8

Trying to create a regex pattern for email address check. That will allow a dot (.) but not if there are more than one next to each other.

Should match: test.test@test.com

Should not match: test..test@test.com

Now I know there are thousands of examples on internet for e-mail matching, so please don't post me links with complete solutions, I'm trying to learn here.

Actually the part that interests me the most is just the local part: test.test that should match and test..test that should not match. Thanks for helping out.

4

5 回答 5

6

You may allow any number of [^\.] (any character except a dot) and [^\.])\.[^\.] (a dot enclosed by two non-dots) by using a disjunction (the pipe symbol |) between them and putting the whole thing with * (any number of those) between ^ and $ so that the entire string consists of those. Here's the code:

$s1 = "test.test@test.com";
$s2 = "test..test@test.com";
$pattern = '/^([^\.]|([^\.])\.[^\.])*$/';
echo "$s1: ", preg_match($pattern, $s1),"<p>","$s2: ", preg_match($pattern, $s2);

Yields:

test.test@test.com: 1
test..test@test.com: 0
于 2012-05-20T13:56:58.360 回答
5

This seams more logical to me:

/[^.]([\.])[^.]/

And it's simple. The look-ahead & look-behinds are indeed useful because they don't capture values. But in this case the capture group is only around the middle dot.

于 2012-05-20T13:50:35.217 回答
1
strpos($input,'..') === false

strpos function is more simple, if `$input' has not '..' your test is success.

于 2012-05-20T13:53:48.947 回答
0
^([^.]+\.?)+@$

That should do for the what comes before the @, I'll leave the rest for you. Note that you should optimise it more to avoid other strange character setups, but this seems sufficient in answering what interests you

Don't forget the ^ and $ like I first did :(

Also forgot to slash the . - silly me

于 2012-05-20T13:48:37.303 回答
0

To answer the question in the title, I'd update the RegExp by Junuxx and allow dots in the beginning and end of the string:

'/^\.?([^\.]|([^\.]\.))*$/'

which is optional . in the beginning followed by any number of non-. or [non-. followed by .].

于 2018-02-26T16:54:00.090 回答