有人问我如何\tomorrow
在 TeX/LaTeX 中实现宏。虽然我可以想出一个相当好的解决方案,适用于一个文档类,但我不知道如何以与文档类无关的方式实现它。
我发现日期但是,即使日期算术可能有用,它也会假定日期的特定格式(这对于我所居住的地区来说也是错误的)。
有人有合理的便携解决方案吗?这可能吗?
答案1
这是 LaTeX3 中的原型;有关明天的数据可在整数变量中找到
\l_tomorrow_day_int
\l_tomorrow_month_int
\l_tomorrow_year_int
代码如下;最后的宏仅仅是数据如何使用的一个例子,可能与有关datetime
。
\documentclass{article}
\ExplSyntaxOn
\prop_new:N \g_tomorrow_months_prop
\bool_new:N \l_tomorrow_leap_bool
\tl_new:N \l_tomorrow_daysinmonth_tl
\int_new:N \l_tomorrow_day_int
\int_new:N \l_tomorrow_month_int
\int_new:N \l_tomorrow_year_int
\prop_gput:Nnn \g_tomorrow_months_prop { 1 } { 31 }
\prop_gput:Nnn \g_tomorrow_months_prop { 2 } { \bool_if:NTF \l_tomorrow_leap_bool { 29 } { 28 } }
\prop_gput:Nnn \g_tomorrow_months_prop { 3 } { 31 }
\prop_gput:Nnn \g_tomorrow_months_prop { 4 } { 30 }
\prop_gput:Nnn \g_tomorrow_months_prop { 5 } { 31 }
\prop_gput:Nnn \g_tomorrow_months_prop { 6 } { 30 }
\prop_gput:Nnn \g_tomorrow_months_prop { 7 } { 31 }
\prop_gput:Nnn \g_tomorrow_months_prop { 8 } { 31 }
\prop_gput:Nnn \g_tomorrow_months_prop { 9 } { 30 }
\prop_gput:Nnn \g_tomorrow_months_prop { 10 } { 31 }
\prop_gput:Nnn \g_tomorrow_months_prop { 11 } { 30 }
\prop_gput:Nnn \g_tomorrow_months_prop { 12 } { 31 }
\cs_new_protected:Npn \tomorrow_check_leap:n #1
{
\int_compare:nTF { 0 = \int_mod:nn { #1 } { 4 } }
{% possibly a leap year
\int_compare:nTF { 0 = \int_mod:nn { #1 } { 100 } }
{% possibly not a leap year
\int_compare:nTF { 0 = \int_mod:nn { #1/100 } { 4 } }
{% leap year
\bool_set_true:N \l_tomorrow_leap_bool
}
{% not leap year
\bool_set_false:N \l_tomorrow_leap_bool
}
}
{% leap year
\bool_set_true:N \l_tomorrow_leap_bool
}
}
{% not leap year
\bool_set_false:N \l_tomorrow_leap_bool
}
}
\cs_new_protected:Npn \tomorrow_set_tomorrow:nnn #1 #2 #3
{
\int_compare:nT { #2 = 2 } { \tomorrow_check_leap:n { #3 } }
\int_set:Nn \l_tomorrow_day_int { #1 }
\int_set:Nn \l_tomorrow_month_int { #2 }
\int_set:Nn \l_tomorrow_year_int { #3 }
\__tomorrow_incr_day:
}
\cs_new_protected:Npn \__tomorrow_incr_day:
{
\int_incr:N \l_tomorrow_day_int
\prop_get:NVN \g_tomorrow_months_prop \l_tomorrow_month_int \l_tomorrow_daysinmonth_tl
\int_compare:nT
{ \l_tomorrow_day_int > \tl_use:N \l_tomorrow_daysinmonth_tl }
{
\int_set:Nn \l_tomorrow_day_int { 1 }
\__tomorrow_incr_month:
}
}
\cs_new_protected:Npn \__tomorrow_incr_month:
{
\int_incr:N \l_tomorrow_month_int
\int_compare:nT { \l_tomorrow_month_int > 12 }
{
\int_set:Nn \l_tomorrow_month_int { 1 }
\int_incr:N \l_tomorrow_year_int
}
}
\cs_generate_variant:Nn \prop_get:Nn { NV,Ne }
\NewDocumentCommand{\printtomorrowof}{mmm}
{
\tomorrow_set_tomorrow:nnn { #1 } { #2 } { #3 }
Today~it~is~
\int_to_arabic:n { #3 }/
\int_to_arabic:n { #2 }/
\int_to_arabic:n { #1 },~
tomorrow~it~is~
\int_to_arabic:n { \l_tomorrow_year_int }/
\int_to_arabic:n { \l_tomorrow_month_int }/
\int_to_arabic:n { \l_tomorrow_day_int }
\par
}
\ExplSyntaxOff
\begin{document}
\printtomorrowof{\day}{\month}{\year}
\printtomorrowof{30}{10}{2012}
\printtomorrowof{31}{10}{2012}
\printtomorrowof{31}{12}{2012}
\printtomorrowof{28}{2}{2012}
\printtomorrowof{28}{2}{2013}
\printtomorrowof{28}{2}{1900}
\printtomorrowof{28}{2}{2000}
\end{document}
如您所见,闰年被正确识别。当然,仅限公历。
为了定义合适的\tomorrow
命令,您可以添加(之前\ExplSyntaxOn
)一个babel
版本
\NewDocumentCommand{\tomorrow}{}
{
\tomorrow_set_tomorrow:nnn { \day } { \month } { \year }
\group_begin:
\day = \l_tomorrow_day_int
\month = \l_tomorrow_month_int
\year = \l_tomorrow_year_int
\today
\group_end:
}
或者一个datetime
版本(datetime
当然需要包)
\NewDocumentCommand{\tomorrow}{}
{
\tomorrow_set_tomorrow:nnn { \day } { \month } { \year }
\formatdate { \l_tomorrow_day_int }
{ \l_tomorrow_month_int }
{ \l_tomorrow_year_int }
}
当然,如果只想要明天的日期,这样做就有点过了。宏实际上允许根据给定的日期计算任何日期,给定间隔(正或负)。也可以将儒略日期的“反转”扩展为“日/月/年”形式,但这样会非常慢的。
\documentclass{article}
\usepackage{xparse}
\ExplSyntaxOn
\DeclareExpandableDocumentCommand{\juliandate}{ m m m }
{
\juliandate_calc:nnnn { #1 } { #2 } { #3 } { \use:n }
}
\NewDocumentCommand{\storejuliandate}{ s m m m m }
{
\IfBooleanTF{#1}
{
\juliandate_calc:nnnn { #3 } { #4 } { #5 } { \cs_set:Npx #2 }
}
{
\juliandate_calc:nnnn { #3 } { #4 } { #5 } { \cs_new:Npx #2 }
}
}
\cs_new:Npn \juliandate_calc:nnnn #1 #2 #3 #4 % #1 = day, #2 = month, #3 = year, #4 = what to do
{
#4
{
\int_eval:n
{
#1 +
\int_div_truncate:nn { 153 * (#2 + 12 * \int_div_truncate:nn { 14 - #2 } { 12 } - 3) + 2 } { 5 } +
365 * (#3 + 4800 - \int_div_truncate:nn { 14 - #2 } { 12 } ) +
\int_div_truncate:nn { #3 + 4800 - \int_div_truncate:nn { 14 - #2 } { 12 } } { 4 } -
\int_div_truncate:nn { #3 + 4800 - \int_div_truncate:nn { 14 - #2 } { 12 } } { 100 } +
\int_div_truncate:nn { #3 + 4800 - \int_div_truncate:nn { 14 - #2 } { 12 } } { 400 } -
32045
}
}
}
\tl_new:N \l__juliandate_g_tl
\tl_new:N \l__juliandate_dg_tl
\tl_new:N \l__juliandate_c_tl
\tl_new:N \l__juliandate_dc_tl
\tl_new:N \l__juliandate_b_tl
\tl_new:N \l__juliandate_db_tl
\tl_new:N \l__juliandate_a_tl
\tl_new:N \l__juliandate_da_tl
\tl_new:N \l__juliandate_y_tl
\tl_new:N \l__juliandate_m_tl
\tl_new:N \l__juliandate_d_tl
\int_new:N \l_juliandate_day_int
\int_new:N \l_juliandate_month_int
\int_new:N \l_juliandate_year_int
\cs_new:Npn \__juliandate_set:nn #1 #2
{
\tl_set:cx { l__juliandate_#1_tl } { \int_eval:n { #2 } }
}
\cs_new:Npn \__juliandate_use:n #1
{
\tl_use:c { l__juliandate_#1_tl }
}
\cs_new_protected:Npn \juliandate_reverse:n #1
{
\__juliandate_set:nn { g }
{ \int_div_truncate:nn { #1 + 32044 } { 146097 } }
\__juliandate_set:nn { dg }
{ \int_mod:nn { #1 + 32044 } { 146097 } }
\__juliandate_set:nn { c }
{ \int_div_truncate:nn { ( \int_div_truncate:nn { \__juliandate_use:n { dg } } { 36524 } + 1) * 3 } { 4 } }
\__juliandate_set:nn { dc }
{ \__juliandate_use:n { dg } - \__juliandate_use:n { c } * 36524 }
\__juliandate_set:nn { b }
{ \int_div_truncate:nn { \__juliandate_use:n { dc } } { 1461 } }
\__juliandate_set:nn { db }
{ \int_mod:nn { \__juliandate_use:n { dc } } { 1461 } }
\__juliandate_set:nn { a }
{ \int_div_truncate:nn { ( \int_div_truncate:nn { \__juliandate_use:n { db } } { 365 } + 1) * 3 } { 4 } }
\__juliandate_set:nn { da }
{ \__juliandate_use:n { db } - \__juliandate_use:n { a } * 365 }
\__juliandate_set:nn { y }
{
\__juliandate_use:n { g } * 400 +
\__juliandate_use:n { c } * 100 +
\__juliandate_use:n { b } * 4 +
\__juliandate_use:n { a }
}
\__juliandate_set:nn { m }
{ \int_div_truncate:nn { \__juliandate_use:n { da } * 5 + 308 } { 153 } - 2 }
\__juliandate_set:nn { d }
{ \__juliandate_use:n { da } - \int_div_truncate:nn { (\__juliandate_use:n { m } + 4) * 153 } { 5 } + 122 }
\int_set:Nn \l_juliandate_year_int
{ \__juliandate_use:n { y } - 4800 + \int_div_truncate:nn { \__juliandate_use:n { m } + 2 } { 12 } }
\int_set:Nn \l_juliandate_month_int
{ \int_mod:nn { \__juliandate_use:n { m } + 2 } { 12 } + 1 }
\int_set:Nn \l_juliandate_day_int
{ \__juliandate_use:n { d } + 1 }
}
\cs_generate_variant:Nn \juliandate_reverse:n { x }
\NewDocumentCommand{\showday}{ m }
{
\juliandate_reverse:n { #1 }
\int_to_arabic:n { \l_juliandate_day_int }-
\int_to_arabic:n { \l_juliandate_month_int }-
\int_to_arabic:n { \l_juliandate_year_int }
}
\NewDocumentCommand{\tomorrow}{ }
{
\group_begin:
\juliandate_reverse:x { \juliandate_calc:nnnn { \day + 1 } { \month } { \year } { \use:n } }
\day = \l_juliandate_day_int
\month = \l_juliandate_month_int
\year = \l_juliandate_year_int
\today
\group_end:
}
\NewDocumentCommand{\tomorrowof}{ m m m }
{
\group_begin:
\juliandate_reverse:x { \juliandate_calc:nnnn { #1 + 1 } { #2 } { #3 } { \use:n } }
\day = \l_juliandate_day_int
\month = \l_juliandate_month_int
\year = \l_juliandate_year_int
\today
\group_end:
}
\ExplSyntaxOff
\begin{document}
\juliandate{18}{12}{2012}
\storejuliandate*{\x}{18}{12}{2012}\x
\storejuliandate*{\x}{1}{1}{1900}\x
\showday{2456280}
\showday{2415021}
\tomorrow
\tomorrowof{31}{12}{2012}
\tomorrowof{28}{2}{2012}
\tomorrowof{29}{2}{2012}
\tomorrowof{28}{2}{2013}
\tomorrowof{28}{2}{1900}
\end{document}
答案2
只需快速烹饪
\documentclass{article}
\usepackage{pgfkeys,pgfcalendar}
\newcount\pgfdatecount
\newcommand{\tomorrow}{%
\pgfcalendardatetojulian{\year-\month-\day+1}{\pgfdatecount}
\pgfcalendarjuliantodate{\the\pgfdatecount}{\myyear}{\mymonth}{\myday}
\pgfcalendarmonthname{\mymonth}\space\myday,\space\myyear%
}
\begin{document}
\today~ is the day before tomorrow.\par
\tomorrow~ is the day after today.
\end{document}
还可以添加星期名称等。您可以查看手册。据我所知,这与班级无关,但月份名称是固定的。如果您愿意,也可以引入它们,但 ISO 日期就是因为这个而变得混乱。
答案3
我建议使用datenumber
包。
序言
\usepackage{datenumber}
\setdatetoday
\addtocounter{datenumber}{1}%
\setdatebynumber{\thedatenumber}%
在文档中
今天是
\today
明天是\datedate
答案4
这个问题已经有一段时间了,但我还没有看到一个简单的 TeX 解决方案。希望它对某些人有用。
\let\mydatesep=-
\newcount\advancedday
\newcount\daysinthismonth
\newcount\tempyear
\newcount\ttempyear
\def\advancedatebydays#1{%
\advancedday=\number\day%
\advance\advancedday by #1%
\tempyear=\number\year%
\ttempyear=\tempyear%
\loop%
\loop%
\ifnum\ttempyear>4 %
\advance\ttempyear by-4 %
\repeat%
\ifnum\month=2 %
\ifnum\ttempyear=4 %
\daysinthismonth=29 %
\else%
\daysinthismonth=28 %
\fi%
\else%
\ifodd\month%
\ifnum\month<8 %
\daysinthismonth=30 %
\else%
\daysinthismonth=31 %
\fi%
\else%
\ifnum\number\month>7 %
\daysinthismonth=31 %
\else%
\daysinthismonth=30 %
\fi%
\fi%
\fi%
\ifnum\advancedday>\daysinthismonth%
\ifnum\month=12 %
\advance \tempyear by 1 %
\advance \ttempyear by 1 %
\advance \advancedday by -\daysinthismonth%
\advance \month by -11 %
\else%
\advance \advancedday by -\daysinthismonth%
\advance \month by 1 %
\fi%
\repeat%
\number\advancedday\mydatesep\number\month\mydatesep\number\tempyear%
}%
\advancedatebydays{302}
\bye