从日期中提取YearWeek

Question

我正在尝试在Xqery中实现Java代码，以便以最有效的方式将数据字符串转换为年周组合。

我在下面

<ValidFromDttm>2017-02-13T00:00:00.001+00:00</ValidFromDttm>

输出应该是 - > 201707

另一个例子 -

<ValidToDttm>2019-12-08T23:59:59.001+00:00</ValidToDttm>

输出应该是 - > 201949

如果Xquery中有任何此类选项，有人可以建议我

Answer 1

format-dateTime本周有一个[W]说明符，所以如果你有权访问XQuery支持的版本处理器，你可以使用例如format-dateTime(xs:dateTime(.), '[Y0001][W01]')，一个更完整的例子是

root/date/format-dateTime(xs:dateTime(.), '[Y0001][W01]')

哪个用于输入文档

<root>
    <date>2017-02-13T00:00:00.001+00:00</date>
    <date>2019-12-08T23:59:59.001+00:00</date>
</root>

输出201707 201949

https://xqueryfiddle.liberty-development.net/b4GWVd

Answer 2

来自Jakob Fix的这篇旧帖可能会帮到你：http://x-query.com/pipermail/talk/2010-November/003298.html

xquery version "1.0";
declare namespace exslt = "http://exslt.org/dates-and-times";
(: for those implementations without exslt extensions :)
declare namespace date = "http://noexslt.org/dates-and-times";
declare variable $date-time := "2010-01-02T00:00:00Z";
declare variable $month-lengths := (0, 31, 28, 31, 30, 31, 30, 31, 31,
30, 31, 30, 31);

(:
 : returns the week of the year as a number.
 :)
declare function date:week-in-year( $date-time as xs:dateTime ) as xs:integer
{
    let $year := fn:year-from-dateTime( $date-time )
    let $day := fn:day-from-dateTime( $date-time )
    let $month := fn:month-from-dateTime( $date-time )

    let $days := sum( subsequence( $month-lengths, 1, $month ) )
    let $is-leap := ($year mod 4 = 0 and $year mod 100 != 0) or $year
mod 400 = 0
    return date:_week-in-year($year, $days + $day + (if ($is-leap and
$month > 2) then 1 else 0))
};

declare function date:_week-in-year( $year as xs:integer, $month-days
as xs:integer) as xs:integer
{
    let $previous-year := $year - 1
    let $is-leap := ($year mod 4 = 0 and $year mod 100 != 0) or $year
mod 400 = 0
    let $dow := ($previous-year + floor($previous-year div 4) -
        floor($previous-year div 100) + floor($previous-year div 400) +
        $month-days) mod 7
    let $day-of-week := if ($dow > 0) then $dow else 7
    let $start-day := ($month-days - $day-of-week + 7) mod 7
    let $week-number := floor(($month-days - $day-of-week + 7) div 7)
cast as xs:integer
    return
        if ($start-day >= 4) then $week-number + 1
        else if ($week-number = 0) then
            let $leap-day := if ((not($previous-year mod 4) and
$previous-year mod 100) or not($previous-year mod 400)) then 1 else 0
            return date:_week-in-year( $previous-year, 365 + $leap-day )
            else $week-number
};

exslt:week-in-year( $date-time ),
date:week-in-year( $date-time cast as xs:dateTime )