RegEx for jQuery Input Validation - javascript

I am using jquery validation plugin for validating my input fields.
I have a field that should accept:
a) letters [a-zA-Z]
b) letters with numbers [a-zA-Z0-9]
c) no special characters
So:
ADFad1334 or 43545SFDDFdf454fgf : is correct, since we have letters and numbers
asdadadASD : is correct, since we have only letters
12312342 : NOT correct, since its only numbers
sdff23424#$ : NOT correct, since there are special characters (in this example # and $)
The code i used is the one below:
$.validator.addMethod("atLeastAletter", function(value) {
return /^[a-zA-Z]*$/img.test(value) || /^[a-zA-Z0-9]*$/img.test(value);
},"Must contain at least a letter and/or number(s). No other characters are allowed");
And then:
$('#RegistrationForm').validate({
rules: {
fieldname: {
atLeastAletter: true
},
.....
The problem with this regular expression is that if the input is only numbers (ex. 3434224), it will accept it and pass validation.
What is the problem?
Thanks in advance.

or this pattern
^(?=\S*[a-zA-Z])[a-zA-Z0-9]+$
Demo
^ # Start of string/line
(?= # Look-Ahead
\S # <not a whitespace character>
* # (zero or more)(greedy)
[a-zA-Z] # Character Class [a-zA-Z]
) # End of Look-Ahead
[a-zA-Z0-9] # Character Class [a-zA-Z0-9]
+ # (one or more)(greedy)
$ # End of string/line

/^\w*[a-zA-Z]+\w*$/
That should match any string with only letters and numbers, but it must contain at least one letter.
sadAddsa // Pass
98463298 // Fail
jdsH98sd // Pass
987Fkjd89 // Pass
jfk!jhj // Fail
kjhgh8768!# // Fail
A8 // Pass
8S // Pass
B // Pass
7 // Fail
B_bkfd86jh // Fail (Using the regex in the edit)
Edit: As Alpha Bravo pointed out. This will match underscores too. If you don't want underscores /^[a-zA-Z0-9]*[a-zA-Z]+[a-zA-Z0-9]*$/will only match letters and numbers, only if it contains one letter.

If you want want to go with all HTML5 look at this fiddle: http://jsfiddle.net/bud68oho/2/
This fiddle uses the required and pattern attributes to validate the form. The required attribute does not allow the form to post if the input is not filled in. The pattern attribute allows you to implement regex for matching. If the input does not meet the requirements of the regex then the form will not post.
The only gotcha with this approach is browser compatibility. Make sure to test this method with your target browsers. The above example works in the latest version of IE and Chrome, I am unable to do any further testing.

/^[a-zA-Z0-9]*$/img.test("1") is true, so just take with two regexp for both:
return /^[a-zA-Z]*$/img.test(value) || (/[a-zA-Z]/img.test(value) && /^[a-zA-Z0-9]*$/img.test(value))
or like this:
return /^[a-zA-Z]*$/img.test(value) || (/^[a-zA-Z0-9]*$/img.test(value) && !/^[0-9]*$/img.test(value))
Notice: If you put i after the regexp (img) you just have to put a-z or A-Z, it's not case sensitive.
Ohter notice: you allow empty strings "" with those regexps.

I think you may need to do a positive alphanumeric check and a negative check for the numeric only case.
$.validator.addMethod("atLeastAletter", function(value) {
return /^[a-zA-Z0-9]+$/img.test(value) && ! /^[0-9]+$/img.test(value)
},"Must contain at least a letter and/or number(s). No other characters are allowed");

Related

regex - don't allow name to finish with hyphen

I'm trying to create a regex using javascript that will allow names like abc-def but will not allow abc-
(hyphen is also the only nonalpha character allowed)
The name has to be a minimum of 2 characters. I started with
^[a-zA-Z-]{2,}$, but it's not good enough so I'm trying something like this
^([A-Za-z]{2,})+(-[A-Za-z]+)*$.
It can have more than one - in a name but it should never start or finish with -.
It's allowing names like xx-x but not names like x-x. I'd like to achieve that x-x is also accepted but not x-.
Thanks!
Option 1
This option matches strings that begin and end with a letter and ensures two - are not consecutive so a string like a--a is invalid. To allow this case, see the Option 2.
^[a-z]+(?:-?[a-z]+)+$
^ Assert position at the start of the line
[a-z]+ Match any lowercase ASCII letter one or more times (with i flag this also matches uppercase variants)
(?:-?[a-z]+)+ Match the following one or more times
-? Optionally match -
[a-z]+ Match any ASCII letter (with i flag)
$ Assert position at the end of the line
var a = [
"aa","a-a","a-a-a","aa-aa-aa","aa-a", // valid
"aa-a-","a","a-","-a","a--a" // invalid
]
var r = /^[a-z]+(?:-?[a-z]+)+$/i
a.forEach(function(s) {
console.log(`${s}: ${r.test(s)}`)
})
Option 2
If you want to match strings like a--a then you can instead use the following regex:
^[a-z]+[a-z-]*[a-z]+$
var a = [
"aa","a-a","a-a-a","aa-aa-aa","aa-a","a--a", // valid
"aa-a-","a","a-","-a" // invalid
]
var r = /^[a-z]+[a-z-]*[a-z]+$/i
a.forEach(function(s) {
console.log(`${s}: ${r.test(s)}`)
})
You can use a negative lookahead:
/(?!.*-$)^[a-z][a-z-]+$/i
Regex101 Example
Breakdown:
// Negative lookahead so that it can't end with a -
(?!.*-$)
// The actual string must begin with a letter a-z
[a-z]
// Any following strings can be a-z or -, there must be at least 1 of these
[a-z-]+
let regex = /(?!.*-$)^[a-z][a-z-]+$/i;
let test = [
'xx-x',
'x-x',
'x-x-x',
'x-',
'x-x-x-',
'-x',
'x'
];
test.forEach(string => {
console.log(string, ':', regex.test(string));
});
The problem is that the first assertion accepts 2 or more [A-Za-z]. You will need to modify it to accept one or more character:
^[A-Za-z]+((-[A-Za-z]{1,})+)?$
Edit: solved some commented issues
/^[A-Za-z]+((-[A-Za-z]{1,})+)?$/.test('xggg-dfe'); // Logs true
/^[A-Za-z]+((-[A-Za-z]{1,})+)?$/.test('x-d'); // Logs true
/^[A-Za-z]+((-[A-Za-z]{1,})+)?$/.test('xggg-'); // Logs false
Edit 2: Edited to accept characters only
/^[A-Za-z]+((-[A-Za-z]{1,})+)?$/.test('abc'); // Logs true
Use this if you want to accept such as A---A as well :
^(?!-|.*-$)[A-Za-z-]{2,}$
https://regex101.com/r/4UYd9l/4/
If you don't want to accept such as A---A do this:
^(?!-|.*[-]{2,}.*|.*-$)[A-Za-z-]{2,}$
https://regex101.com/r/qH4Q0q/4/
So both will accept only word starting from two characters of the pattern [A-Za-z-] and not start or end (?!-|.*-$) (negative lookahead) with - .
Try this /([a-zA-Z]{1,}-[a-zA-Z]{1,})/g
I suggest the following :
^[a-zA-Z][a-zA-Z-]*[a-zA-Z]$
It validates :
that the matched string is at least composed of two characters (the first and last character classes are matched exactly once)
that the first and the last characters aren't dashes (the first and last character classes do not include -)
that the string can contain dashes and be greater than 2 characters (the second character class includes dashes and will consume as much characters as needed, dashes included).
Try it online.
^(?=[A-Za-z](?:-|[A-Za-z]))(?:(?:-|^)[A-Za-z]+)+$
Asserts that
the first character is a-z
the second is a-z or hyphen
If this matches
looks for groups of one or more letters prefixed by a hyphen or start of string, all the way to end of string.
You can also use the I switch to make it case insensitive.

catastrophic backtracking error validation string

^([a-zA-Z0-9]+[._-]?)+[a-zA-Z0-9]+$
I have used above regex to perform validation on input string for below scenarios.
I want to allow only hyphen ('-'), period ('.') & underscore ('_')
name should not start or end with hyphen ('-'), period ('.') & underscore ('_')
name should not contain Spaces
Two consecutive special characters (from set of (._-)) are not allowed
And I have validated it through javascript.
But, when we add name having special characters at the end, browser halts/ hangs instead of returning false.
var regex = new RegExp("^([a-zA-Z0-9]+[._-]?)+[a-zA-Z0-9]+$");
if (regex.test($('#txtBox1').val())) {//success}
Don't make those special delimiters optional in your repeated group:
^([a-zA-Z0-9]+[._-])*[a-zA-Z0-9]+$
# ^ ^
That still matches the same, but it can't backtrack to apply the optional character in positions where it doesn't appear.
Try this as well
var isValid = !!str.match(/[^\w.-]/i)
&& !str.split(/[._-]/).filter( s => s.length == 0 ).length;
Explanation
str.match(/[^\w.-]/i) checks if there is any character matching which is neither alphanumeric, underscore, dot nor hypen.
str.split(/[._-]/) splits the input by these three characters [._-] and then check if there is any empty string. If these characters are at the beginning or end or consecutively placed, then there will be an empty string in the resultant array.

I want to ignore square brackets when using javascript regex [duplicate]

This question already has answers here:
Why is this regex allowing a caret?
(3 answers)
Closed 1 year ago.
I am using javascript regex to do some data validation and specify the characters that i want to accept (I want to accept any alphanumeric characters, spaces and the following !&,'\- and maybe a few more that I'll add later if needed). My code is:
var value = userInput;
var pattern = /[^A-z0-9 "!&,'\-]/;
if(patt.test(value) == true) then do something
It works fine and excludes the letters that I don't want the user to enter except the square bracket and the caret symbols. From all the javascript regex tutorials that i have read they are special characters - the brackets meaning any character between them and the caret in this instance meaning any character not in between the square brackets. I have searched here and on google for an explanation as to why these characters are also accepted but can't find an explanation.
So can anyone help, why does my input accept the square brackets and the caret?
The reason is that you are using A-z rather than A-Za-z. The ascii range between Z (0x5a) and a (0x61) includes the square brackets, the caret, backquote, and underscore.
Your regex is not in line with what you said:
I want to accept any alphanumeric characters, spaces and the following !&,'\- and maybe a few more that I'll add later if needed
If you want to accept only those characters, you need to remove the caret:
var pattern = /^[A-Za-z0-9 "!&,'\\-]+$/;
Notes:
A-z also includesthe characters: [\]^_`.
Use A-Za-z or use the i modifier to match only alphabets:
var pattern = /^[a-z0-9 "!&,'\\-]+$/i;
\- is only the character -, because the backslash will act as special character for escaping. Use \\ to allow a backslash.
^ and $ are anchors, used to match the beginning and end of the string. This ensures that the whole string is matched against the regex.
+ is used after the character class to match more than one character.
If you mean that you want to match characters other than the ones you accept and are using this to prevent the user from entering 'forbidden' characters, then the first note above describes your issue. Use A-Za-z instead of A-z (the second note is also relevant).
I'm not sure what you want but I don't think your current regexp does what you think it does:
It tries to find one character is not A-z0-9 "!&,'\- (^ means not).
Also, I'm not even sure what A-z matches. It's either a-z or A-Z.
So your current regexp matches strings like "." and "Hi." but not "Hi"
Try this: var pattern = /[^\w"!&,'\\-]/;
Note: \w also includes _, so if you want to avoid that then try
var pattern = /[^a-z0-9"!&,'\\-]/i;
I think the issue with your regex is that A-z is being understood as all characters between 0x41 (65) and 0x7A (122), which included the characters []^_` that are between A-Z and a-z. (Z is 0x5A (90) and a is 0x61 (97), which means the preceding characters take up 0x5B thru 0x60).

Javascript match function for special characters

I am working on this code and using "match" function to detect strength of password. how can I detect if string has special characters in it?
if(password.match(/[a-z]+/)) score++;
if(password.match(/[A-Z]+/)) score++;
if(password.match(/[0-9]+/)) score++;
If you mean !##$% and ë as special character you can use:
/[^a-zA-Z ]+/
The ^ means if it is not something like a-z or A-Z or a space.
And if you mean only things like !#$&$ use:
/\W+/
\w matches word characters, \W matching not word characters.
You'll have to whitelist them individually, like so:
if(password.match(/[`~!##\$%\^&\*\(\)\-=_+\\\[\]{}/\?,\.\<\> ...
and so on. Note that you'll have to escape regex control characters with a \.
While less elegant than /[^A-Za-z0-9]+/, this will avoid internationalization issues (e.g., will not automatically whitelist Far Eastern Language characters such as Chinese or Japanese).
you can always negate the character class:
if(password.match(/[^a-z\d]+/i)) {
// password contains characters that are *not*
// a-z, A-Z or 0-9
}
However, I'd suggest using a ready-made script. With the code above, you could just type a bunch of spaces, and get a better score.
Just do what you did above, but create a group for !##$%^&*() etc. Just be sure to escape characters that have meaning in regex, like ^ and ( etc....
EDIT -- I just found this which lists characters that have meaning in regex.
if(password.match(/[^\w\s]/)) score++;
This will match anything that is not alphanumeric or blank space. If whitespaces should match too, just use /[^\w]/.
As it look from your regex, you are calling everything except for alphanumeric a special character. If that is the case, simply do.
if(password.match(/[\W]/)) {
// Contains special character.
}
Anyhow how why don't you combine those three regex into one.
if(password.match(/[\w]+/gi)) {
// Do your stuff.
}
/[^a-zA-Z0-9 ]+/
This will accept only special characters and will not accept a to z & A to Z 0 to 9 digits

validating variable in javascript

Hi i have a field in php that will be validated in javascript using i.e for emails
var emailRegex = /^[\w-\.]+#([\w-]+\.)+[\w-]{2,4}$/;
What i'm after is a validation check which will look for the
first letter as a capital Q
then the next letters can be numbers only
then followed by a .
then two numbers only
and then an optional letter
i.e Q100.11 or Q100.11a
I must admit i look at the above email validation check and i have no clue how it works but it does ;)
many thanks for any help on this
Steve
The ^ marks the beginning of the string, $ matches the end of the string. In other words, the whole string should exactly match this regular expression.
[\w-\.]+: I think you wanted to match letters, digits, dots and - only. In that case, the - should be escaped (\-): [\w\-\.]+. The plus-sign makes is match one or more times.
#: a literal # match
([\w-]+\.)+ letters, digits and - are allowed one or more times, with a dot after it (between the parentheses). This may occur several times (at least once).
[\w-]{2,4}: this should match the TLD, like com, net or org. Because a TLD can only contain letters, it should be replaced by [a-z]{2,4}. This means: lowercase letters may occur two till four times. Note that the TLD can be longer than 4 characters.
An regular expression which should follow the next rules:
a capital Q (Q)
followed by one or more occurrences of digits (\d+)
a literal dot (.)
two digits (\d{2})
one optional letter ([a-z]?)
Result:
var regex = /Q\d+\.\d{2}[a-z]?/;
If you need to match strings case-insensitive, add the i (case-insensitive) modifier:
var regex = /Q\d+\.\d{2}[a-z]?/i;
Validating a string using a regexp can be done in several ways, one of them:
if (regex.test(str)) {
// success
} else {
// no match
}
var emailRegex = /^Q\d+\.\d{2}[a-zA-Z]?#([\w-]+\.)+[a-zA-Z]+$/;
var str = "Q100.11#test.com";
alert(emailRegex.test(str));
var regex = /^Q[0-9]+\.[0-9]{2}[a-z]?$/;
+ means one or more
the period must be escaped - \.
[0-9]{2} means 2 digits, same as \d{2}
[a-z]? means 0 or 1 letter
You can check your regex at http://regexpal.com/

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