SQL 查询其中日期 = 今天减去 7 天

Question

我的网站有一个名为“ExternalHits”的 SQL 命中表。我将 URL 跟踪为 URLx，并将访问页面的日期跟踪为 Datex。我每周运行此查询以获取前一周的总点击数，并且每周我都必须手动更改“之间”日期。有什么方法可以更改我的查询，以便“之间”日期类似于 TODAY AND TODAY-7 吗？我只是不想每周手动更改日期。

    SELECT URLX, COUNT(URLx) AS Count
    FROM ExternalHits
    WHERE datex BETWEEN '02/27/2017' AND '03/05/2017'    
    GROUP BY URLx
    ORDER BY Count DESC; 

Answer 1

declare @lastweek datetime
declare @now datetime
set @now = getdate()
set @lastweek = dateadd(day,-7,@now)

SELECT URLX, COUNT(URLx) AS Count
FROM ExternalHits
WHERE datex BETWEEN @lastweek AND @now
GROUP BY URLx
ORDER BY Count DESC; 

Answer 2

使用 dateadd 从当前日期中删除一周。

datex BETWEEN DATEADD(WEEK,-1,GETDATE()) AND GETDATE()

Answer 3

您可以用以下方法从当前日期中减去 7：

WHERE datex BETWEEN DATEADD(day, -7, GETDATE()) AND GETDATE()

Answer 4

使用内置函数：

SELECT URLX, COUNT(URLx) AS Count
FROM ExternalHits
WHERE datex BETWEEN DATE_SUB(NOW(), INTERVAL 7 DAY) AND NOW()
GROUP BY URLx
ORDER BY Count DESC; 

Answer 5

使用以下内容：

WHERE datex BETWEEN GETDATE() AND DATEADD(DAY, -7, GETDATE())

希望这有帮助。

Answer 6

您可以使用

CURDATE()

和

DATE_SUB()

函数来实现此目的：

SELECT URLX, COUNT(URLx) AS Count
FROM ExternalHits
WHERE datex BETWEEN DATE_SUB(CURDATE(), INTERVAL 7 DAY) AND CURDATE()    
GROUP BY URLx
ORDER BY Count DESC; 

Answer 7

DECLARE @Daysforward int
SELECT @Daysforward = 25 (no of days required)
Select * from table name

where CAST( columnDate AS date) < DATEADD(day,1+@Daysforward,CAST(GETDATE() AS date))

Answer 8

对于 MariaDB：

WHERE datex BETWEEN DATE_ADD(CURDATE(), INTERVAL -7 DAY) AND CURDATE()

它对我有帮助。

Answer 9

纯Mysql：

WHERE datex BETWEEN adddate(curdate(), -7)) AND curdate();