Question

我有一个包含多个共享路径的表我希望得到这些路径中前N个不同的Servernames的所有路径，例如这里的整个表格。

+----+--------------------------------+
| ID | BACKUPPATH                     |
+----+--------------------------------+
|  1 | //server.domain/share/folder   |
|  2 | //server.domain/share/folder3  |
|  3 | //server.domain/share/folder2  |
|  4 | //server2.domain/share/folder1 |
|  5 | //server2.domain/share/folder2 |
|  6 | //server3.domain/share/folder1 |
|  7 | //server3.domain/share/folder2 |
|  8 | //server3.domain/share/folder3 |
+----+--------------------------------+

每次Servernames都会有所不同，不同Servenames的数量也会有所不同。作为示例，我想获得前两个不同的Servernames的所有路径，我期望作为结果：

+----+--------------------------------+
| ID | BACKUPPATH                     |
+----+--------------------------------+
|  1 | //server.domain/share/folder   |
|  2 | //server.domain/share/folder3  |
|  3 | //server.domain/share/folder2  |
|  4 | //server2.domain/share/folder1 |
|  5 | //server2.domain/share/folder2 |
+----+--------------------------------+

作为子查询我使用以下查询来获取Servernames的行集：

select SUBSTRING_INDEX(BACKUPPATH,'/',3) as SERVERNAMES from(select BACKUPPATH from Backuppaths GROUP BY SUBSTRING_INDEX(BACKUPPATH,'/',3))as NUMEROFSERVERS LIMIT 2;

+------------------+
| SERVERNAMES      |
+------------------+
| //server.domain  |
| //server2.domain |
+------------------+

我现在被困在如何使用这个子查询来获得我期望的结果。

感谢您提供的任何帮助

Answer 1

您可以加入内联视图：

select s.*
  from servernames s
  join (select substring_index(backuppath, '/', 3) as servername
          from servernames
         group by servername
         order by min(id) limit 2) v
    on substring_index(backuppath, '/', 3) = v.servername
 order by id

Fiddle: http://sqlfiddle.com/#!2/b6a16/1/0

Answer 2

使用Distinct

SELECT DISTINCT SUBSTRING_INDEX(BACKUPPATH,'/',3) AS SERVERNAMES FROM Backuppaths LIMIT 2;

获取n组的所有记录

2 个答案: