Question

我有2个表，“联系人组”和“联系人”订户，我们正在尝试联接和计数类似的值。对于没有任何值的对应字段，我需要一个值“ 0”。表格示例如下：

Contact Group Table:

id | name      | Hash | 

1  | January   | asld | 
2  | February  | asld | 
3  | March     | asld | 
4  | April     | asld | 
5  | May       | asld | 

Contact Subscriber Table:

id | name            | group_id | subscribed at |  unsubscribed at |

1  | Luke Skywalker  |     1    | 20190211      |     NULL         |
1  | Darth Vader     |     2    | 20190211      |     NULL         |
1  | Tony Stark      |     1    | 20190211      |     NULL         |
1  | Bruce Wayne     |     3    | 20190211      |     NULL         |
1  | Connor MacLeod  |     5    | 20190211      |     NULL         |

联系人组表中的“ id”字段与联系人订户表中的“ group_id”字段匹配。

使用上面的示例，我们正在寻找以下输出：

id | name      | Value

1  | January   | 2
2  | February  | 1
3  | March     | 1
4  | April     | 0
5  | May       | 1

我觉得我们接近了，我们能够提供下面显示的以下代码，该代码显示了ID和月份，但我们无法使count函数正常工作。这是我们目前拥有的：

SELECT 
contact_group.id,
contact_group.name,
COALESCE(group_id, 0) AS value FROM
contact_group
LEFT JOIN
contact_subscriber ON contact_group.id = contact_subscriber.group_id;

这提供了以下结果：

+----+---------------+-------+
| id | name          | value |
+----+---------------+-------+
|  1 | January       |     1 |
|  1 | January       |     1 |
|  2 | February      |     2 |
|  3 | March         |     3 |
|  4 | April         |     0 |
|  5 | May           |     5 |
+----+---------------+-------+
6 rows in set (0.00 sec)

在此方面的任何帮助将不胜感激，我一直在努力寻找解决办法，大约一个星期。

Answer 1

使用聚合函数count()

SELECT 
    contact_group.id    
    contact_group.name,
    count(contact_subscriber.group_id) AS value FROM
    contact_group 
    LEFT JOIN
    contact_subscriber ON contact_group.id = contact_subscriber.group_id
    group by contact_group.name, contact_group.id

Answer 2

SELECT
contact_group.id, contact_group.name, COUNT(contact_subscriber.id) AS value
FROM contact_group
LEFT JOIN contact_subscriber ON contact_group.id = contact_subscriber.group_id
GROUP BY contact_group.id, contact_group.name

Answer 3

一种方法使用子查询：

SELECT cg.*,
       (SELECT COUNT(*)
        FROM contact_subscriber cs
        WHERE cg.id = cs.group_id 
       ) as value
FROM contact_group cg;

因为这可以利用contact_subscriber(group_id)上的索引，所以它可能比其他方法（特别是外部查询中使用GROUP BY的方法）要快得多。

从两个表MySql

3 个答案: