MySQL select 只有唯一值

Question

我有兩張桌子。

客戶：

+--------------+-------------+
|   CLIENT_ID  |    LABEL    |
+--------------+-------------+
|          123 |      label1 |
+--------------+-------------+
|          123 |      label3 |
+--------------+-------------+
|          456 |      label1 |
+--------------+-------------+
|          789 |      label2 |
+--------------+-------------+
|          987 |      label2 |
+--------------+-------------+
|          987 |      label4 |
+--------------+-------------+

經理：

+----+--------------+
| ID |   CLIENT_ID  |
+----+--------------+
|  1 |          123 |
+----+--------------+
|  1 |          456 |
+----+--------------+
|  2 |          456 |
+----+--------------+
|  3 |          789 |
+----+--------------+
|  3 |          987 |
+----+--------------+
|  4 |          789 |
+----+--------------+

我需要來自經理的 select ID，這些經理只有帶有標簽“label1”或“label2”的客戶，而沒有帶有其他標簽的客戶。 生成的 output 應該類似於 2 和 4。

我試着做

select m.id
from managers m
    join clients c on m.client_id = c.client_id
where c.label in ('label1', 'label2');

但它返回所有ID。

Answer 1

關。 只需添加聚合：

select m.id
from managers s join
     clients c
     on m.client_id = c.client_id
where c.label in ('label1', 'label2')
group by m.id
having count(distinct c.label) = 2;

如果您只想要具有這兩個標簽的客戶端，請使用：

select m.id
from managers s join
     clients c
     on m.client_id = c.client_id
group by m.id
having count(distinct c.label) = 2;

select m.id
from managers s join
     clients c
     on m.client_id = c.client_id
group by m.id
having count(distinct case when c.label in ('label1', 'label2') then c.label end) = 2 and
       count(distinct c.label) = 2;

或者，更有效地：

having sum(c.label = 'label1') > 0 and
       sum(c.label = 'label2') > 0 and
       sum(c.label not in ('label1', 'label2')) = 0;

Answer 2

使用group by如下：

select m.id
from managers m
    join clients c on m.client_id = c.client_id
where c.label in ('label1', 'label2')
group by m.id
having count(*) = 1;

Answer 3

請嘗試group by並having此處聲明以獲取唯一記錄

 SELECT ID FROM `Clients` c 
  left join Managers m on (c.client_id = m.CLIENT_ID) 
  where LABEL in ('label1', 'label2') 
  group by ID having count(*) = 1

謝謝

Answer 4

您可以使用 CTE 獲取與特定經理關聯的客戶標簽有效的出現次數：

with r as (select m.id id, sum(c.label in ('label1', 'label2')) c1, count(*) c2 
           from managers m join clients c on m.client_id = c.client_id group by m.id)
select id from r where c1 = c2;

Output：

ID
2
4

Answer 5

加入表，按經理分組並在HAVING子句中為您的條件使用條件聚合：

SELECT m.id
FROM managers m INNER JOIN clients c 
ON m.client_id = c.client_id
GROUP BY m.id
HAVING MAX(c.label IN ('label1', 'label2')) = 1
   AND MAX(c.label NOT IN ('label1', 'label2')) = 0

如果客戶端同時具有標簽'label1'和'label2'並且可以擴展到更多標簽，這也將起作用。

請參閱演示。
結果：

ID
2
4