | Number of Users  |
+-----------------------+-----------------+-----------------+------------------+
| Marks of Awesomeness  | Basic           | 2               | 3                |
| John Hopkins Hospital | Pro             | 2               | 0                |
| Teddy and the Wailers | Premium         | 0               | 0                |
| Lawrence Movers       | Basic           | 0               | 0                |
+-----------------------+-----------------+-----------------+------------------+

编辑：

如果您将 users 表中的 id 33 更改为 38 以匹配所需的结果，那么您将得到以下结果：QUERY

输出：

+-----------------------+-----------------+-----------------+------------------+
| Organisation          | Package Name    | No of Children  | Number of Users  |
+-----------------------+-----------------+-----------------+------------------+
| Marks of Awesomeness  | Basic           | 2               | 3                |
| John Hopkins Hospital | Pro             | 2               | 0                |
| Teddy and the Wailers | Premium         | 0               | 2                |
| Lawrence Movers       | Basic           | 0               | 0                |
+-----------------------+-----------------+-----------------+------------------+

Answer 2

代替您现在使用的隐式 INNER JOIN，尝试使用 LEFT JOINs 和 COALESCE 函数来处理 NULL 值，如下所示：

SELECT 
    SQL_CALC_FOUND_ROWS `id_org`, 
    `org_name`, 
    `package_name`,
    COALESCE(COUNT(id_child), 0) AS child_count, 
    COALESCE(COUNT(idu), 0) AS user_count 
FROM organisations
INNER JOIN packages ON organisations.org_package_id=packages.id_package
LEFT JOIN children ON organisations.id_org=children.org_id
LEFT JOIN system_users ON organisations.id_org=system_users.org_id
GROUP BY SQL_CALC_FOUND_ROWS, `org_name`, `package_name`;

Answer 3

您需要使用连接：

select org.org_name, pack.package_name, count(child.id_child) as NoOfChildren,     count(user.idu) as NoOfUsers from Organisation as org
join Packages as pack
on pack.package_id = org.package_id
left outer join Children as child on child.id_org = org.org_id
left outer join Users as user on user.org_id = org.org_id
group by org.org_name, pack.package_name

还有更新的小提琴： http://sqlfiddle.com/#!2/cfe16/14/0

虽然仍然不完全正确，计数已关闭，但似乎已经做出了更好的答案。