每天的总计数排序,结果按周分组

发布于 2024-12-02 17:00:25 字数 338 浏览 4 评论 0原文

我的查询现在看起来非常简单:

select 
    count(*), 
    date(visit_date), 
    DATE_FORMAT(visit_date,"%a") 
from visits 
group by date(visit_date)

这是结果:

http://d.pr/FmMg

我想要发生的是:

  1. 每周对计数进行排序

您可以修改我的查询以使其满足条件吗?

My query looks like this right now pretty straightforward:

select 
    count(*), 
    date(visit_date), 
    DATE_FORMAT(visit_date,"%a") 
from visits 
group by date(visit_date)

Here is the result:

http://d.pr/FmMg

what I want to happen is:

  1. for each week the count is sorted

Can you modify my query so it satisfies the criteria?

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评论(2

千鲤 2024-12-09 17:00:25

假设您不再需要按天计数,而只需要按周计数:

SELECT
   count(*),
   yearweek(visit_date)
FROM visits
GROUP BY yearweek(visit_date)
ORDER BY yearweek(visit_date) ASC;

Assuming, you no longer need the count by day and are ONLY looking for count by week:

SELECT
   count(*),
   yearweek(visit_date)
FROM visits
GROUP BY yearweek(visit_date)
ORDER BY yearweek(visit_date) ASC;
不乱于心 2024-12-09 17:00:25

你想这样做吗?
- 您可以使用日期部分来获取周数并按其排序。

select
    count(*),
    date(visit_date),
    DATE_FORMAT(visit_date,"%a")
from
    visits
group by
    date(visit_date)
order by 
    datepart(yyyy,visit_date),
    datepart(wk,visit_date),
    count(*)

Are you trying to do like this?
- you can use Datepart to get week number and sort by that.

select
    count(*),
    date(visit_date),
    DATE_FORMAT(visit_date,"%a")
from
    visits
group by
    date(visit_date)
order by 
    datepart(yyyy,visit_date),
    datepart(wk,visit_date),
    count(*)
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