带有 Order By 的 SQL Group By
我有一个标签表,想从列表中获取计数最高的标签.
I have a table of tags and want to get the highest count tags from the list.
示例数据如下
id (1) tag ('night')
id (2) tag ('awesome')
id (3) tag ('night')
使用
SELECT COUNT(*), `Tag` from `images-tags`
GROUP BY `Tag`
让我完美地找回了我正在寻找的数据.但是,我想对它进行组织,以便标签计数最高的在前,并将其限制为仅向我发送前 20 个左右.
gets me back the data I'm looking for perfectly. However, I would like to organize it, so that the highest tag counts are first, and limit it to only send me the first 20 or so.
我试过了...
SELECT COUNT(id), `Tag` from `images-tags`
GROUP BY `Tag`
ORDER BY COUNT(id) DESC
LIMIT 20
并且我不断收到无效使用组功能 - ErrNr 1111"
and I keep getting an "Invalid use of group function - ErrNr 1111"
我做错了什么?
我使用的是 MySQL 4.1.25-Debian
I'm using MySQL 4.1.25-Debian
推荐答案
在所有版本的 MySQL 中,只需为 SELECT 列表中的聚合取别名,并按别名排序:
In all versions of MySQL, simply alias the aggregate in the SELECT list, and order by the alias:
SELECT COUNT(id) AS theCount, `Tag` from `images-tags`
GROUP BY `Tag`
ORDER BY theCount DESC
LIMIT 20
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