Question

目前我有两张包含一些数据的表格。第一个表格如下：

+----------------+-----------+
| name           | member_id |
+----------------+-----------+
| Juice Box      |     49432 |
| Rainsurge      |     49631 |
| spiderpigrider |     50482 |
+----------------+-----------+

第二个表格如下：

+------------+-----------+
| recruit_id | bin(refs) |
+------------+-----------+
|      49432 | 1         |
|      49631 | 1         |
|      49432 | 1         |
|      49631 | 1         |
|      49432 | 1         |
|      49631 | 1         |
|      49432 | 1         |
|      49631 | 1         |
|      49432 | 1         |
|      49631 | 1         |
+------------+-----------+

我想像这样返回名称，总refs和member_id / recruit_id（仅列出至少有1个参考的用户）

+------------+-----------+------------+
| recruit_id | name      | total_refs |
+------------+-----------+------------+
|      49631 | Rainsurge |          5 |
|      49432 | Juice Box |          5 |
+------------+-----------+------------+

select r.recruit_id,bin(r.refs),ipb.name from refs as r 
inner join syndicate_ipb.core_members as ipb on ipb.member_id=r.recruit_id;

这返回了我的数据，但显然没有总计数和重复的名称/ ID

select r.recruit_id,count(bin(r.refs)),ipb.name from refs as r 
inner join syndicate_ipb.core_members as ipb on ipb.member_id=r.recruit_id;

此返回的数据包含所有人的总数，但只有一个ID /名称

+------------+--------------------+-----------+
| recruit_id | count(bin(r.refs)) | name      |
+------------+--------------------+-----------+
|      49432 |                 10 | Juice Box |
+------------+--------------------+-----------+

这会返回数据，但不会再次计数

select distinct r.recruit_id,bin(r.refs),ipb.name from refs as r 
inner join syndicate_ipb.core_members as ipb on ipb.member_id=r.recruit_id;

+------------+-------------+-----------+
| recruit_id | bin(r.refs) | name      |
+------------+-------------+-----------+
|      49432 | 1           | Juice Box |
|      49631 | 1           | Rainsurge |
+------------+-------------+-----------+

非常感谢任何帮助或指导。我觉得我已经离这里很近但只是没有足够的能力使用SQL来获得它。谢谢！

Answer 1

您必须使用group by子句：

select r.recruit_id, ipb.name, count(bin(refs)) as total_refs
  from refs as r 
         inner join syndicate_ipb.core_members as ipb 
                 on ipb.member_id=r.recruit_id
 group by r.recruit_id, ipb.name
 having count(bin(refs)) >= 1

此group by r.recruit_id, ipb.name会对结果进行分组，而having count(bin(refs)) >= 1会认为它只返回至少有一个参考的成员

不要只根据您想要的列对列进行分组。即使MySql允许它，它也不是SQL Ansi模式，甚至MySql现在也遵守它。在select语句中使用与整列相关的聚合函数分组。

Answer 2

你快到了。你最后错过了GROUP BY条款。

<强>查询：

SELECT
    r.recruit_id,
    count(bin(r.refs)),
    ipb.name
FROM refs AS r
INNER JOIN syndicate_ipb.core_members AS ipb 
ON ipb.member_id = r.recruit_id
GROUP BY r.recruit_id;

注意：

如果bin(refs)列始终包含值1，那么实际上您不需要保留该列。在这种情况下，您可以使用count(*)或count(r.recruit_id)来获取计数。

如果bin(refs)列包含任何值，那么count将无法为您提供正确的答案。在这种情况下，您需要使用sum之类的Sum( bin(refs))。

Answer 3

SELECT ipb.*, COUNT(`r`.`recruit_id`) AS cid FROM `ipb` 
INNER JOIN `r` ON `r`.`join_id` = ipb.`member_id`
GROUP BY ipb.`member_id`

具有连接的每个不同用户的Mysql计数

3 个答案: