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I need to fetch distinct values of a table, which has an auto-incremented id field. I need the id to identify, but I don't care which id when all values are the same except the id.

I came up with the following, which seems to work. Is there a better way to do this though?

SELECT id, date_f, date_t, num_n, num_d, mn, is 
FROM t
GROUP BY date_f, date_t, num_n, num_d, mn, is

Another concern is, would this always return the same ids if the query is executed more than once?

EDIT:

Sample db:

id date_f date_t num_n num_d mn is
1   10      10      10  10   10 10
2   10      10      10  10   10 10
3   10      10      10  10   10 10

I want to store all of the columns of one row out of these. I don't care if the id is 1, 2 or 3 as long as it's the same the next time I execute the query (without adding/deleting any rows between). So far, two answers suggested using min(id) which seems like a good idea.

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3 回答 3

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如果您根本没有在结果集中使用 id，那么上面的答案就可以了。如果您出于任何目的需要在结果集中至少有一个 Id，那么您应该稍微更改您的语句：

SELECT min(id), date_f, date_t, num_n, num_d, mn, is 
FROM t
GROUP BY date_f, date_t, num_n, num_d, mn, is

这将返回与不同集匹配的第一个 Id（或者如果您想要最后一个，则使用 (Max) - 这可能会随着时间的推移而改变，尽管随着数据被添加到表中。

于 2013-11-07T13:49:02.270 回答
1

你所拥有的基本上是正确的 - 我会补充说，如果你想要 id （身份），那么你需要添加一个聚合，例如 -

SELECT min(id), date_f, date_t, num_n, num_d, mn, is 
FROM t
GROUP BY date_f, date_t, num_n, num_d, mn, is

这将确保您每次都获得第一个/相同的（当然，除非它被删除）

于 2013-11-07T13:49:56.297 回答
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SELECT DISTINCT date_f, date_t, num_n, num_d, mn, is FROM t

于 2013-11-07T13:43:54.917 回答