如何仅选择SQL中具有不同版本的文档列表的最大版本? [英] How do you select only the maximum version of a list of documents that have different versions in SQL?

问题描述

我有一个看起来像这样的数据库表:

I have a database table that looks like this:

| ID | TITLE | VERSION | 
| 1  | file1 |    1    |
| 2  | file2 |    1    |
| 3  | file1 |    2    |
| 4  | file2 |    2    |

我需要一个SQL查询，该查询将返回第3行和第4行，因为它们是file1和file 2的最新版本.

I need an SQL query that will return rows 3 and 4 because they are the latest versions of file1 and file 2.

如果我在如下所示的表上运行查询:

If I run the query on a table that looks like this:

| ID | TITLE | VERSION | 
| 1  | file1 |    1    |
| 2  | file2 |    1    |
| 3  | file1 |    2    |
| 4  | file2 |    2    |
| 5  | file3 |    1    |

它应该返回第3、4和5行，因为"file3" version1是file3的最新版本.

It should return rows 3,4 and 5 because "file3" version1 is the latest version of file3.

我知道我需要在SQL中使用"MAX"函数，但是我被"GROUP BY"关键字抛弃了.我对如何使用它不是很熟悉.

I know I need to use "MAX" function in SQL, however I'm getting throw off with the "GROUP BY" keyword. I'm not very familiar with how to use it.

将感谢所有/任何建议！

Would appreciate all / any advice!

我们正在使用Oracle 11g.

We are using Oracle 11g.

推荐答案

的确，使用子查询获取由TITLE分组的MAX版本，然后将其结果与表连接以获得ID:

Indeed, use a subquery to obtain the MAX version, grouped by TITLE, and then join the result of it with your table to obtain the ID:

SELECT t.*
FROM tbl t INNER JOIN 
     (SELECT title, MAX(version) version
      FROM tbl
      GROUP BY title
     ) max_t ON (t.version = max_t.version AND t.title = max_t.title);

DEMO .

DEMO.

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