I'm trying to create a validation for a password field which allows only the a-zA-Z0-9 characters and .!##$%^&*()_+-=
I can't seem to get the hang of it.
What's the difference when using regex = /a-zA-Z0-9/g and regex = /[a-zA-Z0-9]/ and which chars from .!##$%^&*()_+-= are needed to be escaped?
What I've tried up to now is:
var regex = /a-zA-Z0-9!##\$%\^\&*\)\(+=._-/g
but with no success
var regex = /^[a-zA-Z0-9!##\$%\^\&*\)\(+=._-]+$/g
Should work
Also may want to have a minimum length i.e. 6 characters
var regex = /^[a-zA-Z0-9!##\$%\^\&*\)\(+=._-]{6,}$/g
a sleaker way to match special chars:
/\W|_/g
\W Matches any character that is not a word character (alphanumeric & underscore).
Underscore is considered a special character so
add boolean to either match a special character or _
What's the difference?
/[a-zA-Z0-9]/ is a character class which matches one character that is inside the class. It consists of three ranges.
/a-zA-Z0-9/ does mean the literal sequence of those 9 characters.
Which chars from .!##$%^&*()_+-= are needed to be escaped?
Inside a character class, only the minus (if not at the end) and the circumflex (if at the beginning). Outside of a charclass, .$^*+() have a special meaning and need to be escaped to match literally.
allows only the a-zA-Z0-9 characters and .!##$%^&*()_+-=
Put them in a character class then, let them repeat and require to match the whole string with them by anchors:
var regex = /^[a-zA-Z0-9!##$%\^&*)(+=._-]*$/
You can be specific by testing for not valid characters. This will return true for anything not alphanumeric and space:
var specials = /[^A-Za-z 0-9]/g;
return specials.test(input.val());
Complete set of special characters:
/[\!\#\#\$\%\^\&\*\)\(\+\=\.\<\>\{\}\[\]\:\;\'\"\|\~\`\_\-]/g
To answer your question:
var regular_expression = /^[A-Za-z0-9\!\#\#\$\%\^\&\*\)\(+\=\._-]+$/g
How about this:-
var regularExpression = /^(?=.*[0-9])(?=.*[!##$%^&*])[a-zA-Z0-9!##$%^&*]{6,}$/;
It will allow a minimum of 6 characters including numbers, alphabets, and special characters
There are some issue with above written Regex.
This works perfectly.
^[a-zA-Z\d\-_.,\s]+$
Only allowed special characters are included here and can be extended after comma.
// Regex for special symbols
var regex_symbols= /[-!$%^&*()_+|~=`{}\[\]:\/;<>?,.##]/;
This regex works well for me to validate password:
/[ !"#$%&'()*+,-./:;<=>?#[\\\]^_`{|}~]/
This list of special characters (including white space and punctuation) was taken from here: https://www.owasp.org/index.php/Password_special_characters. It was changed a bit, cause backslash ('\') and closing bracket (']') had to be escaped for proper work of the regex. That's why two additional backslash characters were added.
Regex for minimum 8 char, one alpha, one numeric and one special char:
/^(?=.*[A-Za-z])(?=.*\d)(?=.*[!##$%^&*])[A-Za-z\d!##$%^&*]{8,}$/
this is the actual regex only match:
/[-!$%^&*()_+|~=`{}[:;<>?,.##\]]/g
You can use this to find and replace any special characters like in Worpress's slug
const regex = /[`~!##$%^&*()-_+{}[\]\\|,.//?;':"]/g
let slug = label.replace(regex, '')
function nameInput(limitField)
{
//LimitFile here is a text input and this function is passed to the text
onInput
var inputString = limitField.value;
// here we capture all illegal chars by adding a ^ inside the class,
// And overwrite them with "".
var newStr = inputString.replace(/[^a-zA-Z-\-\']/g, "");
limitField.value = newStr;
}
This function only allows alphabets, both lower case and upper case and - and ' characters. May help you build yours.
This works for me in React Native:
[~_!##$%^&*()\\[\\],.?":;{}|<>=+()-\\s\\/`\'\]
Here's my reference for the list of special characters:
https://owasp.org/www-community/password-special-characters
If we need to allow only number and symbols (- and .) then we can use the following pattern
const filterParams = {
allowedCharPattern: '\\d\\-\\.', // declaring regex pattern
numberParser: text => {
return text == null ? null : parseFloat(text)
}
}
Related
Attempting to get all alphanumeric chars after : symbol unless a space exists, in which case the space will be the terminating mark.
// the following should all return foo
text = 'a :foo bar';
text = 's:foo';
text = ':foo, test';
Tried this, but doesn't get stuff unless there's a space. I'll probably need a regex, but not sure how that would be constructed
var t = following.substring(following.lastIndexOf(":")+1,following.lastIndexOf(' '));
How about: /:([a-zA-Z0-9]+)/
':foo, test'.match(/:([a-zA-Z0-9]+)/)[1] //returns foo
's:foo'.match(/:([a-zA-Z0-9]+)/)[1] //returns foo
':foo, test'.match(/:([a-zA-Z0-9]+)/)[1] //returns foo
These RegEx uses 2 parts:
: finds the : followed by
[a-zA-Z0-9]+ any alphanumeric character
Note:
Since this Regulars Expressions matches the types of characters I specified there is not need in this case to "break on space" since it will only match the alphanumeric characters after : it will automatically exclude space, comma, etc. any other character included within the square brackets will make that character a possible match.
I'd suggest using regular expressions, such as (albeit untested):
var result = text.match(/:([a-zA-Z0-9]*)\s/)[1];
References:
javascript regular expressions.
String.match().
You could use regex like this
/:([\w]+)[\W]/g
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).
I have a username field in my form. I want to not allow spaces anywhere in the string. I have used this regex:
var regexp = /^\S/;
This works for me if there are spaces between the characters. That is if username is ABC DEF. It doesn't work if a space is in the beginning, e.g. <space><space>ABC. What should the regex be?
While you have specified the start anchor and the first letter, you have not done anything for the rest of the string. You seem to want repetition of that character class until the end of the string:
var regexp = /^\S*$/; // a string consisting only of non-whitespaces
Use + plus sign (Match one or more of the previous items),
var regexp = /^\S+$/
If you're using some plugin which takes string and use construct Regex to create Regex Object i:e new RegExp()
Than Below string will work
'^\\S*$'
It's same regex #Bergi mentioned just the string version for new RegExp constructor
This will help to find the spaces in the beginning, middle and ending:
var regexp = /\s/g
This one will only match the input field or string if there are no spaces. If there are any spaces, it will not match at all.
/^([A-z0-9!##$%^&*().,<>{}[\]<>?_=+\-|;:\'\"\/])*[^\s]\1*$/
Matches from the beginning of the line to the end. Accepts alphanumeric characters, numbers, and most special characters.
If you want just alphanumeric characters then change what is in the [] like so:
/^([A-z])*[^\s]\1*$/
I am trying to do a basic string replace using a regex expression, but the answers I have found do not seem to help - they are directly answering each persons unique requirement with little or no explanation.
I am using str = str.replace(/[^a-z0-9+]/g, ''); at the moment. But what I would like to do is allow all alphanumeric characters (a-z and 0-9) and also the '-' character.
Could you please answer this and explain how you concatenate expressions.
This should work :
str = str.replace(/[^a-z0-9-]/g, '');
Everything between the indicates what your are looking for
/ is here to delimit your pattern so you have one to start and one to end
[] indicates the pattern your are looking for on one specific character
^ indicates that you want every character NOT corresponding to what follows
a-z matches any character between 'a' and 'z' included
0-9 matches any digit between '0' and '9' included (meaning any digit)
- the '-' character
g at the end is a special parameter saying that you do not want you regex to stop on the first character matching your pattern but to continue on the whole string
Then your expression is delimited by / before and after.
So here you say "every character not being a letter, a digit or a '-' will be removed from the string".
Just change + to -:
str = str.replace(/[^a-z0-9-]/g, "");
You can read it as:
[^ ]: match NOT from the set
[^a-z0-9-]: match if not a-z, 0-9 or -
/ /g: do global match
More information:
https://developer.mozilla.org/en-US/docs/JavaScript/Guide/Regular_Expressions
Your character class (the part in the square brackets) is saying that you want to match anything except 0-9 and a-z and +. You aren't explicit about how many a-z or 0-9 you want to match, but I assume the + means you want to replace strings of at least one alphanumeric character. It should read instead:
str = str.replace(/[^-a-z0-9]+/g, "");
Also, if you need to match upper-case letters along with lower case, you should use:
str = str.replace(/[^-a-zA-Z0-9]+/g, "");
str = str.replace(/\W/g, "");
This will be a shorter form
We can use /[a-zA-Z]/g to select small letter and caps letter sting in the word or sentence and replace.
var str = 'MM-DD-yyyy'
var modifiedStr = str.replace(/[a-zA-Z]/g, '_')
console.log(modifiedStr)
I need a regex which check the string contains only A-Z, a-z and special characters but not digits i.e. (0-9).
Any help is appreciated.
You can try with this regex:
^[^\d]*$
And sample:
var str = 'test123';
if ( str.match(/^[^\d]*$/) ) {
alert('matches');
}
Simple:
/^\D*$/
It means, any number of not-a-digit characters. See it in action…
The alternative is to reverse your test. Just check if there's a digit present, using the trivial:
/\d/
…and if that matches, your string fails.
You're looking for a character class: ^[A-Za-z.,!##$%^&*()=+_-]+$.
The ^ and $ anchor the regex by marching the beginning and end of the string, respectively.
what about:
var re = /^[a-zA-Z!#$%]+$/;
Fell free to add any special character you need inside the character class