Use xstring to substitute reformatted names in strings

Question

I would like to bold-face occurrences of strings conditionally on them occurring within longer strings, using the \IfSubStr and \StrSubstitute macros from the xstring package. The issue that I'm running into is related to expansion of the \makebold macro, which depends on the work of the \surnamefirst and \surnamelast macros. How can I get the \makebold macro to work correctly?

\documentclass{article}

\usepackage{xstring}

\newcommand{\name}{John Doe}

\makeatletter

% Macro to extract and store "last name" from full-name string
\newcommand{\lastname}{%
    \IfSubStr{\name}{.}%
        {\StrBehind{\name}{. }}%
        {\StrBehind{\name}{ }}%
        [\@lastname]
}

% Extract and store first name from full-name string
\newcommand{\firstname}{%
    \StrBefore{\name}{ }[\@firstname]
}

% Extract first initial from full-name string
\newcommand{\firstinitial}{%
    \StrLeft{\name}{1}[\@firstinitial]
}

% Extract middle initial (if any) from full-name string
\newcommand{\middleinitial}{%
    \StrBetween{\name}{ }{.}[\@middleinitial]
}

\newcommand{\testxstring}{%
    \middleinitial % Find middle initial
    \IfStrEq{\@middleinitial}{}%
        {No middle initial!}%
        {Middle initial: \@middleinitial}
}

% Surname first
\newcommand{\surnamefirst}{%
    \firstinitial
    \middleinitial % Find middle initial
    \lastname
    \IfStrEq{\@middleinitial}{}%
        {\@lastname, \@firstinitial.}%
        {\@lastname, \@firstinitial.~\@middleinitial.}
}

% Surname last
\newcommand{\surnamelast}{%
    \firstinitial
    \middleinitial % Find middle initial
    \lastname
    \IfStrEq{\@middleinitial}{}%
        {\@firstinitial.~\@lastname}%
        {\@firstinitial.~\@middleinitial.~\@lastname}
}

% Render full name in bold
\newcommand{\makeboldname}[1]{%
    \expandarg
    \IfSubStr{#1}{\getvalue{name}}%
        {\StrSubstitute{#1}{\name}{\textbf{\name}}\par}%
        {#1}
}

% Render only last name in bold
\newcommand{\makeboldlastname}[1]{%
    \lastname
    \expandarg
    \IfSubStr{#1}{\@lastname}%
        {\StrSubstitute{#1}{\@lastname}{\textbf{\@lastname}}\par}%
        {#1}
}

% Render surname-first and surname-last in bold
\newcommand{\makebold}[1]{%
    \expandarg
    \IfSubStr{#1}{\surnamefirst}%
        {\StrSubstitute{#1}{\surnamefirst}{\textbf{\surnamefirst}}\par}%
        {\IfSubStr{#1}{\surnamelast}%
            {\StrSubstitute{#1}{\surnamelast}{\textbf{\surnamelast}}\par}%
            {#1}%
        }
}

\makeatother

\begin{document}

    \name

    \surnamefirst

    \surnamelast

    \makeboldname{His name was John Doe.}

    \makeboldlastname{Again, Doe was his last name.}

    \makebold{I heard that Doe, J. was from Kentucky.}

    \makebold{There was also a J. Doe from New York.}

\end{document}

Expected output:

John Doe

Doe, J.

J. Doe

His name was John Doe.

Again, Doe was his last name.

I heard that Doe, J. was from Kentucky.

There was also a J. Doe from New York.

Answer 1

This seems to be a job for regular expressions:

\documentclass{article}
\usepackage{xparse}

\ExplSyntaxOn
\NewDocumentCommand{\definename}{m}
 {
  \adam_name_define:n { #1 }
 }

\NewDocumentCommand{\makebold}{m}
 {
  \adam_name_makebold:n { #1 }
 }

\NewDocumentCommand{\makeboldlastname}{m}
 {
  \adam_name_makebold_lastname:n { #1 }
 }


\tl_new:N \l_adam_name_full_tl
\tl_new:N \l_adam_name_first_tl
\tl_new:N \l_adam_name_last_tl
\tl_new:N \l_adam_name_initials_tl
\tl_new:N \l__adam_name_input_tl

\cs_new_protected:Nn \adam_name_define:n
 {
  % store the full name
  \tl_set:Nn \l_adam_name_full_tl { #1 }
  % duplicate it in two token lists to be modified
  \tl_set_eq:NN \l_adam_name_first_tl \l_adam_name_full_tl
  \tl_set_eq:NN \l_adam_name_last_tl \l_adam_name_full_tl
  % remove everything from the last space to the end (first name)
  \regex_replace_once:nnN { (.*)\s[^\s]*\Z } { \1 } \l_adam_name_first_tl
  % remove everything up to the last space (surname)
  \regex_replace_once:nnN { .*\s([^\s]*)\Z } { \1 } \l_adam_name_last_tl
  % duplicate the token list with the first name
  \tl_set_eq:NN \l_adam_name_initials_tl \l_adam_name_first_tl
  % remove letters after the first in the first name part, replacing them with a period (initials)
  \regex_replace_all:nnN { ([[:alpha:]])[[:alpha:]]+ } { \1. } \l_adam_name_initials_tl
 }

\cs_new_protected:Nn \adam_name_makebold:n
 {
  \tl_set:Nn \l__adam_name_input_tl { #1 }
  % \u{<token list name>} stands for the contents of the token list
  % we'll search for “full name” or “initials surname” or
  % “surname, first name” or “surname, initials”
  % and replace the match with \textbf{\match}
  \regex_replace_once:nnN
   {
    (
     \u{l_adam_name_full_tl}                                   % name surname
     |
     \u{l_adam_name_initials_tl} \s \u{l_adam_name_last_tl}    % initials surname
     |
     \u{l_adam_name_last_tl} , \s \u{l_adam_name_first_tl}     % surname, name
     |
     \u{l_adam_name_last_tl} , \s \u{l_adam_name_initials_tl}  % surname, initials
    )
   }
   { \c{textbf} \cB\{ \1 \cE\} }
   \l__adam_name_input_tl
  % print
  \tl_use:N \l__adam_name_input_tl
 }

\cs_new_protected:Nn \adam_name_makebold_lastname:n
 {
  \tl_set:Nn \l__adam_name_input_tl { #1 }
  % search for the surname and replace it with \textbf{<match>}
  \regex_replace_once:nnN
   { (\u{l_adam_name_last_tl}) }
   { \c{textbf} \cB\{ \1 \cE\} }
   \l__adam_name_input_tl
  \tl_use:N \l__adam_name_input_tl
 }

\ExplSyntaxOff

\begin{document}

\definename{John Doe}

\makebold{His name was John Doe.}

\makebold{His name was J. Doe.}

\makebold{His name was Doe, John.}

\makebold{His name was Doe, J. again.}

\makeboldlastname{Again, Doe was his last name.}

\definename{John T. Doe}

\makebold{His name was John T. Doe.}

\makebold{His name was J. T. Doe.}

\makebold{His name was Doe, John T. again.}

\makebold{His name was Doe, J. T. again.}

\makeboldlastname{Again, Doe was his last name.}

\end{document}

Answer 2

As promised, below is an answer using xstring by its author, Christian Tellechea.

\documentclass{article}

\usepackage[T1]{fontenc}

\usepackage{xstring}

\def\makebold#1#2{%
    \begingroup\expandarg
    \StrCount{#1}{ }[\nbspaces]%
    \StrCut[\nbspaces]{#1}{ }\firstname\lastname
    \StrSplit\firstname1\firstnameinitial\fisrtnameremain
    \IfSubStr\fisrtnameremain{ }
        {\StrBehind\fisrtnameremain{ }[\fisrtnameremain]}
        {\let\fisrtnameremain\empty}%
    \edef\firstnameinitial{\firstnameinitial.\ifx\fisrtnameremain\empty\else\space\fi\fisrtnameremain}%
    \edef\stringA{#1}%
    \edef\stringB{\firstnameinitial\space\lastname}%
    \edef\stringC{\lastname, \firstname}%
    \edef\stringD{\lastname, \firstnameinitial}%
    \StrSubstitute{#2}\stringA{\expandafter\textbf\expandafter{\stringA}}[\finaltext]%
    \StrSubstitute\finaltext\stringB{\expandafter\textbf\expandafter{\stringB}}[\finaltext]%
    \StrSubstitute\finaltext\stringC{\expandafter\textbf\expandafter{\stringC}}[\finaltext]%
    \StrSubstitute\finaltext\stringD{\expandafter\textbf\expandafter{\stringD}}[\finaltext]%
    \StrSubstitute\finaltext\lastname{\expandafter\textbf\expandafter{\lastname}}[\finaltext]%
    \finaltext
    \endgroup
}

\begin{document}

    \makebold{John Doe}{His name was John Doe.}

    \makebold{John Doe}{His name was J. Doe.}

    \makebold{John Doe}{His name was Doe, John.}

    \makebold{John Doe}{Again, Doe was his last name.}
    \bigskip

    \makebold{John T. Doe}{His name was John T. Doe.}

    \makebold{John T. Doe}{His name was J. T. Doe.}

    \makebold{John T. Doe}{His name was Doe, John T. again.}

    \makebold{John T. Doe}{His name was Doe, J. T. again.}

    \makebold{John T. Doe}{Again, Doe was his last name.}

\end{document}

The output is as expected. It's definitely less verbose, though regex is more universal.