In python, how do I regex match the "." char in, say, a string representing an address?

Question

I am getting an address string from the client (first I check that it is a string under a certain length), and I have this so far: ^[a-zA-Z0-9 ]+$ but it will fail if the address has a "." char in it.

I want a single regex that also matches if a "." char shows up: the address can only contain spaces, numbers and letters.

Obviously, one can do if "." in address_string, but I am trying to do my check in one regex match.

How do I also escape/search for a "." char in a string?

Add `\.` to your regex: `^[a-zA-Z0-9\. ]+$` Of course this will allow things like `a..........b`, which maybe you don't really want. — John Gordon, Oct 17 '19 at 15:57

score -1 · Accepted Answer · answered Oct 17 '19 at 15:57

-1

Just put in an escape character:

^[a-zA-Z0-9 \.]+$

answered Oct 17 '19 at 15:57

RightmireM

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You don't need to escape `.`s inside of character classes. – 0x5453 Oct 17 '19 at 15:59

In python, how do I regex match the "." char in, say, a string representing an address?

1 Answers1