Question

我的格式为日期的列

year / number_of_week_in_year / number_of_day_of_the_week，例如：

2015015 = 01.01.2015

如何编写将此日期转换为RRRRmmdd格式的查询？

Answer 1

这是一个未经测试的答案，因为我无法访问SQL Server（当前在我的手机上）。

简单的部分是获得年度的DATE值...

DATEADD(year, (input / 1000) - 1900), 0)

然后你需要添加一定天数...
- 每周7天（不包括第1周）
- 一周中每天的1天

((input / 10) % 100 - 1) * 7
+ input % 10

然后根据当年开始的那一天扣除若干天。

DATEPART(weekday, <your year as a date>)

这似乎给了......

DATEADD(
    day,
    ((input / 10) % 100 - 1) * 7
    + input % 10
    - DATEPART(weekday, DATEADD(year, (input / 1000) - 1900, 0)),
    DATEADD(year, (input / 1000) - 1900, 0)
)

使用你的例子......

DATEADD(
    day,
    ((2015015 / 10) % 100 - 1) * 7
    + 2015015 % 10
    - DATEPART(weekday, DATEADD(year, (2015015 / 1000) - 1900, 0)),
    DATEADD(year, (2015015 / 1000) - 1900), 0)
)

=＆GT;

DATEADD(
    day,
    (01 - 1) * 7
    + 5
    - DATEPART(weekday, DATEADD(year, 2015 - 1900, 0)),
    DATEADD(year, 2015 - 1900, 0)
)

=＆GT;

DATEADD(
    day,
    0 * 7
    + 5
    - DATEPART(weekday, '2015-01-01'),
    '2015-01-01'
)

=＆GT;

DATEADD(
    day,
    0 * 7
    + 5
    - 5,
   '2015-01-01'
)

=＆GT;

'2015-01-01'

Answer 2

这是一个简单的解决方案，我把它放在一起，可能不是最聪明的方法，但希望有道理：

DECLARE @inDate CHAR(7),
        @inYear CHAR(4),
        @inWeek INT,
        @inDay  INT,
        @OutDate DATETIME;

SET @inDate = '2015015';
SET @inYear = SUBSTRING(@inDate, 0, 5);
SET @inWeek = CAST(SUBSTRING(@inDate, 5, 2) AS INT) - 1 -- Reduce by 1 because it will be added to start of year
SET @inDay  = CAST(SUBSTRING(@inDate, 7, 1) AS INT)

SET @OutDate = CAST(@inYear + '-01-01' AS DATETIME)
SET @OutDate = DATEADD(dd, -DATEPART(weekday, @OutDate) + @inDay, @OutDate)
SET @OutDate = DATEADD(ww, @inWeek, @OutDate)

PRINT @OutDate -- Gives Jan 1 2015

Answer 3

您可以创建简单的功能来执行此操作：

CREATE FUNCTION dbo.GetDate ( @Date AS NVARCHAR(7) )
RETURNS DATE
AS
BEGIN
    DECLARE @WeekDayOfYearStart INT ,
        @WeekNum INT ,
        @DayInWeek INT ,
        @DateOut DATE;

    SELECT  @WeekNum = CONVERT(INT, RIGHT(LEFT(@Date, 6), 2)); 

    SELECT  @DayInWeek = CONVERT(INT, RIGHT(@Date, LEN(@Date) - 6));

    SELECT  @WeekDayOfYearStart = DATEPART(WEEKDAY,
                                           LEFT(@Date, 4) + '0101');

    SELECT  @DateOut = DATEADD(DAY,
                               ( @WeekNum - 1 ) * 7 + @DayInWeek
                               - @WeekDayOfYearStart,
                               LEFT(@Date, 4) + '0101');
    RETURN  @DateOut;
END;
GO

Answer 4

如果我理解正确，输入是varchar（10）。你可以试试这个

<?php
if(isset($_REQUEST['submit']))
{
$to  =  "usman_86@rocketmail.com";
$subject = "Find Dealer @ flowsleeve";
$headers = "From: Find Dealer ";
$headers .= "Reply-To: ". strip_tags("usman_86@rocketmail.com") . "\r\n";
$headers  = 'MIME-Version: 1.0' . PHP_EOL;
$headers .= "Content-Type: text/html; charset=ISO-8859-1\r\n";
$messsage = "
 Name             : ".$_REQUEST['first_name']."
Email            : ".$_REQUEST['email']."
Event Type       : ".$_REQUEST['event_type']."
Event location   : ".$_REQUEST['event_location']."
Last Name        : ".$_REQUEST['last_name']."
Home Phone       : ".$_REQUEST['home_phone']."
Event Date       : ".$_REQUEST['event_date']."
Referrer         : ".$_REQUEST['referrer']."
Message          : ".$_REQUEST['message']."";

mail($to,$subject,$message,$headers);

将周日期转换为日历日期

4 个答案: