Regular Expression - Match only 7 chars?

Question

I'm trying to match a SEDOL (exactly 7 chars: 6 alpha-numeric chars followed by 1 numeric char)

My regex

([A-Z0-9]{6})[0-9]{1}

matches correctly but strings greater than 7 chars that begin with a valid match also match (if you see what I mean :)). For example:

B3KMJP4

matches correctly but so does:

B3KMJP4x

which shouldn't match.

Can anyone show me how to avoid this?

Answer 1

Dollar sign at the end of the regex (called an anchor) signifies end of string:

^([A-Z0-9]{6})\d$

I also added "^" at the start which signifies start of string and prevents matching xB3KMJP4 I also simplified the original regex.

By the way, as per Wikipedia, for the first character, vowels are not used. I'm not quite sure if that's a rule or a convention.

Answer 2

You need to use both start and end anchors like this:

^([A-Z 0-9]{6})[0-9]{1}$

This will match a string which has 6 alphanumeric+space char followed by one digit. It does not match if such a string is found as a suffix or prefix of a bigger string.

Also you you can get rid of {1} because [0-9] matches a single digit by itself.

Also \d represents a single digit. So you can shorten your regex as follows:

^([A-Z \d]{6})\d$

Answer 3

 ^([A-Z\d]{6})\d$

• Use ^ for start of string
• $ for end of string
• Remove extra space,just noticed that one
• Swapped out 0-9 with \d
• Removed {1} since this is redundant

Answer 4

You're forgetting that regex matches anywhere in the string. To fix it, try this.

^([A-Z 0-9]{6})[0-9]{1}$

The ^ means to match the beginning of the string, and the $ means to match the end of the string.