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确定给定月份的最后一天

tag-icon • tag-icon macros datetime
确定给定月份的最后一天

我想确定某个月有多少天。

使用该datenumber包，我想我可以简单地将日期设置为下一个月，然后向后推一天，即月末，然后提取日期。那当然似乎当我想到这个时很简单，但后来却无法真正提取月份的日期。

那么我需要更换什么

\xdef\LastDayOfMonth{\value{dateday}}

获取该月的天数。

笔记：

  • 我并不反对额外的方案，所以任何方案解决方案都可以。

代码：

\documentclass{article}
\usepackage{datenumber}
\usepackage{pgffor}

\newcounter{GivenDateCounter}
\newcommand{\LastDayOfMonth}{XX}%
\newcommand{\NumberOfDaysInMonth}[2]{%
    % #1 = YYYY
    % #2 = MM
    \ifnum#2<12\relax
        % Since this is NOT Dec we can just go to the 1st of the next
        % month and back up one day.
        \setmydatenumber{GivenDateCounter}%
                {#1}%
                {\numexpr#2+1\relax}% Next month
                {01}% First of the month
    \else
        % Since this IS Dec we can just go to Jan 1st of the next
        % year and back up one day.
        \setmydatenumber{GivenDateCounter}%
                {\numexpr#1+1\relax}% Next Year
                {01}% Next day is Jan next year since current month is 12
                {01}% First of the month
    \fi
    \addtocounter{GivenDateCounter}{-1}% Now we are the end of the required month
    \xdef\LastDayOfMonth{\value{dateday}}%  <---- Stuck here.
    \typeout{**** \string\LastDayOfMonth for #1-#2=\LastDayOfMonth}%
}%

\begin{document}

\foreach \YearNumber in {2013, 2014} {%
    \foreach \MonthNumber in {01, 02, ..., 12} {%
        \par
        Number of Days in \YearNumber-\MonthNumber{} is 
        \NumberOfDaysInMonth{\YearNumber}{\MonthNumber}.%
    }%
}%


\end{document}

答案1

我想说使用\arabic{dateday}，因为\value{dateday}不可扩展（它扩展到寄存器名称，而不是值）。但是它不起作用，因为dateday没有设置。

由于您不受 的约束datetime，这里有一个带有 的版本xparse。

\documentclass{article}
\usepackage{xparse,pgffor,multicol}

\ExplSyntaxOn
\DeclareExpandableDocumentCommand\NumberOfDaysInMonth { m m }
 {
  \grill_numb_days_month:nn { #1 } { #2 }
 }

\cs_new:Npn \grill_numb_days_month:nn #1 #2
 {
  \int_case:nn { #2 }
   {
    {1}{31}
    {2}{ \grill_if_leap_year:nTF { #1 } { 29 } { 28 } }
    {3}{31}
    {4}{30}
    {5}{31}
    {6}{30}
    {7}{31}
    {8}{31}
    {9}{30}
    {10}{31}
    {11}{30}
    {12}{31}
   }
 }

\prg_new_conditional:Npnn \grill_if_leap_year:n #1 { p,T,F,TF }
 {
  \bool_if:nTF
   {
    \int_compare_p:n { \int_mod:nn { #1 } { 4 } != 0 }
    ||
    (
     \int_compare_p:n { \int_mod:nn { #1 } { 4 } == 0 }
     &&
     \int_compare_p:n { \int_mod:nn { \int_div_truncate:nn { #1 } { 100 } } { 4 } != 0 }
    ) 
   }
   { \prg_return_false: }
   { \prg_return_true: }
 }

\ExplSyntaxOff

\begin{document}
\begin{multicols}{2}

\foreach \YearNumber in {1900, 2000, 2012, 2013} {%
    \foreach \MonthNumber in {1, 2, ..., 12} {%
        \par
        Number of Days in \YearNumber-\MonthNumber{} is 
        \NumberOfDaysInMonth{\YearNumber}{\MonthNumber}.%
    }
}

\end{multicols}

\end{document}

注意\NumberOfDaysInMonth是完全可扩展的，因此

\edef\MDCCFebruary{\NumberOfDaysInMonth{1700}{2}}

将定义要扩展为的宏28。

在此处输入图片描述

条件命令\grill_if_leap_year:nTF还可以用于其他场合，有谓词、真、假、真/假四种形式。

答案2

您可以使用捆绑包pgfcalendar中的包pgf。实用程序宏接受日期输入，如的<year>-<month>-last最后一天。<month><year>

\documentclass{article}
\usepackage{pgfkeys,pgfcalendar}

\begin{document} 

\newcount\datecount
\newcommand\daysinmonth[2]{%
  \pgfcalendardatetojulian{#1-#2-last}{\datecount}%
  \pgfcalendarjuliantodate{\the\datecount}{\dyear}{\dmonth}{\dday}%
  \dday
}%

\daysinmonth{2012}{2}\par
\daysinmonth{2014}{2}\par
\daysinmonth{2014}{5}

\end{document}

答案3

一种老式的方法，使用\ifnum和 mod 计算intcalc：

在此处输入图片描述

\documentclass{article}
\usepackage{multicol,pgffor,intcalc}
\newcommand{\NumberOfDaysInMonth}[2]{% \NumberOfDaysInMonth{<year>}{<month>}
  \ifcase#2
  \or 31% January
  \or% February
    \ifnum\intcalcMod{#1}{4}=0
      \ifnum\intcalcMod{#1}{100}=0
        \ifnum\intcalcMod{#1}{400}=0
          29% Year divisible by 4 and 100 and 400
        \else
          28% Year divisible by 4 and 100 (not 400)
        \fi
      \else
        29% Year divisible by 4 (not 100 nor 400)
      \fi
    \else
      28% Year not divisible by 4
    \fi
  \or 31% March
  \or 30% April
  \or 31% May
  \or 30% June
  \or 31% July
  \or 31% August
  \or 30% September
  \or 31% October
  \or 30% November
  \or 31% December
  \else
    999% Invalid month...
  \fi
}

\begin{document}

\begin{multicols}{2}

\foreach \YearNumber in {1900, 2000, 2012, 2013} {%
    \foreach \MonthNumber in {1, 2, ..., 12} {%
        \par
        Number of Days in \YearNumber-\MonthNumber{} is 
        \NumberOfDaysInMonth{\YearNumber}{\MonthNumber}.%
    }
}

\end{multicols}

\end{document}

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