Regular expression not able to read complete string, only working correct with single character.
var abc = "ab";
var patter = /^([a-z0-9A-Z])$/;
if (patter.test(abc)) {
console.log('yes');
} else {
console.log('no');
}
You must set a quantifier when you don't want just one character.
Add a * to match zero or more character (or a + if you want to be sure there's at least one character);
var patter = /^[a-z0-9A-Z]*$/;
Note that I removed the parentheses : they're useless with the test method.
Related
I am currently replacing all non-letter characters using
var stringwithoutspecialCharacter = "testwordwithpunctiuation.".replace(/[^\w\s!?]/g, '');
The problem is that I do not know which special character will appear (that needs removing). However I do need to be able to access the removed special character after I've run some code with the word without the special character.
Example inputs:
"test".
(temporary)
foo,
Desired output:
['"','test','"',"."]
['(','temporary',')']
['foo',',']
How could this be achieved in javascript?
Edit: To get both valid and invalid characters, change the regular expression
Quick solution is to define an array to collect the matches.
Then pass in a function into your replace() call
var matches = [];
var matcher = function(match, offset, string) {
matches.push(match);
return '';
}
var stringwithoutspecialCharacter = "testwordwithpunctiuation.".replace(/[^\w\s!?]|[\w\s!?]+/g, matcher);
console.log("Matches: " + matches);
I have a little code snippet where I use Regular Expressions to rip off punctuation, numbers etc from a string. I am getting undefined along with output of my ripped string. Can someone explain whats happening? Thanks
var regex = /[^a-zA-z\s\.]|_/gi;
function ripPunct(str) {
if ( str.match(regex) ) {
str = str.replace(regex).replace(/\s+/g, "");
}
return str;
}
console.log(ripPunct("*#£#__-=-=_+_devide-00000110490and586#multiply.edu"));
You should pass a replacement pattern to the first replace method, and also use A-Z, not A-z, in the pattern. Also, there is no point to check for a match before replacing, just use replace directly. Also, it seems the second chained replace is redundant as the first one already removes whitespace (it contains \s). Besides, the |_ alternative is also redundant since the [^a-zA-Z\s.] already matches an underscore as it is not part of the symbols specified by this character class.
var regex = /[^a-zA-Z\s.]/gi;
function ripPunct(str) {
return str.replace(regex, "");
}
console.log(ripPunct("*#£#__-=-=_+_devide-00000110490and586#multiply.edu"));
I'm trying to build a regex which allows the following characters:
A-Z
a-z
1234567890
!##$%&*()_-+={[}]|\:;"'<,>.?/~`
All other characters are invalid. This is the regex I built, but it is not working as I expect it to. I expect the .test() to return false when an invalid character is present:
var string = 'abcd^wyd';
function isValidPassword () {
var regex = /[0-9A-Za-z!##$%&*()_\-+={[}\]|\:;"'<,>.?\/\\~`]+[0-9A-Za-z!##$%&*()_\-+={[}\]|\:;"'<,>.?\/\\~`]*/g
return regex.test(string);
}
In this case, the test is always returning "true", even when "^" is present in the string.
Your regex only checks that at least one of the allowed characters is present. Add start and end anchors to your regex - /^...$/
var string = 'abcd^wyd';
function isValidPassword () {
var regex = /^[0-9A-Za-z!##$%&*()_\-+={[}\]|\:;"'<,>.?\/\\~`]+[0-9A-Za-z!##$%&*()_\-+={[}\]|\:;"'<,>.?\/\\~`]*$/g
return regex.test(string);
}
... another approach, is instead of checking all characters are good, to look for a bad character, which is more efficient as you can stop looking as soon as you find one...
// return true if string does not (`!`) match a character that is not (`^`) in the set...
return !/[^0-9A-Za-z!##$%&*()_\-+={[}\]|\:;"'<,>.?\/\\~`]/.test()
Instead of searching allowed characters search forbidden ones.
var string = 'abcd^wyd';
function regTest (string) {//[^ == not
var regex = /[^0-9A-Za-z!##$%&*()_\-+={[}\]|\:;"'<,>.?\/\\~`]/g
return !regex.test(string);//false if found
}
console.log(regTest(string));
The regex, as you've written is checking for the existence of the characters in the input string, regardless of where it appears.
Instead you need to anchor your regex so that it checks the entire string.
By adding ^ and $, you are instructing your regex to match only the allowed characters for the entire string, rather than any subsection.
function isValidPassword (pwd) {
var regex = /^[0-9A-Za-z!##$%&*()_\-+={[}\]|\:;"'<,>.?\/\\~`]+[0-9A-Za-z!##$%&*()_\-+={[}\]|\:;"'<,>.?\/\\~`]*$/g\;
return regex.test(pwd);
}
alert(isValidPassword('abcd^wyd'));
Your regexp is matching the first part of o=your string i.e. "abcd" so it is true . You need to anchor it to the start (using ^ at the beginning) and the end of the string (using $ at the end) so your regexp should look like:
^[0-9A-Za-z!##$%&*()_\-+={[}\]|\:;"'<,>.?\/\\~`]+[0-9A-Za-z!##$%&*()_\-+={[}\]|\:;"'<,>.?\/\\~`]$
That way it will need to match the entire string.
You can visualize it in the following link:
regexper_diagram
This regex will work.
var str = 'eefdooasdc23432423!##$%&*()_-+={[}]|:;"\'<,>.?/~\`';
var reg = /.|\d|!|#|#|\$|%|&|\*|\(|\)|_|-|\+|=|{|\[|}|]|\||:|;|"|'|<|,|>|\.|\?|\/|~|`/gi;
// test it.
reg.test(str); //true
I use this site to test my regex.
Regex 101
I know there are several question like this on Stack-overflow, but I can't seem to get a straight answer out of the questions already posted.Looking forward if someone can help me.
I want to validate a string & return TRUE if it satisfies below condition
String contains only one special character i.e _ (underscore)
& this special character should not appear at beginning or end of the string
Example:
var demo1="23dsfXXXa32_XXXX" // Valid, should returns TRUE
var demo2="_23dsfXXXa32_XXXX" // Invalid,should returns FALSE
var demo3= "23dsfXXXa32XXXX_" //invalid,should returns FALSE
var demo4= "_" //invalid,should returns FALSE
var demo5= "&sdfsa_XX";// returns false
Tried: FIDDLE
if(/^[a-zA-Z0-9_ ]*$/.test(demo1) == true) {
alert('Valid String');
}
Result: Not functioning as per expected
Since you've said you require that the character be there, one way is to do it with a regular expression is with a negated character class at each end with a non-negated one in the middle:
var rex = /^[^_]+[_][^_]+$/;
That only handle underscores; add other "special" characters to all three character classes there.
How that works:
^ matches start of string
[^_]+ requires one or more characters not in the class
[_] requires exactly one character in the class
[^_]+ requires one or more characters not in the class
$ matches end of string
You could simplify it by using indexOf to verify that its not in the first or last position and split to see if its there only once. This is usually faster than a regex pattern.
function checkString(str){
return str.indexOf("_") !== 0 && str.indexOf("_") !== str.length-1 && str.split("_") === 2;
}
I am trying to validate a string, that should contain letters numbers and special characters &-._ only. For that I tried with a regular expression.
var pattern = /[a-zA-Z0-9&_\.-]/
var qry = 'abc&*';
if(qry.match(pattern)) {
alert('valid');
}
else{
alert('invalid');
}
While using the above code, the string abc&* is valid. But my requirement is to show this invalid. ie Whenever a character other than a letter, a number or special characters &-._ comes, the string should evaluate as invalid. How can I do that with a regex?
Add them to the allowed characters, but you'll need to escape some of them, such as -]/\
var pattern = /^[a-zA-Z0-9!##$%^&*()_+\-=\[\]{};':"\\|,.<>\/?]*$/
That way you can remove any individual character you want to disallow.
Also, you want to include the start and end of string placemarkers ^ and $
Update:
As elclanrs understood (and the rest of us didn't, initially), the only special characters needing to be allowed in the pattern are &-._
/^[\w&.\-]+$/
[\w] is the same as [a-zA-Z0-9_]
Though the dash doesn't need escaping when it's at the start or end of the list, I prefer to do it in case other characters are added. Additionally, the + means you need at least one of the listed characters. If zero is ok (ie an empty value), then replace it with a * instead:
/^[\w&.\-]*$/
Well, why not just add them to your existing character class?
var pattern = /[a-zA-Z0-9&._-]/
If you need to check whether a string consists of nothing but those characters you have to anchor the expression as well:
var pattern = /^[a-zA-Z0-9&._-]+$/
The added ^ and $ match the beginning and end of the string respectively.
Testing for letters, numbers or underscore can be done with \w which shortens your expression:
var pattern = /^[\w&.-]+$/
As mentioned in the comment from Nathan, if you're not using the results from .match() (it returns an array with what has been matched), it's better to use RegExp.test() which returns a simple boolean:
if (pattern.test(qry)) {
// qry is non-empty and only contains letters, numbers or special characters.
}
Update 2
In case I have misread the question, the below will check if all three separate conditions are met.
if (/[a-zA-Z]/.test(qry) && /[0-9]/.test(qry) && /[&._-]/.test(qry)) {
// qry contains at least one letter, one number and one special character
}
Try this regex:
/^[\w&.-]+$/
Also you can use test.
if ( pattern.test( qry ) ) {
// valid
}
let pattern = /^(?=.*[0-9])(?=.*[!##$%^&*])(?=.*[a-z])(?=.*[A-Z])[a-zA-Z0-9!##$%^&*]{6,16}$/;
//following will give you the result as true(if the password contains Capital, small letter, number and special character) or false based on the string format
let reee =pattern .test("helLo123#"); //true as it contains all the above
I tried a bunch of these but none of them worked for all of my tests. So I found this:
^(?=.*\d)(?=.*[a-z])(?=.*[A-Z])(?=.*[^a-zA-Z0-9])(?!.*\s).{8,15}$
from this source: https://www.w3resource.com/javascript/form/password-validation.php
Try this RegEx: Matching special charecters which we use in paragraphs and alphabets
Javascript : /^[a-zA-Z]+(([\'\,\.\-_ \/)(:][a-zA-Z_ ])?[a-zA-Z_ .]*)*$/.test(str)
.test(str) returns boolean value if matched true and not matched false
c# : ^[a-zA-Z]+(([\'\,\.\-_ \/)(:][a-zA-Z_ ])?[a-zA-Z_ .]*)*$
Here you can match with special char:
function containsSpecialChars(str) {
const specialChars = /[`!##$%^&*()_+\-=\[\]{};':"\\|,.<>\/?~]/;
return specialChars.test(str);
}
console.log(containsSpecialChars('hello!')); // 👉️ true
console.log(containsSpecialChars('abc')); // 👉️ false
console.log(containsSpecialChars('one two')); // 👉️ false