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Use Whitelist RegEx in Javascript to validate a string

I'm trying to prevent an action based on wether a string passes the whiteList Regex in Javascript
const whiteList = /[#A-Za-z0-9.,-]/g // Regex from external source. It will be difficult to modify.
const str= 'ds%d';
console.log(str.replace(whiteList, '').length === 0) // Expected false - WORKS
// How can I make this statement return false ?
console.log(whiteList.test(str)) //Expected false Actual true
Tried using replace command to check if a string passed a validation based on a whitelist. It works but I believe there could be a better way of solving this problem.

You get true from test because there is a character in the string that matches the expression. The expression has no anchors, so it's not that it requires all characters in the string to match, just one. To require all characters to match, you'd need a "start of input" assertion (^) at the beginning, an "end of input" assertion ($) at the end, and either a "zero or more" (*) or "one or more" (+) quantifier on the character class (depending on whether an empty string should pass).
If you're getting the expression from elsewhere, you can add those to it after the fact:
const whiteList = /[#A-Za-z0-9.,-]/g // Regex from external source. It will be difficult to modify.
const str= 'ds%d';
console.log(str.replace(whiteList, '').length === 0);
const improvedList = new RegExp("^" + whiteList.source + "+$");
console.log(improvedList.test(str)); // Now shows false
That does make the assumption that the original regex has the problem described. You might check first, but it would be easy to construct regular expressions that seemed like they needed modifying but didn't.
Alternatively, just use the replace check you have, since it works as well. It's not that much more expensive.

validate string which has only one special character using javascript

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;
}

Matching special characters and letters in regex

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

Regex grouping and context

I've been messing around with something I saw on twitter which to me makes sense but doesn't work as expected.
Why doesn't the first example work? (My regex could be wrong I suppose but it looks ok)
I expected the context of my toUpperCase call to be the group and be a shorthand version of the second example.
var output = "james".replace(/(^.+?)/,"".toUpperCase.call("$1"));
var output2 = "james".replace(/(^.+?)/,function(a){
return "".toUpperCase.call(a);
});
console.log(output); // outputs james
console.log(output2); // outputs James
Edit I fixed the regex from a M42 comment. Bad pasting on my part.

#M42 is correct, in that there's no reason that your first regex would match "james". But fixing it won't work either:
var output = "james".replace(/(^.+?)/,"".toUpperCase.call("$1"));
console.log(output); // outputs james
That's because there are two options for the second argument to replace(): a string or a function. If it's a string, then "$1" will be replaced with the first match (and so on). If it's a function, then the first argument will be the first match (and so on).
In your second example, you're using the function parameter, and correctly getting the first match as an argument. But in your first example, you're passing in the result of the function call "".toUpperCase.call("$1") - which, when you run it, returns the string "$1". So the first example is actually using the string argument "$1" for .replace(), which does nothing but replace the first match in the string with itself:
"james".replace(/(^.+?)/,"$1"); // "james"
That's why this won't work as a shorthand - you're not actually passing in a function.

The first example doesn't work because there is no match (i.e. there're no parens around the name). And more the hanchor ^ must not be placed after some char.
your regex:
/ : delimiter
\( : open parens
( : begining of group 1
^ : start of string
.+? : one or more char non greedy
) : end of group 1
\) : close parens
/ : delimiter
this doesn't match james, so there is no replace or upperCased

Split string in JavaScript using a regular expression

I'm trying to write a regex for use in javascript.
var script = "function onclick() {loadArea('areaog_og_group_og_consumedservice', '\x26roleOrd\x3d1');}";
var match = new RegExp("'[^']*(\\.[^']*)*'").exec(script);
I would like split to contain two elements:
match[0] == "'areaog_og_group_og_consumedservice'";
match[1] == "'\x26roleOrd\x3d1'";
This regex matches correctly when testing it at gskinner.com/RegExr/ but it does not work in my Javascript. This issue can be replicated by testing ir here http://www.regextester.com/.
I need the solution to work with Internet Explorer 6 and above.
Can any regex guru's help?

Judging by your regex, it looks like you're trying to match a single-quoted string that may contain escaped quotes. The correct form of that regex is:
'[^'\\]*(?:\\.[^'\\]*)*'
(If you don't need to allow for escaped quotes, /'[^']*'/ is all you need.) You also have to set the g flag if you want to get both strings. Here's the regex in its regex-literal form:
/'[^'\\]*(?:\\.[^'\\]*)*'/g
If you use the RegExp constructor instead of a regex literal, you have to double-escape the backslashes: once for the string literal and once for the regex. You also have to pass the flags (g, i, m) as a separate parameter:
var rgx = new RegExp("'[^'\\\\]*(?:\\\\.[^'\\\\]*)*'", "g");
while (result = rgx.exec(script))
print(result[0]);

The regex you're looking for is .*?('[^']*')\s*,\s*('[^']*'). The catch here is that, as usual, match[0] is the entire matched text (this is very normal) so it's not particularly useful to you. match[1] and match[2] are the two matches you're looking for.
var script = "function onclick() {loadArea('areaog_og_group_og_consumedservice', '\x26roleOrd\x3d1');}";
var parameters = /.*?('[^']*')\s*,\s*('[^']*')/.exec(script);
alert("you've done: loadArea("+parameters[1]+", "+parameters[2]+");");
The only issue I have with this is that it's somewhat inflexible. You might want to spend a little time to match function calls with 2 or 3 parameters?
EDIT
In response to you're request, here is the regex to match 1,2,3,...,n parameters. If you notice, I used a non-capturing group (the (?: ) part) to find many instances of the comma followed by the second parameter.
/.*?('[^']*')(?:\s*,\s*('[^']*'))*/

Maybe this:
'([^']*)'\s*,\s*'([^']*)'