Regular expression to match whole words surrounded by spaces
"\b\.?\s+" & yourword & "\b\s+"
\b\ stands for word boundary
\s+ stands for one or more spaces
Here is an improved regular expression without word boundaries that also looks for optional punctuation marks:
"([\.!?]?\s+" & yourword & "\s+)"
The optional group is enclosed in brackets, without them the group becomes mandatory.
Regular expression to match words with flexions
"((?i)\s+" & yourword & "[en|e|er|s|es\b]{0,}[\.\,!?]?\s+)"
?i sets ignore case
en|e|er|s|es\b sets word flexions before word boundary
{0,} gets all forms
[\.\,!?] optional punctuation after word boundary
\s+ mandatory space
Please let me know how it works in your applications.
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