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How to remove % from the string only if the % is present in first or last of a string

I want to remove the % character from my string if the % character present in the string, then it should check whether it is in the beginning or end then it should trim the value then the provide the result.
Eg: var str = "Value%" or "%Value" or "%Value%"
The result should be = Value.
Eg: var str="Va%ue"
The result should be =Va%ue.
Eg: var str= "Value"
The result should be = Value.
Thanks in Advance

str = (str[0] == '%' || str.endsWith('%')? str.replace(/%/g, '') : str);
Check if the string starts or ends with % before replacing

Your regex basically needs to have two alternatives combined with an 'or' sign.
You use ^ to signal beginning and $ to signal ending, then combine them with |.
The regex for percent lookup: ^%|%$.
If you put that in to the replace() function and add the global lookup flag g, you can easily achieve what you're looking for:
const percentLookupRegex = /^%|%$/g;
str.replace(percentLookupRegex, '');
Here's a live example:
https://regex101.com/r/nAK64n/2

If you want to use regex then you should look into the anchors ^ and $
Eg.
str.replace(/^%/, '');
Will replace % in the beginning of the line with nothing.
An alternative approach is to use startsWith and endsWith and then slice the string appropriately:
if (str.startsWith('%') {
str = str.slice(1);
}
Removing a % at the end of the string is left as an exercise for the reader

str = (str.indexOf('%') === 0) ? str.substring(1) : str; // check for first character
str = (str.lastIndexOf('%') === (str.length - 1)) ? str.substring(0, str.length - 1) : str; //check for last character
This will only check for first and last character and remove only those

Regular Expression Returns Undefined

I'm attempting one of the beginner coderByte challenges, Simple Symbols. Challenge summary below.
"Using the JavaScript language, have the function SimpleSymbols(str) take the str parameter being passed and determine if it is an acceptable sequence by either returning the string true or false. The str parameter will be composed of + and = symbols with several letters between them (ie. ++d+===+c++==a) and for the string to be true each letter must be surrounded by a + symbol. So the string to the left would be false. The string will not be empty and will have at least one letter."
function SimpleSymbols(str){
var RegExp = /\+\w\+/gi;
var regexp1 = /^\w/gi;
var regexp2 = /\w$/g;
if(regexp1.test(str) == true){
return false
} else if(regexp2.test(str) == true){
return false
} else if(RegExp == true){
return true
}
}
console.log(SimpleSymbols('+d+=3=+s+'));
console.log(SimpleSymbols('f++d+'));
The first regular expression I'm testing, /^\w/gi, comes back undefined, and I can't figure out why?
https://regex101.com/ is a great tool I've used before, and my expression does identify f as the first character in the string, but when I test it in codepen, it comes back undefined in the console.

Code
See regex in use here
^[+=\d]*\+(?:[a-z]\+[+=\d]*)+$
Alternatively, using the opposite logic (catching invalid strings instead of valid ones), you can use (?:^|[^+])[a-z]|[a-z](?:[^+]|$)
Usage
Please note the valid/invalid strings below have been created according to the OP's explanation of valid and invalid strings: That each letter must be surrounded by a + symbol. and that the plus sign + may be shared between characters such that +a+a+ is valid (specified in comments below the question).
var a = [
// valid
"++d+===+c++==+a++",
"+a+a+a+",
"+a++a+",
"+a+",
// invalid
"++d+===+c++==a",
"+=d+",
"+dd+",
"+d=+",
"+d+d",
"d+d+"
];
var r = /^[+=\d]*\+(?:[a-z]\+[+=\d]*)+$/mi;
a.forEach(function(s){
console.log(r.test(s));
});
Explanation
^ Assert position at the start of the line
[+=\d]* Match any number of characters in the set (+, =, or digit)
\+ Match a literal plus sign +
(?:[a-z]\+[+=\d]*)+ Match one or more of the following
[a-z] Match a lowercase ASCII letter
\+ Match a literal plus sign +
[+=\d]* Match any number of characters in the set (+, =, or digit)
$ Assert position at the end of the line

It's returning undefined because your expression does not meet any of the criteria. Since you have no else {} defined, than nothing gets returned. Thus you get undefined. Try this:
function SimpleSymbols(str){
var RegExp = /\+\w\+/gi;
var regexp1 = /^\w/gi;
var regexp2 = /\w$/g;
if(regexp1.test(str) == true){
return false
} else if(regexp2.test(str) == true){
return false
} else if(RegExp == true){
return true
} else {
return "catch all here";
}
}
console.log(SimpleSymbols('+d+=3=+s+'));
console.log(SimpleSymbols('f++d+'));

You can use a single regex and the string.test() method (which returns just true/false).
Below are 2 different ways (regex) to do it .
First requires a separate + between word chars. Example +a++b+ (true)
^
(?: [+=]* \+ \w \+ [+=]* )+
$
Second can take a common + between word chars. Example +a+b+ (true)
^
(?:
[+=]* \+ \w
(?= \+ )
)+
[+=]*
$
var patt1 = new RegExp("^(?:[+=]*\\+\\w\\+[+=]*)+$");
function SimpleSymbols_1(str){
return patt1.test(str);
}
var patt2 = new RegExp("^(?:[+=]*\\+\\w(?=\\+))+[+=]*$");
function SimpleSymbols_2(str){
return patt2.test(str);
}
console.log(SimpleSymbols_1('+d+=3=+s+'));
console.log(SimpleSymbols_1('f++d+'));
console.log(SimpleSymbols_1('+a+b+c+'));
console.log(SimpleSymbols_2('+a+b+c+'));
console.log(SimpleSymbols_2('+a+=+c+'));
console.log(SimpleSymbols_2('+a++c+'));

Thank you all for throwing some support/comments my way. Again, I am new to JavaScript and Regular Expressions are fairly foreign to me, though I am gaining some traction in understanding them. Here is the updated solution I posted. It's quite convoluted and perhaps a more inelegant and non-simple way to come to the right answer, but it worked.
function SimpleSymbols(str){
var RegExp = /\+[a-z]\+/gi;
var regexp1 = /^[a-z]/gi;
var regexp2 = /[a-z]$/g;
var regexp3 = /[a-z]\=/gi;
var regexp4 = /\=[a-z]/gi;
if(regexp1.test(str) === true){
return false
} else if(regexp2.test(str) === true){
return false
} else if(regexp3.test(str) === true){
return false
} else if(regexp4.test(str) === true){
return false
} else {
return true
}
}
console.log(SimpleSymbols('+d+=3=+s+'));
console.log(SimpleSymbols('f++d+'));
console.log(SimpleSymbols('+d===+a+'));
console.log(SimpleSymbols('+a='));
console.log(SimpleSymbols('2+a+a+'));
console.log(SimpleSymbols('==a+'));
I was sure there had to be a way to use only one regular expression, but again, I'm still very much a novice.
Thanks again everyone.

How do I check for brackets in a specific place in the string?

I have this code and it needs to returns true or false based on the string you give it.
This is the only example on which it doesn't work. How can I check if brackets exist in a specific index of the string?
function telephoneCheck(str) {
var newStr = str.replace(/-/g,'').replace(/ /g,'').replace(/\(|\)/g,'');
var valid = true;
var re = /\([^()]*\)/g;
while (str.match(re))
str = str.replace(re, '');
if (str.match(/[()]/)){
valid = false;
}
if(newStr.length === 10 && valid === true && str.indexOf()){
return true;
}else if(newStr.length === 11 && str[0] != "-" && newStr[0] == 1 && valid === true){
return true;
}else{
return false;
}
}
telephoneCheck("(6505552368)");

Based on your code I think you might be looking for something like this:
'(6505552368)'.replace(/^\((\d+)\)$/, '$1');
The ^ and $ in the RegExp will match the start and the end of the string. The \d+ will match one or more numbers. The extra parentheses form a capture group that is then used in the replacement as $1.
You might be better off doing more work using RegExps rather than doing all that replacing but without knowing the exact requirements it's difficult to be more precise. I highly suggest learning more about RegExp syntax.
If you literally just want to know whether 'brackets exist in a specific index' then you can just use:
str.charAt(index) === '(';

To check if there are brackets at a specific index in the string:
/[()]/.test(str[index])
To check if there are any brackets in the string at all:
/[()]/.test(str)
If you want to test for a specific bracket type (e.g. opening but not closing) remove the other one (e.g. closing) from the regex.

javascript variable contains instead of equals

in a javascript if...else statement, instead of checking if a variable equals (==) a value, is it possible to check if a variable includes a value?
var blah = unicorns are pretty;
if(blah == 'unicorns') {}; //instead of doing this,
if(blah includes 'unicorns') {}; //can i do this?
also, the word it includes should be the FIRST word of the variable. Thanks!!!

If by "first word", you mean a character sequence from the beginning of the string to the first space, then this will do it:
if ((sentence + ' ').indexOf('unicorns ') === 0) {
// note the trailing space ^
}
If instead of a space it can be any white-space character, you should use a regular expression:
if (/^unicorns(\s|$)/.test(sentence)) {
// ...
}
// or dynamically
var search = 'unicorns';
if (RegExp('^' + search + '(\\s|$)').test(sentence)) {
// ...
}
You can also use the special word-boundary character, depending on the language you want to match:
if (/^unicorns\b/.test(sentence)) {
// ...
}
More about regular expressions.
Related question:
How to check if a string "StartsWith" another string?

if(blah.indexOf('unicorns') == 0) {
// the string "unicorns" was first in the string referenced by blah.
}
if(blah.indexOf('unicorns') > -1) {
// the string "unicorns" was found in the string referenced by blah.
}
indexOf
To remove the first occurrence of a string:
blah = blah.replace('unicorns', '');

You can also use a quick regex test:
if (/unicorns/.test(blah)) {
// has "unicorns"
}

Remove all special characters except space from a string using JavaScript

I want to remove all special characters except space from a string using JavaScript.
For example,
abc's test#s
should output as
abcs tests.

You should use the string replace function, with a single regex.
Assuming by special characters, you mean anything that's not letter, here is a solution:
const str = "abc's test#s";
console.log(str.replace(/[^a-zA-Z ]/g, ""));

You can do it specifying the characters you want to remove:
string = string.replace(/[&\/\\#,+()$~%.'":*?<>{}]/g, '');
Alternatively, to change all characters except numbers and letters, try:
string = string.replace(/[^a-zA-Z0-9]/g, '');

The first solution does not work for any UTF-8 alphabet. (It will cut text such as Привіт). I have managed to create a function which does not use RegExp and use good UTF-8 support in the JavaScript engine. The idea is simple if a symbol is equal in uppercase and lowercase it is a special character. The only exception is made for whitespace.
function removeSpecials(str) {
var lower = str.toLowerCase();
var upper = str.toUpperCase();
var res = "";
for(var i=0; i<lower.length; ++i) {
if(lower[i] != upper[i] || lower[i].trim() === '')
res += str[i];
}
return res;
}
Update: Please note, that this solution works only for languages where there are small and capital letters. In languages like Chinese, this won't work.
Update 2: I came to the original solution when I was working on a fuzzy search. If you also trying to remove special characters to implement search functionality, there is a better approach. Use any transliteration library which will produce you string only from Latin characters and then the simple Regexp will do all magic of removing special characters. (This will work for Chinese also and you also will receive side benefits by making Tromsø == Tromso).

search all not (word characters || space):
str.replace(/[^\w ]/, '')

I don't know JavaScript, but isn't it possible using regex?
Something like [^\w\d\s] will match anything but digits, characters and whitespaces. It would be just a question to find the syntax in JavaScript.

I tried Seagul's very creative solution, but found it treated numbers also as special characters, which did not suit my needs. So here is my (failsafe) tweak of Seagul's solution...
//return true if char is a number
function isNumber (text) {
if(text) {
var reg = new RegExp('[0-9]+$');
return reg.test(text);
}
return false;
}
function removeSpecial (text) {
if(text) {
var lower = text.toLowerCase();
var upper = text.toUpperCase();
var result = "";
for(var i=0; i<lower.length; ++i) {
if(isNumber(text[i]) || (lower[i] != upper[i]) || (lower[i].trim() === '')) {
result += text[i];
}
}
return result;
}
return '';
}

const str = "abc's#thy#^g&test#s";
console.log(str.replace(/[^a-zA-Z ]/g, ""));

Try to use this one
var result= stringToReplace.replace(/[^\w\s]/g, '')
[^] is for negation, \w for [a-zA-Z0-9_] word characters and \s for space,
/[]/g for global

With regular expression
let string = "!#This tool removes $special *characters* /other/ than! digits, characters and spaces!!!$";
var NewString= string.replace(/[^\w\s]/gi, '');
console.log(NewString);
Result //This tool removes special characters other than digits characters and spaces
Live Example : https://helpseotools.com/text-tools/remove-special-characters

dot (.) may not be considered special. I have added an OR condition to Mozfet's & Seagull's answer:
function isNumber (text) {
reg = new RegExp('[0-9]+$');
if(text) {
return reg.test(text);
}
return false;
}
function removeSpecial (text) {
if(text) {
var lower = text.toLowerCase();
var upper = text.toUpperCase();
var result = "";
for(var i=0; i<lower.length; ++i) {
if(isNumber(text[i]) || (lower[i] != upper[i]) || (lower[i].trim() === '') || (lower[i].trim() === '.')) {
result += text[i];
}
}
return result;
}
return '';
}

Try this:
const strippedString = htmlString.replace(/(<([^>]+)>)/gi, "");
console.log(strippedString);

const input = `#if_1 $(PR_CONTRACT_END_DATE) == '23-09-2019' #
Test27919<alerts#imimobile.com> #elseif_1 $(PR_CONTRACT_START_DATE) == '20-09-2019' #
Sender539<rama.sns#gmail.com> #elseif_1 $(PR_ACCOUNT_ID) == '1234' #
AdestraSID<hello#imimobile.co> #else_1#Test27919<alerts#imimobile.com>#endif_1#`;
const replaceString = input.split('$(').join('->').split(')').join('<-');
console.log(replaceString.match(/(?<=->).*?(?=<-)/g));

Whose special characters you want to remove from a string, prepare a list of them and then user javascript replace function to remove all special characters.
var str = 'abc'de#;:sfjkewr47239847duifyh';
alert(str.replace("'","").replace("#","").replace(";","").replace(":",""));
or you can run loop for a whole string and compare single single character with the ASCII code and regenerate a new string.