Php MySQL单次查询以获取任何活动或非活动记录

我有一个场景，其中有两个表，如下所示

groups

id| name
1 | Group 1
2 | Group 2
3 | Group 3


users

id| name | group_id | sub_status
1 | John | 1        | Active
2 | Doe  | 1        | Inactive
3 | Simon| 2        | Active
4 | David| 3        | Active

现在，我需要选择具有活动或非活动订阅状态的组。

---

就像我在寻找活动组一样，任何一个我都应该得到Group1、Group2和Group3，因为在Group1中有一个活动订阅。如果我查询非活动组，我应该得到null，因为Group1有一个活动订阅

您可以按每个组统计活动和非活动记录。之后，您可以按该计数筛选组。例如，您可以通过以下查询查找活动组

SELECT 
 groups.id,
 groups.name,
 sum(IF(users.sub_status = 'Active', 1, 0)) as active_count 
 sum(IF(users.sub_status = 'Inactive', 1, 0)) as inactive_count  
FROM groups
LEFT JOIN users ON users.group_id = group.id
GROUP BY groups.id, groups.name
HAVING active_count > 0

和非活动组：

SELECT 
 groups.id,
 groups.name,
 sum(IF(users.sub_status = 'Active', 1, 0)) as active_count 
 sum(IF(users.sub_status = 'Inactive', 1, 0)) as inactive_count  
FROM groups
LEFT JOIN users ON users.group_id = group.id
GROUP BY groups.id, groups.name
HAVING active_count = 0

select g.* from groups g
where exists (
  select 1 from users
  where group_id = g.id and sub_status = 'Active'
);

select g.* from groups g
where not exists (
  select 1 from users
  where group_id = g.id and sub_status = 'Active'
);

select u.group_id
from users u
group by u.group_id
having min(u.sub_status) = 'active';    -- any active member

---

你不需要加入表格。 对于活动组，存在以下项：

SELECT 
 groups.id,
 groups.name,
 sum(IF(users.sub_status = 'Active', 1, 0)) as active_count 
 sum(IF(users.sub_status = 'Inactive', 1, 0)) as inactive_count  
FROM groups
LEFT JOIN users ON users.group_id = group.id
GROUP BY groups.id, groups.name
HAVING active_count = 0

select g.* from groups g
where exists (
  select 1 from users
  where group_id = g.id and sub_status = 'Active'
);

select g.* from groups g
where not exists (
  select 1 from users
  where group_id = g.id and sub_status = 'Active'
);

select u.group_id
from users u
group by u.group_id
having min(u.sub_status) = 'active';    -- any active member

并且不存在非活动组：

SELECT 
 groups.id,
 groups.name,
 sum(IF(users.sub_status = 'Active', 1, 0)) as active_count 
 sum(IF(users.sub_status = 'Inactive', 1, 0)) as inactive_count  
FROM groups
LEFT JOIN users ON users.group_id = group.id
GROUP BY groups.id, groups.name
HAVING active_count = 0

select g.* from groups g
where exists (
  select 1 from users
  where group_id = g.id and sub_status = 'Active'
);

select g.* from groups g
where not exists (
  select 1 from users
  where group_id = g.id and sub_status = 'Active'
);

select u.group_id
from users u
group by u.group_id
having min(u.sub_status) = 'active';    -- any active member

---

请参阅。

我想您需要这样的查询：

select g.id, 
       case when count(case when u.sub_status = 'Inactive' then 1 end)=1 and 
                 count(case when u.sub_status = 'Active' then 1 end) > 0
            then
         null
       else  
         g.name
       end inactive_case,
       case when count(case when u.sub_status = 'Active' then 1 end) = 1 then
         g.name
       end active_case
  from groups g
  left join users u on u.group_id = g.id
 group by g.id, g.name;

 id inactive_case   active_case
 1                  Group 1
 2  Group 2         Group 2
 3  Group 3         Group 3

---

您可以为此使用聚合：

对于活动组：

SELECT 
 groups.id,
 groups.name,
 sum(IF(users.sub_status = 'Active', 1, 0)) as active_count 
 sum(IF(users.sub_status = 'Inactive', 1, 0)) as inactive_count  
FROM groups
LEFT JOIN users ON users.group_id = group.id
GROUP BY groups.id, groups.name
HAVING active_count = 0

select g.* from groups g
where exists (
  select 1 from users
  where group_id = g.id and sub_status = 'Active'
);

select g.* from groups g
where not exists (
  select 1 from users
  where group_id = g.id and sub_status = 'Active'
);

select u.group_id
from users u
group by u.group_id
having min(u.sub_status) = 'active';    -- any active member

对于非活动：

select u.group_id
from users u
group by u.group_id
having min(u.sub_status) = 'inactive';  -- all inactive

---

这使用了“active”>“inactive”这一事实。为此，我建议使用0/1值而不是字符串。

问题是什么？有些东西不行吗？我想我理解你的要求，但是，如果您将基于输入和条件的预期输出作为所需行的列表显示给我们，则会使其更为复杂obvious@kerbholz在此处查找查询这将生成非活动组：从id不在其中的组中选择名称从sub_status='Active'中的用户中选择组\u id；我不认为这应该是一个答案，tho。它不应该是SUM而不是COUNT吗？@Pepper是的，这是个错误。我修好了。谢谢你有什么理由要数一数吗？如果存在“活动”或不存在“非活动”的记录，那么仅仅查找这些记录是否就足够了？这似乎是一个更正确的答案。由于您指出的原因，我删除了我的。字段状态来自哪里？