小编典典

MySQL SELECT按组最常出现

mysql

如何获得MySQL中每个标签最常出现的类别?理想情况下,我想模拟一个聚合函数来计算列的模式

SELECT 
  t.tag 
  , s.category 
FROM tags t 
LEFT JOIN stuff s 
USING (id) 
ORDER BY tag;

+------------------+----------+
| tag              | category |
+------------------+----------+
| automotive       |        8 |
| ba               |        8 |
| bamboo           |        8 |
| bamboo           |        8 |
| bamboo           |        8 |
| bamboo           |        8 |
| bamboo           |        8 |
| bamboo           |       10 |
| bamboo           |        8 |
| bamboo           |        9 |
| bamboo           |        8 |
| bamboo           |       10 |
| bamboo           |        8 |
| bamboo           |        9 |
| bamboo           |        8 |
| banana tree      |        8 |
| banana tree      |        8 |
| banana tree      |        8 |
| banana tree      |        8 |
| bath             |        9 |
+-----------------------------+

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2020-05-17

共1个答案

小编典典

SELECT t1.*
FROM (SELECT tag, category, COUNT(*) AS count
      FROM tags INNER JOIN stuff USING (id)
      GROUP BY tag, category) t1
LEFT OUTER JOIN 
     (SELECT tag, category, COUNT(*) AS count
      FROM tags INNER JOIN stuff USING (id)
      GROUP BY tag, category) t2
  ON (t1.tag = t2.tag AND (t1.count < t2.count 
      OR t1.count = t2.count AND t1.category < t2.category))
WHERE t2.tag IS NULL
ORDER BY t1.count DESC;

我同意这对于单个SQL查询来说实在太多了。GROUP BY子查询内部的任何使用都会使我畏缩。您可以通过使用视图使它 看起来 更简单:

CREATE VIEW count_per_category AS
    SELECT tag, category, COUNT(*) AS count
    FROM tags INNER JOIN stuff USING (id)
    GROUP BY tag, category;

SELECT t1.*
FROM count_per_category t1
LEFT OUTER JOIN count_per_category t2
  ON (t1.tag = t2.tag AND (t1.count < t2.count 
      OR t1.count = t2.count AND t1.category < t2.category))
WHERE t2.tag IS NULL
ORDER BY t1.count DESC;

但它基本上是在幕后做同样的工作。

您评论说,您可以在应用程序代码中轻松地执行类似的操作。那你为什么不那样做呢?进行更简单的查询以获取每个类别的计数:

SELECT tag, category, COUNT(*) AS count
FROM tags INNER JOIN stuff USING (id)
GROUP BY tag, category;

并在应用程序代码中对结果进行排序。

2020-05-17