This is some code.
\documentclass{article}
\usepackage｛xstring}
\begin{document}
\def\xloop#1{%
\ifx\relax#1
\else
**\IFInteger{#1}{first integer is here.}** % need to be modified
\expandafter\xloop%
\fi}
\def\test#1{\xloop#1\relax}
\test{abc3de5f｝
\end{document}
On the original question of how to terminate your loop:
In case of your \xloop you have a terminating \relax. Provided that your input never contains a \relax token (which is already a prerequisite of your own loop) you can simply terminate your loop by swallowing everything up to that \relax, you just have to pay attention, since you're swallowing a \fi in the process you'll need to reinsert one. The following does that (but is then equivalent to the, imho, more clean expl3 based solution of my other answer):
\documentclass{article}
\usepackage{xstring}
\begin{document}
\def\xloop#1{%
\ifx\relax#1%
\else
\IfInteger{#1}{\xloopDone}{}%
\expandafter\xloop
\fi}
\def\xloopDone#1\relax{\fi first integer was here}
\def\test#1{\xloop#1\relax}
\test{abc3de5f}
\end{document}
Yields first integer was here and nothing else.
This answer relies on the fact that the digits 0-9 fall sequentially on the ASCII chart between slots 48 and 57, inclusive. The slot just prior to 0 contains / and the slot following 9 contains :.
We make use of TeX syntax that a grave accent (backtick) prior to a symbol produces the slot number (ASCII value) for the symbol. So I set up two \ifnum tests...the first looks to see if the slot of the current symbol is beyond the slot of /, and the second test checks if the slot of the current symbol is prior to the slot of :.
If both these conditions are true, the slot of the symbol falls in the range of the integer slots. Otherwise, it falls outside that range.
\documentclass{article}
\def\xloop#1{%
\ifx\relax#1
\else
%
\ifnum`#1>`/
\ifnum`#1<`:
\def\next{#1 is first integer\\}%
\else
\def\next{#1 is not an integer\\\expandafter\xloop}
\fi
\else
\def\next{#1 is not an integer\\\expandafter\xloop}
\fi
\next%
\fi}
\def\test#1{\xloop#1\relax}
\begin{document}
\test{abc3de5f}
\end{document}
Skillmon suggests a faster alternative which, as he notes, is equally robust. In this case the comparison 1<1\noexpand#1 will prove false, unless #1 is a numerical digit.
\documentclass{article}
\def\xloop#1{%
\ifx\relax#1
\else
%
\ifnum1<1\noexpand#1
\def\next{#1 is first integer\\}%
\else
\def\next{#1 is not an integer\\\expandafter\xloop}
\fi
\next%
\fi}
\def\test#1{\xloop#1\relax}
\begin{document}
\test{abc3de5f}
\end{document}
\ifnum1<1\noexpand#1.
\if-like expression, the grave accent before a character indicates that the character-slot number of that character is to be given (as an integer), rather than the character itself. Since the slot (ASCII code) of a / is 47, the test is to determine if the slot of the given character #1 is greater than 47. The second such test determines if the slot of the given character is less than 58 (the slot of :). If you look at an ASCII chart, the only such character slots for #1 that qualify are the digits 0 through 9.
#1 indicate ? As a pair of accent is used around #1 . oh. I understand, it's for /.
Here is an expandable solution using TeX primitives only:
\def\xloop#1{%
\ifx\relax#1\else
\ifnum 2<1#1 \loopE \else non-integer:#1 \fi
\expandafter \xloop
\fi
}
\def\loopE #1\relax{\fi\fi}
\xloop abc3de0f\relax
\bye
The test if #1 is an integer is done by \ifnum. There are two situations:
\ifnum 2<1x \loopE \else non-integer:x\fi
this test is false because 2<1 is not true. The else-text is procesed.
\ifnum 2<10 \loopE \else non-integer:0\fi
this test is true because 2<10 (or 11, or 12, etc.) is true. The macro \loopE does final work: ignores the next text to \relax and closes two opened \if branches.
Why just doing only digits?
Here \defineclass provides a form of \str_case:nnF, so the current item in the loop can be tested for being in the class or not.
With \loopbreak you can stop the loop. See the last examples.
\documentclass{article}
\ExplSyntaxOn
\NewExpandableDocumentCommand{\stringloop}{mmmmm}
{% #1 = string to test
% #2 = class name (CN)
% #3 = what to do if the item is in the class (YES)
% #4 = what to do if the item is not in the class (NO}
% #5 = what to do at the end if not stopped (END)
\xcn_stringloop:ennnn {\tl_to_str:n {#1}} {#2} {#3} {#4} {#5}
}
\cs_new:Nn \xcn_stringloop:nnnnn
{
\__xcn_stringloop:w {#2} {#3} {#4} {#5} #1 \q_nil
}
\cs_generate_variant:Nn \xcn_stringloop:nnnnn {e}
\cs_new:Npn \__xcn_stringloop:w #1 #2 #3 #4 #5
{% #1 = CN, #2 = YES, #3 = NO, #4 = END, #5 = current item in string
\quark_if_nil:nTF {#5}
{
#4
}
{
\__xcn_stringloop_item:nnnnn {#1} {#5} {#2} {#3} {#4}
}
}
\cs_new:Nn \__xcn_stringloop_item:nnnnn
{% #1 = CN, #2 = current item in string, #3 = YES, #4 = NO, #5 = END
\use:c {__xcn_stringloop_class_#1:nTF} {#2} {#3} {#4}
\__xcn_stringloop:w {#1} {#3} {#4} {#5}
}
\NewDocumentCommand{\defineclass}{mm}
{% #1 = class name, #2 = members
\cs_new:ce {__xcn_stringloop_class_#1:nTF}
{
\exp_not:N \str_case:nnF {##1}
{ \tl_map_function:nN {#2} \__xcn_stringloop_make_case:n }
{ ##3 }
}
}
\cs_new:Nn \__xcn_stringloop_make_case:n { {#1}{##2} }
\NewExpandableDocumentCommand{\loopbreak}{m}
{
#1 \__xcn_stringloop_break:w
}
\cs_new:Npn \__xcn_stringloop_break:w #1 \q_nil {}
\NewExpandableDocumentCommand{\checkforTF}{mmmm}
{
\xcn_stringloop:ennnn {\tl_to_str:n {#1}} {#2} {\loopbreak{#3}} {} {#4}
}
\ExplSyntaxOff
\defineclass{digits}{0123456789}
\defineclass{unknowns}{xyz}
\begin{document}
\raggedright
\stringloop{abc3def5g}{digits}{DIGIT }{NONDIGIT }{END}
\bigskip
\stringloop{abc3def5g}{digits}{\loopbreak{DIGIT, STOP}}{NONDIGIT }{END}
\bigskip
\checkforTF{abc3def5g}{digits}{There's a digit}{There's no digit}
\bigskip
\checkforTF{abcdefg}{digits}{There's a digit}{There's no digit}
\bigskip
\edef\test{\checkforTF{abc3def5g}{digits}{There's a digit}{There's no digit}}
\texttt{\meaning\test}
\bigskip
\stringloop{abxcy}{unknowns}{(unknown)}{(parameter)}{}
\bigskip
\stringloop{abxc}{unknowns}{\loopbreak{(unknown)}}{(parameter)}{end}
\bigskip
\stringloop{abc}{unknowns}{\loopbreak{(unknown)}}{(parameter)}{end}
\end{document}
If this is only about checking whether there was a number contained, the following is a rather fast approach for that:
\documentclass{article}
\ExplSyntaxOn
\NewExpandableDocumentCommand\ifcontainsnum{mmm}
{ \xcn_if_contains_num:nTF {#1} {#2} {#3} }
\prg_new_conditional:Npnn \xcn_if_contains_num:n #1 { TF, T, F, p }
{
\str_map_function:nN {#1} \__xcn_if_contains_num:n
\prg_return_false:
}
\cs_new:Npn \__xcn_if_contains_num:n #1
{ \xcn_if_digit:nT {#1} { \str_map_break:n { \use_i:nn \prg_return_true: } } }
\prg_new_conditional:Npnn \xcn_if_digit:n #1 { TF, T, F, p }
{
\if_int_compare:w 1 < 1\exp_not:N#1 \exp_stop_f:
\prg_return_true:
\else:
\prg_return_false:
\fi:
}
\ExplSyntaxOff
\begin{document}
\ifcontainsnum{abc}{yes}{no}
\ifcontainsnum{abc1}{yes}{no}
\ifcontainsnum{ab{c1}}{yes}{no}
\end{document}
This yields no, yes, and yes.