Ok getting a little clearer. Still bamboozled by the following



.Replacement.Text = "\3"
.Execute Replace:=wdReplaceAll

How does the replace work? what is \3 why does Greg use \2 and is this where the original tags are deleted?


.Text = "(\<" & arrTerms(i) & ")(\>)(*)\1/\2"

What is (*)\1/\2 doing it looks like jibberish?

A wildcard Find can use brackets () to delineate certain segments of the expression for re-use. The Find expression:
.Text = "(\<" & arrTerms(i) & ")(\>)(*)\1/\2"
has 3 such delineated segments and it is the 3rd of these that gets re-used for the replacement. In Greg's code, there were just 2 such delineations and the 2nd gets reused.
Not jibberish; just clever.

The:
(\<" & arrTerms(i) & ")(\>)
defines 2 segments of the Find expression for re-use: the string comprising the '<' and array entry for the 1st segment; and the '>' for the 2nd segment. A '\' is required before certain Find characters, including < and >.

The:
(*)
defines as a 3rd segment whatever follows the 2nd segment that precedes a repeat of the 1st segment, as indicated by the \1.

The:
\1/\2
says to repeat the 1st segment, insert the /, then repeat the 2nd segment.

FWIW:
.Text = "(\<" & arrTerms(i) & ")(\>)(*)\1/\2"
is the same as:
.Text = "\<" & arrTerms(i) & "\>(*)\<" & arrTerms(i) & "/\>"
for which the replacement would be \1.

__________________
Cheers,
Paul Edstein
[Fmr MS MVP - Word]

Far out it's like a puzzle. Thanks heaps for the break down. I feel like I'm getting close to getting a grasp on it but I still need to re-read the explanation another couple of times sleep on it and muck around with both your codes. This example just showed me how many ways there are to do the same thing, I hadn't realized. I thought it would be alot more constrained but it's more like pick your own adventure. Thanks for the help.