Question

I need to write a regular expression for an ‘option symbol’ for my company so we can validate these symbols across our site.

An option symbol is composed of two parts:

Part1         Part2
_ _ _ _ _ _ | _ _ _ _ _ _ _ _ _ 

I can write a regular expression for Part2 as it is fairly simple.

However, Part1 (the first 6 character positions) can be a little complicated.

It boils down to:

• Part1 must be {6} characters in length total.
• There must be between {1,4} alpha characters in the first positions.
• After that, there can optionally be {1} numeric character.
• Finally, the rest of the characters remaining must be spaces, so Part1 totals to 6 characters.

The problem I’m having is that the number of spaces is variable based on the number of characters before it. This makes me think it isn’t easily representable by a regular language.

How can I avoid brute-forcing it like so:

([A-Za-z]{1}[0-9]{1}[ ]{4}|
[A-Za-z]{2}[0-9]{1}[ ]{3}|
[A-Za-z]{3}[0-9]{1}[ ]{2}|
[A-Za-z]{4}[0-9]{1}[ ]{1}|
[A-Za-z]{1}[ ]{5}|
[A-Za-z]{2}[ ]{4}|
[A-Za-z]{3}[ ]{3}|
[A-Za-z]{4}[ ]{2}|
[A-Za-z]{5}[ ]{1})

Here are some example option symbols (remember, ignore everything beyond the first 6 characters):

F     123456P12345678
CMG   123456P12345678
AAPL  123456P12345678
GOOG1 123456C12345678
F5    123456C12345678

Answer 1

You can use a lookbehind assertion:

^[A-Za-z]{1,4}\d? +\b(?<=^.{6})

Explanation:

^: Match the start of the line or string (depending on whether you set the option RegexOptions.Multiline or not).

[A-Za-z]{1,4}\d? +: Match 1 to 4 alpha characters and an optional digit, followed by at least one space

\b: Assert that we are now at a word boundary (i.e., that the next character is alphanumeric)

(?<=^.{6}): Assert that the above match is exactly six characters long.