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*$/
Matches from the beginning of the line to the end. Accepts alphanumeric characters, numbers, and most special characters.
If you want just alphanumeric characters then change what is in the [] like so:
/^([A-z])*[^\s]\1*$/
Related
I have a script line this :
#type1 this is the text of the note
I've tried this bu didn't workout for me :
^\#([^\s]+)
I watch to catch type in other words I to get whats between the hash sign "#" and the next white space, excluding the hash "#" sign, and the string that I want to catch is alphanumeric string.
With the regex functionality provided by Javascript:
exec_result = /#(\w*)/.exec('#whatever string comes here');
I believe exec_result[1] should be the string you want.
The return value of exec() method could be found over here:
https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/RegExp/exec
You're really close:
/^\#(\w+)\s/
The \w matches any letters or numbers (and underscores too). And the space should be outside the matching group since I guess you don't want to capture it.
To get an alphanumeric match (which will get you type1), instead of the negated character class [^\s] which matches not a whitespace character, you could use a character class and specify what you want to match like [A-Za-z0-9].
Then use a negative lookahead to assert what is on the right is not a non-whitespace char:
^#([A-Za-z0-9]+)(?!\S)
Regex demo
Your match is in the first capturing group. Note that you don't have to escape the \#
For example using the case insensitive flag /i
const regex = /^#([A-Za-z0-9]+)(?!\S)/i;
const str = `#type1 this is the text of the note`;
console.log(str.match(regex)[1]);
If you only want to match type, you might use:
^#([a-z]+)[a-z0-9]*(?!\S)
Regex demo
const regex = /^#([a-z]+)[a-z0-9]*(?!\S)/i;
const str = `#type1 this is the text of the note`;
console.log(str.match(regex)[1]);
I've figured it out.
/^\#([^\s]+)+(.*)$/
I need to hyphenate a string in javascript. The string is a url (e.g '/home/about/').
My current regex, is working but the output is not as desired.
If the first/last character of the string is a special character, it should be removed and instead of being changed into a hyphen.
Example:
var string = '/home/about/';
string.replace(/[^a-zA-Z0-9]/g, '-').toLowerCase();
// Returns -home-about- but I need home-about
^\/ means / at begin and \/$ means / at the end. joined them with pipe to handle both removals from the end.
string = string.replace(/^\/|\/$/g, '').toLowerCase();
Then do your regex operation:
string.replace(/[^a-zA-Z0-9]/g, '-').toLowerCase();
You can simply do this:
var s="/home/about/";
s.match(/[^\/]+/g).join('-'); // home-about
Instead of replace use finding groups.
Where You will look for a group of any characters prefixed and postfixed with any of the special characters (its only / or some more?).
Next concatenate '-' with that new string, and You are done.
I'm trying to create a validation for a password field which allows only the a-zA-Z0-9 characters and .!##$%^&*()_+-=
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)
}
}
I'm trying to use the following code with jQuery to validate hex value strings but I get unexpected results:
var a = new RegExp("0x[a-fA-F0-9]+")
var result = a.test('0x1n')
In this case, result actually returns true. What am I missing here?
You need anchors to match the beginning and the end of the string. This will make the regular expression try to match against the entire string instead of just a part of the string:
var a = new RegExp("^0x[a-fA-F0-9]+$")
Otherwise your regular expression matches the 0x1 part and returns true.
On another note, the following would be a little better:
var re = /^0x[a-f0-9]+$/i;
The i flag makes it case insensitive so you don't have to specify a-f and A-F.
Your regex does match that string, because you don't have any anchors on it. If you change your regex to ^0x[a-fA-F0-9]+$, then the string 0x1n will not match.
Edit: To further explain why your string matches, your regular expression is actually trying to match a string that contains 0x followed by one or more characters in the [a-fA-F0-9] character class. The string 0x1n contains 0x followed by 1, which is in the [a-fA-F0-9] character class.
Adding anchors means that your string must start with 0x, then finish with one or more characters in the [a-fA-F0-9] character class. 0x1n would fail to match, since it ends in an n, which is not in that character class.
It returns true because you're not requiring the entire string to match that pattern. Try this:
var a = new RegExp("^0x[a-fA-F0-9]+$")
I've got a string which contains q="AWORD" and I want to replace q="AWORD" with q="THEWORD". However, I don't know what AWORD is.. is it possible to combine a string and a regex to allow me to replace the parameter without knowing it's value? This is what I've got thus far...
globalparam.replace('q="/+./"', 'q="AWORD"');
What you have is just a string, not a regular expression. I think this is what you want:
globalparam.replace(/q=".+?"/, 'q="THEWORD"');
I don't know how you got the idea why you have to "combine" a string and a regular expression, but a regex does not need to exist of wildcards only. A regex is like a pattern that can contain wildcards but otherwise will try to match the exact characters given.
The expression shown above works as follows:
q=": Match the characters q, = and ".
.+?": Match any character (.) up to (and including) the next ". There must be at least one character (+) and the match is non-greedy (?), meaning it tries to match as few characters as possible. Otherwise, if you used .+", it would match all characters up to the last quotation mark in the string.
Learn more about regular expressions.
Felix's answer will give you the solution, but if you actually want to construct a regular expression using a string you can do it this way:
var fullstring = 'q="AWORD"';
var sampleStrToFind = 'AWORD';
var mat = 'q="'+sampleStrToFind+'"';
var re = new RegExp(mat);
var newstr = fullstring.replace(re,'q="THEWORD"');
alert(newstr);
mat = the regex you are building, combining strings or whatever is needed.
re = RegExp constructor, if you wanted to do global, case sensitivity, etc do it here.
The last line is string.replace(RegExp,replacement);