This question already has answers here:
Regular expression to enforce complex passwords, matching 3 out of 4 rules
(4 answers)
Closed 7 years ago.
A regular expression for password
It should have one number
One character compulsorily
Not case sensitive.
Special character is not mandatory.
xyz.match(/^(?:[0-9]+[a-z]|[a-z]+[0-9])[a-z0-9]*$/i)
In this if we give special character it is not accepting
asd12 //working
Asd123 //working
Asd123# //not working
You need 2 look-aheads anchored at the beginning instead of a non-capturing group:
/^(?=[^0-9]*[0-9])(?=[^a-z]*[a-z]).*$/i
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
The .* subpattern will allow any characters, but the look-aheads will demand at least 1 digit and 1 Latin letter.
Note that non-capturing groups still consume characters, and cannot be used to check for presence or absence of specific subpatterns in a string. Only zero-width assertions provide that mechanism.
See this regex demo
Related
This question already has answers here:
How to get the count of only special character in a string using Regex?
(6 answers)
Closed 2 years ago.
I need to form the RegEx to produce the output only if more than two occurrences of special characters exists in the given string.
1) abcd##qwer - Match
2) abcd#dsfsdg#fffj-Match
3) abcd#qwetg- No Match
4) acwexyz - No Math
5) abcd#ds#$%fsdg#fffj-Match
Can anyone help me on this?
Note: I need to use this regular expression in one of the existing tool not in any programming language.
UPDATE after OP edit
The edited OP introduces a small amount of additional complexity that necessitates a different pattern entirely. The keys here are that (a) there is now a significantly limited set of "special characters" and (b) that these characters must appear at least twice (c) in any position in the string.
To implement this, you would use something like:
(?:.*?[##$%].*?){2,}
Asserts a non-capturing group,
Which contains any number of characters, followed by
Any character in the set ##$%
Followed by any number of characters
Ensures this pattern happens twice in a given string.
Original answer
By "special characters", I assume you mean anything outside standard alphanumeric characters. You can use the pattern below in most flavors of Regex:
([^A-Za-z0-9])\1
This (a) creates a set of all characters not including alphanumeric characters and matches a character against it, then (b) checks to see if the same character appears adjacent.
Regex101
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Reference - What does this regex mean?
(1 answer)
Closed 5 years ago.
i wrote this regex for tokenize a text: "\b\w+\b"
but someone suggets me to convert it into \b[^\W\d_]+\b
can anyone explaing to me why this second way (using negation) is better?
thanks
The first one matches all letters, numbers and the underscore. Depending on the regex engine, this may include unicode letters and numbers. (the word boundaries are superfluous in this case btw.)
The second regex matches only letters (excluding non-word-charcters, digits and the underscore). Due to the word boundary, it will only match them, if they are surrounded by non-word-characters or start/end of th string.
If your regex engine supports this, you might want to use [[:alpha:]] or \p{L} (or [A-Za-z] in case of non-unicode) instead to make your intent clearer.
This question already has answers here:
RegEx for allowing alphanumeric at the starting and hyphen thereafter
(4 answers)
Closed 5 years ago.
I want to build a regular expression which only matches [A-Za-z0-9\-] with an additional rule that hyphens (-) are not allowed to appear at the start and at the end.
For example:
my-site is matched.
m is matched.
mysite- is not matched.
-mysite is not matched.
Currently, I've come up with ^[A-Za-z0-9][A-Za-z0-9\-]*[A-Za-z0-9]+$.
But this doesn't match m.
How can I change my regular expression so that it fits my needs?
Use look arounds:
^(?!-)[A-Za-z0-9-]*(?<!-)$
The reason this works is that look arounds don't consume input, so the look ahead and the look behind can both assert on the same character.
Note that you don't need to escape the dash within the character class if it's the first or last character.
This question already has an answer here:
Reference - What does this regex mean?
(1 answer)
Closed 6 years ago.
I came across this regex used for password validation:
(?=.*[a-z])(?=.*[A-Z])(?=.*[\d])(?=.*[^a-zA-Z\d])(?=\S+$).{8,}
There are only two things that are unclear to me about this regex:
what are .* used for and why this regex doesn't work without them?
what is the difference/benefit or using [\d] instead of \d, because the regex works just fine in both cases
.* matches any sequence of characters; . matches any character (other than newline, which is not relevant here) and * matches zero or more of the preceding pattern. This is used in the lookaheads to search for matches anywhere in the password. If you didn't have it,then it would require that you have those types of characters in a specific order: a lowercase letter followed by an uppercase letter followed by a digit. With .*, it means the password must contain at least one of each of them, but they can be anywhere in the password.
There's no difference between \d and [\d]. Whoever write this might just use the brackets out of habit, or perhaps to make it easier to modify it to put other characters into the character class.
This question already has an answer here:
Reference - What does this regex mean?
(1 answer)
Closed 8 years ago.
I have the following regex :
.*(?:(?:(?<!a)cc|string).*number).*
And I am trying to understand what the ? in the beginning of the string between brackets mean. I know the a? means that the previous character 'a' can be repeated zero or one time. But what does it mean when it appears in the beginning of a string ?
The answer requires a little history lesson. When Larry Wall wanted to add new features to regexes in Perl, he couldn't just change the meaning of existing metacharacters, or assign special meanings to characters that didn't have them. That would have broken a lot of regexes that had been working. Instead, he had to look for character sequences that would never appear in a regex.
There was only the one kind of group originally: what we now call capturing groups. The opening parenthesis was a metacharacter, so it would make no sense to follow it with a quantifier. You could match a literal open-paren zero or one time with \(?, or you could match (and capture) a literal question mark with (\?), but if you tried to use (? in regex it would throw an exception.
Larry changed the rule so (? could appear in a regex, but it must form the beginning of a special-group construct, which requires at least one more character. So, to answer your question, the string doesn't start with ?. The sequence (?: forms a single token, representing the beginning of a non-capturing group. We also have (?= and (?! for positive and negative lookaheads, (?<= and (?<! for lookbehinds, and so on.
(?:) is a non-capturing group. It do a matching operation only. It won't capture anything.
(?<!) is a Negative lookbehind.