If alph to numberic in regex

Use pattern:
^[A-Z]{5}\d{5}$

This should do it:

[A-Z]{5}[0-9]{5}

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In code I think it goes something like this...

If Regex.Match(str, "^[A-Z]{5}\\d{5}$").Success Then
     strErr = "Error!"
End If

Oops, that would be:

If Regex.Match(str, "^[A-Z]{5}\\d{5}$").Success Then
     strMsg = "Great!"
End If

My solution was missing the leading ^ to anchor the search at the start of the string.  @TerryAtOpus got it right.

You only need the "$" at the end of the pattern if you do not want any characters to be allowed after the five digits.  (That was not in your problem description.)

^[A-Z]{5}[0-9]{5}

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I can use this, correct? I am not sure what this does  regexoptions.complied?
private reg_str as new regex("^[A-Z]{5}[0-9]{5}$", regexoptions.complied)

if reg_str.ismatch(mystring) then
...
end if

^[A-Z]{5}\\d{5}$")  I don't understand this
^[A-Z]{5}[0-9]{5}  , I understand this

sorry, I am new to regex

\d means a digit. The \ is backslashed because of the way .NET handles strings.

(so \\d and [0-9] are the same thing)

As for the question about regexoptions, you might like to use something like this if you want to ignore case:

Dim options As RegexOptions = RegexOptions.IgnoreCase
If Regex.Match(str, "^[A-Z]{5}\\d{5}$").Success Then
     strMsg = "Great!"
End If

Sorry, should be:

Dim options As RegexOptions = RegexOptions.IgnoreCase
If Regex.Match(str, "^[A-Z]{5}\\d{5}$", RegexOptions).Success Then
     strMsg = "Great!"
End If