Question

考虑以下查询和输出：

MariaDB [cf]> SELECT user_name, target_id 
              FROM table_google_apps 
              WHERE user_name LIKE "%stepha%" OR user_name LIKE "%tobias%" 
              LIMIT 5;

+----------------+----------------------------------------------+
| user_name      | target_id                                    |
+----------------+----------------------------------------------+
| Stephan B      | 1S4H-FfVVkDG2kcJKyaZqnmLhVYSRIrZDlkRGQQ9P0Go |
| Stephan B      | 1qzkKgYKXeZ671dmLswp-BFyF-cs4NtobZlt0KmE6mwg |
| Stephan B      | 0B_qFxWHs9u2naGdRSlktY1dwR0E                 |
| Tobias S       | 1qzkKgYKXeZ671dmLswp-BFyF-cs4NtobZlt0KmE6mwg |
| Stephan        | 0B_qFxWHs9u2nOVdLQW4taXkycG8                 |
+----------------+----------------------------------------------+
5 rows in set (0.00 sec)

可以清楚地看到Stephan和Tobias处理同一文档（target id 1qzkKgYKXeZ671dmLswp-BFyF-cs4NtobZlt0KmE6mwg）。如何编写一个只返回stephan的target_id等于target_id形式的数据的查询？

例如：...... where stephan.target_id = tobias.target_id ..

谢谢！

Answer 1

这将为您提供文件的所有用户：

SELECT 
  GROUP_CONCAT(user_name SEPARATOR ', ') AS users,
  COUNT(*) AS user_count,
  target_id 
FROM 
  table_google_apps
GROUP BY 
  target_id

Answer 2

您正在寻找自我加入：

SELECT DISTINCT t1.user_name, t1.target_id 
FROM table_google_apps AS t1
INNER JOIN table_google_apps AS t2 
   ON t1.target_id = t2.target_id AND t1.user_name <> t2.user_name
WHERE t1.user_name LIKE "%stepha%" OR t1.user_name LIKE "%tobias%" 
LIMIT 5;

查询根据相同的 target_id值和不同的 user_name值将表连接到自身。因此，它返回共享相同user_name值的target_id个值。

注意：只要您只对共享相同user_name值的target_id个值感兴趣，该查询就会有效。

Demo here

Mysql：where子句 - 过滤信息

2 个答案: