Question

我正在尝试做类似

的事情

SELECT `v1`.`date`, COUNT(DISTINCT(`v2`.`id`)) AS `visits` 
FROM `visitor_visits` AS `v1` JOIN `visitor_visits` AS `v2` 
ON (`v1`.`date` = `v2`.`date`) GROUP BY `v1`.`date`, `v2`.`date` 
ORDER BY `v1`.`date

获取不同的日期，以及匹配的所有其他日期的计数。

我很确定这是错误的查询，因为日期列是非唯一的。

Answer 1

出了什么问题：

SELECT v1.date, COUNT(v1.id) AS visits 
FROM visitor_visits AS v1
GROUP BY v1.date
ORDER BY v1.date

PS。我假设id是主键（唯一）。

Answer 2

如果您要查看每个日期的不同访客ID，则应该只是

SELECT v1.date,
       COUNT(DISTINCT(v1.id)) as Visitors
   FROM
      visitor_visits AS v1
   group by
      v1.Date

MySQL中的自联接计数

2 个答案: