EDIT: Thank you all for your inputs. What ever you answered was right.But I thought I didnt explain it clear enough.
I want to check the input value while typing itself.If user is entering any other character that is not in the list the entered character should be rolled back.
(I am not concerning to check once the entire input is entered).
I want to validate a date input field which should contain only characters 0-9[digits], -(hyphen) , .(dot), and /(forward slash).Date may be like 22/02/1999 or 22.02.1999 or 22-02-1999.No validation need to be done on either occurrence or position. A plain validation is enough to check whether it has any other character than the above listed chars.
[I am not good at regular expressions.]
Here is what I thought should work but not.
var reg = new RegExp('[0-9]./-');
Here is jsfiddle.
Your expression only tests whether anywhere in the string, a digit is followed by any character (. is a meta character) and /-. For example, 5x/- or 42%/-foobar would match.
Instead, you want to put all the characters into the character class and test whether every single character in the string is one of them:
var reg = /^[0-9.\/-]+$/
^ matches the start of the string
[...] matches if the character is contained in the group (i.e. any digit, ., / or -).
The / has to be escaped because it also denotes the end of a regex literal.
- between two characters describes a range of characters (between them, e.g. 0-9 or a-z). If - is at the beginning or end it has no special meaning though and is literally interpreted as hyphen.
+ is a quantifier and means "one or more if the preceding pattern". This allows us (together with the anchors) to test whether every character of the string is in the character class.
$ matches the end of the string
Alternatively, you can check whether there is any character that is not one of the allowed ones:
var reg = /[^0-9.\/-]/;
The ^ at the beginning of the character class negates it. Here we don't have to test every character of the string, because the existence of only character is different already invalidates the string.
You can use it like so:
if (reg.test(str)) { // !reg.test(str) for the first expression
// str contains an invalid character
}
Try this:
([0-9]{2}[/\-.]){2}[0-9]{4}
If you are not concerned about the validity of the date, you can easily use the regex:
^[0-9]{1,2}[./-][0-9]{1,2}[./-][0-9]{4}$
The character class [./-] allows any one of the characters within the square brackets and the quantifiers allow for either 1 or 2 digit months and dates, while only 4 digit years.
You can also group the first few groups like so:
^([0-9]{1,2}[./-]){2}[0-9]{4}$
Updated your fiddle with the first regex.
Related
I am new to regex, i have this use case:
Allow characters, numbers.
Zero or one question mark allowed. (? - valid, consecutive question marks are not allowed (??)).
test-valid
?test - valid
??test- invalid
?test?test - valid
???test-invalid
test??test -invalid
Exlcude $ sign.
[a-zA-Z0-9?] - seems this doesn't work
Thanks.
Try the following regular expression: ^(?!.*\?\?)[a-zA-Z0-9?]+$
first we're using Negetive lookahead - which allows us to exclude any character which is followed by double question marks (Negetive lookahaed does not consume characters)
Since question mark has special meaning in regular expressions (Quantifier — Matches between zero and one times), each question mark is escaped using backslash.
The plus sign at the end is a Quantifier — Matches between one and unlimited times, as many times as possible
You can test it here
Your description can be broken down into the regex:
^(?:\??[a-zA-Z0-9])+\??$
You say characters and your description shows letters and numbers only, but it's possible \w (word characters) may be used instead - this includes underscore
It's between ^ and $ meaning the whole field must match (no partial matches, although if you want those you can remove this. The + means there must be at least one match (so empty string won't match). The capturing group ((\??[a-zA-Z0-9])) says I must either see a question mark followed by letters or just letters repeating many times, and the final question mark allows the string to end with a single question mark.
You probably don't want capturing groups here, so we can start that with ?: to prevent capture leading to:
^(?:\??[a-zA-Z0-9])+\??$
Matches
test
?test
?test?test
test?
Doesn't match
??test
???test
test??test
test??
<empty string>
?
I have written a regex that returns true or false depending on whether the text provided is a valid first/last name:
let letters = `a-zA-Z`;
letters += `àáâäãåąčćęèéêëėįìíîïłńòóôöõøùúûüųūÿýżźñçčšž`;
letters += `ÀÁÂÄÃÅĄĆČĖĘÈÉÊËÌÍÎÏĮŁŃÒÓÔÖÕØÙÚÛÜŲŪŸÝŻŹÑßÇŒÆČŠŽ∂ð`;
const re = new RegExp(`^[${letters}][${letters} ,.'’-]+[${letters}.]$`, 'u')
return(name.match(re));
So far, I'm able to ensure it only validates names that actually start with a letter and do not contain numerals or any special characters other than dot, hyphen, or comma. However, it still rejects names like Jo and Xi. I understand it's due to the three separate $-blocks. But the blocks are there to ensure the name doesn't start with a non-letter or end in a non-letter other than dot. How should I modify my expression to accommodate this?
Also, is there any way to shorten this expression without compromising its range? I still need it to cover extended Latin characters.
If the minimum length of the word is 2 chars, you could use a negative lookahead ^(?!.*[ ,'’]$) to assert that the string does not end with the characters that you would not allow and leave out the last [${letters}.]
Regex demo
If the minimum length is 1, you could use another negative lookahead (?![ .,'’]) and add the dot as well so that a single dot is not allowed at the beginning and then use the single character class that contains all allowed characters.
^(?!.*[ ,'’]$)(?![ .,'’])[a-zA-ZàáâäãåąčćęèéêëėįìíîïłńòóôöõøùúûüųūÿýżźñçčšžÀÁÂÄÃÅĄĆČĖĘÈÉÊËÌÍÎÏĮŁŃÒÓÔÖÕØÙÚÛÜŲŪŸÝŻŹÑßÇŒÆČŠŽ∂ð ,.'’-]+$
Regex demo
Say I have a RegEx like the following:
^[a-zA-Z]\w{12}$
And I have the following string:
%7AgTy!5hG^vxWa2#AgW
I would like to "pull" out of that string something that conforms to that regex. In this example we would get the following:
AgTy5hGvxWa2A
Reason: it starts with A because the regex says the first letter must be [a-zA-Z] (so it skips the first 2 characters), and then it pulls successive \ws out until it reaches 12 characters.
Is this sort of thing possible?
Edit: My apologies for being unclear. I'm not looking for a new regular expression that will give the proper output. Rather, I'm looking for a way to use the existing RegEx to extract the proper output. In my program these regular expressions are entered by hand by the user to extract a password from a long base256 hash such that it will conform to these existing password requirement regexes.
Instead of trying to match what you want and reconstructing the string, replace everything you don't want with nothing. This gives the impression that you're extracting what you need, but, in fact, it's doing the opposite; gets rid of everything you don't want to extract. I also dropped $ from the end of your original pattern otherwise it'll never match the string you present in your question.
See regex in use here
^[^a-z]+|\W+
^ Assert position at the start of the line
[^a-z]+ Matches any character that is not in the range a-z one or more times. Since the i flag is specified, this also matches A-Z
\W+ Match any non-word character one or more times
const regex = /^[^a-z]+|\W+/gi
const a = [
`%7AgTy!5hG^vxWa2#AgW`,
`%7AgTy!5hG^vxWa2#`
]
a.forEach(function(s) {
var clean = s.replace(regex, '')
var match = clean.match(/^[a-z]\w{12}/i)
console.log(match)
})
I'm trying to validate a username field in my form via client-side valdiation and I'm having some trouble.
I'm trying to use match them against regexs, which seems to work for my password strength/match. However when I try and change the regular expression to one that is suitable for usernames it doesn't work.
This is the regular expression that works, it checks to see if the length is at least 6 chars long.
var okRegex = new RegExp("(?=.{6,}).*", "g");
This is the other regular expression which does not work:
var okRegex = new RegExp("/^[a-z0-9_-]{3,16}$/");
How do I write a regex that performs username validation? (That it's of a certain length, has only letters and numbers)
You're mixing regex literals with the RegExp constructor. Use one or the other, but not both:
okRegex = new RegExp('^[a-z0-9_-]{3,16}$');
or
okRegex = /^[a-z0-9_-]{3,16}$/;
As #zzzzBow answered you are mixing up two ways of using regular expressions. Choose one or the other. Now, a break down:
^ Matches the beginning of the string (that means that the string must start with whatever follows).
[a-z0-9_-] Matches the charecters a-z, A-Z, digits 0-9 _ (underscore) and - (dash/hyphen).
{3,16} States that there must be 3-16 occurences from the above character class.
$ Matches the end of the string, so the can't be anything after the 16 characters above.
Hope that helps.
I have a regular expression in JavaScript to allow numeric and (,.+() -) character in phone field
my regex is [0-9-,.+() ]
It works for numeric as well as above six characters but it also allows characters like % and $ which are not in above list.
Even though you don't have to, I always make it a point to escape metacharacters (easier to read and less pain):
[0-9\-,\.+\(\) ]
But this won't work like you expect it to because it will only match one valid character while allowing other invalid ones in the string. I imagine you want to match the entire string with at least one valid character:
^[0-9\-,\.\+\(\) ]+$
Your original regex is not actually matching %. What it is doing is matching valid characters, but the problem is that it only matches one of them. So if you had the string 435%, it matches the 4, and so the regex reports that it has a match.
If you try to match it against just one invalid character, it won't match. So your original regex doesn't match the string %:
> /[0-9\-,\.\+\(\) ]/.test("%")
false
> /[0-9\-,\.\+\(\) ]/.test("44%5")
true
> "444%6".match(/[0-9\-,\.+\(\) ]/)
["4"] //notice that the 4 was matched.
Going back to the point about escaping, I find that it is easier to escape it rather than worrying about the different rules where specific metacharacters are valid in a character class. For example, - is only valid in the following cases:
When used in an actual character class with proper-order such as [a-z] (but not [z-a])
When used as the first or last character, or by itself, so [-a], [a-], or [-].
When used after a range like [0-9-,] or [a-d-j] (but keep in mind that [9-,] is invalid and [a-d-j] does not match the letters e through f).
For these reasons, I escape metacharacters to make it clear that I want to match the actual character itself and to remove ambiguities.
You just need to anchor your regex:
^[0-9-,.+() ]+$
In character class special char doesn't need to be escaped, except ] and -.
But, these char are not escaped when:
] is alone in the char class []]
- is at the begining [-abc] or at the end [abc-] of the char class or after the last end range [a-c-x]
Escape characters with special meaning in your RegExp. If you're not sure and it isn't an alphabet character, it usually doesn't hurt to escape it, too.
If the whole string must match, include the start ^ and end $ of the string in your RegExp, too.
/^[\d\-,\.\+\(\) ]*$/