Implementing Switch Case with date or other numbers

Question

I try to use a "Switch Case" macro in my LaTex File. But I want to put in the "Switch Case" macro the month not a number. Unfortunately a get nada as output instead of 31.01.22.

Source:Implementing switch cases

\documentclass{article}
\usepackage{pdftexcmds}
\makeatletter
\newcommand*{\dothis}[1]{%
    \stringcases
    {#1}%
    {%      
        {1}{31.01.\the\year}
        {2}{28.02.\the\year}
        {3}{31.03.\the\year}
        {4}{30.04.\the\year}
        {5}{31.05.\the\year}
        {6}{30.06.\the\year}
        {7}{31.07.\the\year}
        {8}{31.08.\the\year}
        {9}{30.09.\the\year}
        {10}{31.10.\the\year}
        {11}{30.11.\the\year}
        {12}{31.12.\the\year}
    }%
    {[nada]}%
}
\newcommand{\stringcases}[3]{%
    \romannumeral
    \str@case{#1}#2{#1}{#3}\q@stop
}
\newcommand{\str@case}[3]{%
    \ifnum\pdf@strcmp{\unexpanded{#1}}{\unexpanded{#2}}=\z@
    \expandafter\@firstoftwo
    \else
    \expandafter\@secondoftwo
    \fi
    {\str@case@end{#3}}
    {\str@case{#1}}%
}
\newcommand{\str@case@end}{}
\long\def\str@case@end#1#2\q@stop{\z@#1}
\makeatother
\begin{document}
\dothis{\the\month}

\dothis{2}

\dothis{3}

\dothis{4}


\end{document}

If there is a better way to get the last day of the month, I am also open to read your proposals. — IH Pro, Jan 06 '22 at 01:34

egreg · Accepted Answer · 2022-01-06T13:59:20.980

I'd not reinvent the wheel…

\documentclass{article}
\ExplSyntaxOn
\NewExpandableDocumentCommand{\dothis}{m}
  {
    \str_case_e:nnF {#1}
      {
        {1}{31.01.\the\year}
        {2}{28.02.\the\year}
        {3}{31.03.\the\year}
        {4}{30.04.\the\year}
        {5}{31.05.\the\year}
        {6}{30.06.\the\year}
        {7}{31.07.\the\year}
        {8}{31.08.\the\year}
        {9}{30.09.\the\year}
        {10}{31.10.\the\year}
        {11}{30.11.\the\year}
        {12}{31.12.\the\year}
      }
      {[nada]}%
  }
\ExplSyntaxOff
\begin{document}
\dothis{\the\month}
\dothis{2}
\dothis{3}
\dothis{4}
\dothis{15}
\end{document}

An extended version to show the potential of expl3.

\documentclass{article}
\ExplSyntaxOn
\NewExpandableDocumentCommand{\dothis}{om}
  {
    \ihpro_dothis:ee
     {
      \int_eval:n { \IfNoValueTF { #1 } { \c_sys_year_int } { #1 } }
     }
     { \int_eval:n { #2 } }
  }
\cs_new:Nn \ihpro_dothis:nn
  {
    \str_case_e:nnF {#2}
      {
        {1}{31.01.#1}
        {2}{28.02.#1}
        {3}{31.03.#1}
        {4}{30.04.#1}
        {5}{31.05.#1}
        {6}{30.06.#1}
        {7}{31.07.#1}
        {8}{31.08.#1}
        {9}{30.09.#1}
        {10}{31.10.#1}
        {11}{30.11.#1}
        {12}{31.12.#1}
      }
      {[nada]}%
  }
\cs_generate_variant:Nn \ihpro_dothis:nn { ee }
\ExplSyntaxOff
\begin{document}
\dothis{\month}
\dothis{2}
\dothis{3}
\dothis{4}
\dothis{15}
\dothis[2010]{8}
\end{document}

Here's an extended code for leap years.

\documentclass{article}
\ExplSyntaxOn
\NewExpandableDocumentCommand{\dothis}{om}
  {
    \ihpro_dothis:ee
     {
      \int_eval:n { \IfNoValueTF { #1 } { \c_sys_year_int } { #1 } }
     }
     { \int_eval:n { #2 } }
  }
\cs_new:Nn \ihpro_dothis:nn
  {
    \str_case_e:nnF {#2}
      {
        {1}{31.01.#1}
        {2}{\ihpro_if_leap:nTF{#1}{29}{28}.02.#1}
        {3}{31.03.#1}
        {4}{30.04.#1}
        {5}{31.05.#1}
        {6}{30.06.#1}
        {7}{31.07.#1}
        {8}{31.08.#1}
        {9}{30.09.#1}
        {10}{31.10.#1}
        {11}{30.11.#1}
        {12}{31.12.#1}
      }
      {[nada]}%
  }
\cs_generate_variant:Nn \ihpro_dothis:nn { ee }
\prg_new_conditional:Nnn \ihpro_if_leap:n { T, F, TF, p }
 {
  \int_compare:nTF { \int_mod:nn { #1 } { 400 } == 0 }
   {
    \prg_return_true:
   }
   {
    \bool_lazy_and:nnTF
     { \int_compare_p:n { \int_mod:nn { #1 } { 4 } == 0} }     % divisible by 4
     { !\int_compare_p:n { \int_mod:nn { #1 } { 100 } == 0 } } % but not by 100
     { \prg_return_true: }
     { \prg_return_false: }
   }
 }
\ExplSyntaxOff
\begin{document}
\dothis{\month}
\dothis{2}
\dothis{3}
\dothis{4}
\dothis{15}
\dothis[2010]{8}
\dothis[2010]{2}
\dothis[2020]{2}
\dothis[2000]{2}
\dothis[1900]{2}
\end{document}

but is there a way where leap year is considered. For example for the year 2024 I want to get \dothis{2}=> 29.02.24 — IH Pro, Jan 06 '22 at 12:43
If this is difficult to handle then nevermind. Thank you very much for the answer — IH Pro, Jan 06 '22 at 12:45

Ulrich Diez · Answer 2 · 2022-01-06T06:45:57.560

You can omit \unexpanded and for the sake of complicatedness add detecting whether \the\year denotes a leap-year in case the month is 2/February:

\documentclass{article}
\usepackage{pdftexcmds}
\makeatletter
@ifdefinable\stopromannumeral{\chardef\stopromannumeral=`^^00}%
\newcommand{\dothis}[1]{%
    \romannumeral\str@case{#1}%
    {1}{31.01.\the\year}%
    {2}{.02.\the\year}%
    {3}{31.03.\the\year}%
    {4}{30.04.\the\year}%
    {5}{31.05.\the\year}%
    {6}{30.06.\the\year}%
    {7}{31.07.\the\year}%
    {8}{31.08.\the\year}%
    {9}{30.09.\the\year}%
    {10}{31.10.\the\year}%
    {11}{30.11.\the\year}%
    {12}{31.12.\the\year}%
    {#1}{[nada]}%
    \str@case@end
}%
\newcommand{\str@case}[3]{%
    \ifnum\pdf@strcmp{#1}{#2}=0
    \expandafter@firstoftwo\else\expandafter@secondoftwo\fi
    {\str@case@end{#2}{#3}}{\str@case{#1}}%
}%
@ifdefinable\str@case@end{%
  \long\def\str@case@end#1#2#3\str@case@end{%
    \ifnum\pdf@strcmp{#1}{2}=0 \expandafter@firstoftwo\else\expandafter@secondoftwo\fi
    {%
      \ifnum\the\numexpr((\the\numexpr\the\year\relax)/4)4\relax=@firstofone{\the\numexpr\the\year\relax} %
      \expandafter@firstoftwo\else\expandafter@secondoftwo\fi
      {%
        \ifnum\the\numexpr((\the\numexpr\the\year\relax)/25)25\relax=@firstofone{\the\numexpr\the\year\relax} %
        \expandafter@firstoftwo\else\expandafter@secondoftwo\fi
        {%
          \ifnum\the\numexpr((\the\numexpr\the\year\relax)/400)400\relax=@firstofone{\the\numexpr\the\year\relax} %
          \expandafter@firstoftwo\else\expandafter@secondoftwo\fi
        }{@firstoftwo}%
      }{@secondoftwo}%
      {\stopromannumeral29}{\stopromannumeral28}%
    }%
    {\stopromannumeral}%
    #2% 
  }%
}%
\makeatother
\begin{document}
\begingroup
\newcount\tempcnt
\let\year=\tempcnt
\year=1983
\dothis{2}\par
\year=1984
\dothis{2}\par
\year=1700
\dothis{2}\par
\year=1800
\dothis{2}\par
\year=1900
\dothis{2}\par
\year=2000
\dothis{2}\par
\year=2100
\dothis{2}\par
\endgroup    
\dothis{\the\month}\par
\dothis{2}\par
\dothis{3}\par
\dothis{4}\par


\end{document}

Ulrich Diez · Answer 3 · 2022-01-06T19:57:00.463

The following is inspired by egreg's answer:

If you don't need a leap-year conditional, then you can incorporate the leap-year-calculation into the code for calculating and displaying the last day of February—the following code is a wild mixture of \int_eval:n (integers) and \fp_eval:n (floating point)—with \fp_eval:n you can evaluate comparison-operators like = / != and boolean operators like && / ||:

\documentclass{article}
\ExplSyntaxOn
\NewExpandableDocumentCommand{\dothis}{om}
  {
    \ihpro_dothis:ee
     {
      \int_eval:n { \IfNoValueTF { #1 } { \c_sys_year_int } { #1 } }
     }
     { \int_eval:n { #2 } }
  }
\cs_new:Nn \ihpro_dothis:nn
  {
    \str_case_e:nnF {#2}
      {
        {1}{31.01.#1}
        {2}{
             \int_eval:n{
               % the \fp_eval-expression yields 1 for leap-years, 0 otherwise.
               \fp_eval:n {    ( \int_mod:nn {#1}{4}=0 && \int_mod:nn{#1}{100}!=0 )
                            || \int_mod:nn {#1}{400}=0   }  +  28
             }.02.#1
           }
        {3}{31.03.#1}
        {4}{30.04.#1}
        {5}{31.05.#1}
        {6}{30.06.#1}
        {7}{31.07.#1}
        {8}{31.08.#1}
        {9}{30.09.#1}
        {10}{31.10.#1}
        {11}{30.11.#1}
        {12}{31.12.#1}
      }
      {[nada]}%
  }
\cs_generate_variant:Nn \ihpro_dothis:nn { ee }
\ExplSyntaxOff
\begin{document}
\dothis{\month}
\dothis{\the\month}
\dothis{2}
\dothis{3}
\dothis{4}
\dothis{15}
\dothis[2010]{8}
\dothis[2000]{2}
\dothis[1900]{2}
\dothis[1984]{2}
\dothis[1700]{2}
\dothis[1800]{2}
\dothis[1900]{2}
\dothis[2000]{2}
\dothis[2010]{2}
\dothis[2020]{2}
\dothis[2100]{2}
% This yields an error-message 
% ! Missing number, treated as zero :
% \dothis{no number}
\end{document}

Implementing Switch Case with date or other numbers

3 Answers3