I'm trying to create a regex for alias validation:
And I'm allowing letters, numbers and 1 dot.
I have done the following:
/^[a-z0-9\\.]+$/i
However it allows more then 1 dot?
This should do it:
/^(?:\.[a-z0-9]+|[a-z0-9]+(?:\.[a-z0-9]*)?)$/i
This allows the string to either:
start with one dot that is followed by at least one alphanumeric character, or
start with one or more alphanumeric character that may be followed by one dot and zero or more alphanumeric characters.
I think it is not a good idea to allow a dot as a first or last character, in that case:
/^[a-z0-9]+\.?[a-z0-9]+$/i
try this:
^(?:[a-z0-9]+\.?[a-z0-9]*|[a-z0-9]*\.?[a-z0-9]+)$
places the dot in the center, then allows it to be surrounded on either side.
Related
I need a regex that would test if a word is composed of letters (alpha characters), white spaces, and periods (.). I need this to use for validating names that is entered in my database.
This is what I currently use:
preg_match('/^[\pL\s]+$/u',$foo)
It works fine for checking alpha characters and whitespaces, but rejects names with periods as well. I hope you guys can help as I have no idea how to use regex.
Add a dot to the character class so that it would match a literal dot also.
preg_match('/^[\p{L}.\s]+$/u',$foo)
OR
preg_match('/^[\pL.\s]+$/u',$foo)
Explanation:
^ Asserts that we are at the start.
[\pL.\s]+ Matches any character in the list one or more times. \pL matches any Kind of letter from any language.
$ Asserts that we are at the end.
The following regex should satisfy your condition:
preg_match('/^[a-zA-Z\s.]+$/',$foo)
In this link, you will find all the information you need to figure regex out with PHP. PHP Regex Cheat Sheet
Basically, if you want to add the period you add . :
preg_match('/^[\pL\s\.]+$/u',$foo)
Enjoy! :)
I need a small help with regex for the following
Alphanumeric with only lower case alphabets allowed
Starts with number or alphabet
Allows period (.)
Doesn't allow consecutive periods No ..
Doesn't allow any other special characters
Thanks,
-GM
^(?![^.]*\.\.)[a-z0-9][a-z0-9.]*$
The negative lookahead at the beginning covers your 4th requirement, everything else should be pretty straightforward. ^ and $ are beginning and end of string anchors, the character classes enforce the requirement that only lowercase letters, numbers, and . are allowed.
To add the length constraint (between 6 and 16 characters) just change the * to {5,15}. * means "repeat the previous element zero or more times", {n,m} means "repeat the previous element between n and m times (inclusive)". The reason {5,15} is used instead of {6,16} is that one character is already consumed by the first character class. Here is the end result:
^(?![^.]*\.\.)[a-z0-9][a-z0-9.]{5,15}$
Here's some assistance without giving away the answer, as you'll learn the most.
To match from a certain combination of characters, e.g. alphanumeric, use character classes, e.g. [a-z0-9]. Note that this expression matches exactly one character. You must use quantifiers to match more than one, e.g. +.
To "start" or "end" with something, you must use anchors, ^ and $, before the first or after the last character, respectively. (Watch out, though. In a character class, the ^ inverts the character class.)
In regex, . has a special meaning as a wildcard (matching any character besides newline characters). Therefore you have to escape them, \., to select the literal dot. Another way to escape the dot is to put it in a character class: [.].
Non-consecutiveness is trickier. You will need to look up more information about negative lookahead assertions (or lookaround assertions in general).
All the bolded words are terms you can Google to learn.
I'd say something along those lines: /^[a-z0-9]+(\.[a-z0-9]+)*\.?$/ (suppose that the line can end with a period)
Use this if the string may not end with a period:
/^[a-z0-9]+(\.[a-z0-9]+)*$/
or this if it may:
/^[a-z0-9]+(\.[a-z0-9]+)*\.?$/
This should be the best
^([a-z0-9]+\.?)+$
I tried to create regular expression with specification below
any alphabetic character (at least one)
any numeric character (at least one)
no spaces
accept all special characters (except ",;&|')
^(?=.*[0-9])(?=.*[a-z])(?!.*\s)((?!.*[",;&|'])|(?=(.*\W){1,}))(?!.*[",;&|'])$
This is the one I tried.
What I can do with this?
Question is still vague in nature, please provide some examples of accepted strings.
Just to get you started you can use:
character class in a negative lookahead
Don't forget start & end anchors:
Regex:
/^(?=.*?\d)(?=.*?[a-z])(?!.*?[ ",;&|']).+$/i
This regex will match 1 or more characters that are not one of ",;&|' and atleast one digit and a-z alpgabet is required.
Live Demo: http://www.rubular.com/r/nxdi79ZcRx
In PHP use it like this:
'/^(?=.*?\d)(?=.*?[a-z])(?!.*?[ ",;&|\']).+$/i'
Begins with alphanumeric ^[a-z0-9]
Then followed by this optional dot \.?
If there is a dot, then it MUST be followed by 2 to 4 alphabets [a-z]{2,4}
It must be ends with an alphabet [a-z]$
It has to be a dot and only two dots max.
it's like domain names:
yahoo.co.uk or yahoo.com, but you cannot do this yahoo.co.u or this yahoo.co., yes something like that.
You can group the optional dot with the 2-4 characters that must follow it: (\.[a-z]{2,4}). That said, you will have either none, or up to two of these groups of dot + alphabetic characters (\.[a-z]{2,4}){0,2}.
The must end with [a-z] part, you can check with a positive lookbehind (?<=[a-z]) giving this as the full regex:
^[a-z0-9]+(\.[a-z]{2,4}){0,2}(?<=[a-z])$
This will work in Perl and PHP regexes (PCRE), but not in JavaScript, because it does not support lookbehind. In this specific case, you can work around that limitation.
If there is at least one dot, there's already a guarantee that it will end in [a-z], because that test is in the group that the dot is a part of. If there is no dot, you need to force a [a-z] at the end. To do this you can turn the one-or-more quantifier (+) into a zero-or-more (*) and force the end to be an [a-z] when there are no "dot groups". When there are dot groups, you can keep the same pattern, but now with at least one mandatory dot.
^([a-z0-9]*[a-z]|[a-z0-9](+\.[a-z]{2,4}){1,2})$
This checks for a string that begins with [a-z][0-9] and then contains one or two dots followed by 2/4 alphabets. It works (in Python, at least) for the examples you provided (true for yahoo.co.uk and yahoo.com, false for yahoo.co.u and yahoo.co.)
^[a-z0-9]+(\.[a-z]{2,4}){1,2}$
Edit - upon re-reading, I think you may want this instead:
^[a-z0-9]*([a-z0-9](\.[a-z]{2,4}){1,2}$|[a-z]$)
This will match strings (in addition to the above) that do not include dots but end with a letter, such as yahoo, but not yahoo2.
Try this:
^[a-z0-9](\.[a-z]{2,4}|.*[a-z]$)
^[a-z0-9](?=[^.]*\.[^.]+$|[^.]*\.[^.]\.[^.]+$)(\.(?=[a-z][a-z]){1,2}).*[a-z]$
I'm still kinda new to using Regular Expressions, so here's my plight. I have some rules for acceptable usernames and I'm trying to make an expression for them.
Here they are:
1-15 Characters
a-z, A-Z, 0-9, and spaces are acceptable
Must begin with a-z or A-Z
Cannot end in a space
Cannot contain two spaces in a row
This is as far as I've gotten with it.
/^[a-zA-Z]{1}([a-zA-Z0-9]|\s(?!\s)){0,14}[^\s]$/
It works, for the most part, but doesn't match a single character such as "a".
Can anyone help me out here? I'm using PCRE in PHP if that makes any difference.
Try this:
/^(?=.{1,15}$)[a-zA-Z][a-zA-Z0-9]*(?: [a-zA-Z0-9]+)*$/
The look-ahead assertion (?=.{1,15}$) checks the length and the rest checks the structure:
[a-zA-Z] ensures that the first character is an alphabetic character;
[a-zA-Z0-9]* allows any number of following alphanumeric characters;
(?: [a-zA-Z0-9]+)* allows any number of sequences of a single space (not \s that allows any whitespace character) that must be followed by at least one alphanumeric character (see PCRE subpatterns for the syntax of (?:…)).
You could also remove the look-ahead assertion and check the length with strlen.
make everything after your first character optional
^[a-zA-Z]?([a-zA-Z0-9]|\s(?!\s)){0,14}[^\s]$
The main problem of your regexp is that it needs at least two characters two have a match :
one for the [a-zA-Z]{1} part
one for the [^\s] part
Beside this problem, I see some parts of your regexp that could be improved :
The [^\s] class will match any character, except spaces : a dot or semi-colon will be accepted, try to use the [a-zA-Z0-9] class here to ensure the character is a correct one.
You can delete the {1} part at the beginning, as the regexp will match exactly one character by default