Matching both \d and \. in a regular expression - vba

Question

I'm trying to match any combination of float numbers with a regular expression in VBA inside of Microsoft Word.
Example:
<text="">
<text="1">
<text="2">
<text="3">
<text="3.5">
<text="4">
<text="44">
I've tried the following patterns:
regex.Pattern = "<text=" & Chr(34) & "\d*\.*\d*" & Chr(34) & ">"
regex.Pattern = "<text=" & Chr(34) & "[\d\.]*" & Chr(34) & ">"
What's interesting is neither of these patterns will match the 44 either.

Answer 1

from the regex tester site,
Positive Integers --- ^\d+$
Negative Integers --- ^-\d+$
Integer --- ^-{0,1}\d+$
Positive Number --- ^\d*\.{0,1}\d+$
Negative Number --- ^-\d*\.{0,1}\d+$
Positive Number or Negative Number - ^-{0,1}\d*\.{0,1}\d+$
to test for the empty result being valid, and assuming all positive numbers, you would end up with
regex.Pattern = "<text=" & Chr(34) & "\d*\.{0,1}\d*" & Chr(34) & ">"

Answer 2

Let's break it down. A floating point number has two parts, integral and decimal. There are some choices on the format:
Here is pattern for a pretty common notation:
regex.Pattern = "<text=" & Chr(34) & "(-?\d+(\.\d+)?)?" & Chr(34) & ">"
Some choices:
Do you want to support omitting the integral part? (e.g. '.001')
Use * instead of + in the first number match: (-?\d*(\.\d+)?)?
Do you want to support leading zeros? (e.g. 000.)
This is gonna be more tricky :)
Do you want to support decimal point without decimal part? (e.g. '12.')
Then add ? after the decimal point match: (-?\d*(\.?\d+)?)?
Is the decimal part mandatory?
Remove the very last ?: (-?\d*(\.?\d+))?