Question

我在a中有一个表格，其结构如下：

CustID --- DateAdded ---

 396       2012-02-09 
 396       2012-02-09 
 396       2012-02-08 
 396       2012-02-07 
 396       2012-02-07
 396       2012-02-07 
 396       2012-02-06
 396       2012-02-06

我想知道如何计算SQL中过去7天每天的记录数，然后将其作为整数返回。

目前我编写了以下SQL查询：

SELECT * 
  FROM Responses
 WHERE DateAdded >= dateadd(day, datediff(day, 0, GetDate()) - 7, 0)

RETURN

但是，这仅返回过去7天的所有条目。如何计算过去7天内每天的记录？

Answer 1

select DateAdded, count(CustID)
from Responses
WHERE DateAdded >=dateadd(day,datediff(day,0,GetDate())- 7,0)
GROUP BY DateAdded

Answer 2

select DateAdded, count(CustID)
from tbl
group by DateAdded

大约7天的时间间隔是依赖于DB的问题

Answer 3

SELECT DateAdded, COUNT(1) AS NUMBERADDBYDAY
FROM Responses
WHERE DateAdded >= dateadd(day,datediff(day,0,GetDate())- 7,0)
GROUP BY DateAdded

Answer 4

这类似于上面的答案，它使用MySql DATE_FORMAT（）函数。我还选择了1月的一个特定星期。

SELECT 
    DatePart(day, DateAdded) AS date, 
    COUNT(entryhash) AS count 
FROM Responses 
    where DateAdded > '2020-01-25' and DateAdded < '2020-02-01' 
GROUP BY 
    DatePart(day, DateAdded )

Answer 5

SELECT count(*), dateadded FROM Responses
WHERE DateAdded >=dateadd(day,datediff(day,0,GetDate())- 7,0)
group by dateadded

RETURN

这将为您提供每个日期添加值的记录计数。不要错误地在select中添加更多列，期望每天只获得一个计数。 group by子句将为列出的每个列的唯一实例提供一行。

Answer 6

select DateAdded, count(DateAdded) as num_records
from your_table
WHERE DateAdded >=dateadd(day,datediff(day,0,GetDate())- 7,0)
group by DateAdded
order by DateAdded

Answer 7

你也可以试试这个：

SELECT DISTINCT（DATE（dateadded））AS unique_date，COUNT（*）AS amount 从表 GROUP BY unique_date ORDER BY unique_date ASC

Answer 8

如果您的时间戳记包含时间，而不仅包括日期，请使用：

SELECT DATE_FORMAT('timestamp', '%Y-%m-%d') AS date, COUNT(id) AS count FROM table GROUP BY DATE_FORMAT('timestamp', '%Y-%m-%d')

如何计算每天的记录数？

8 个答案: