Develop Reference

Finding ++ in Regular Expression

I want to find ++ or -- or // or ** sign in in string can anyone help me?
var str = document.getElementById('screen').innerHTML;
var res = str.substring(0, str.length);
var patt1 = ++,--,//,**;
var result = str.match(patt1);
if (result)
{
alert("you cant do this :l");
document.getElementById('screen').innerHTML='';
}

This finds doubles of the characters by a backreference:
/([+\/*-])\1/g
[from q. comments]: i know this but when i type var patt1 = /[++]/i; code find + and ++
[++] means one arbitrary of the characters. Normally + is the qantifier "1 or more" and needs to be escaped by a leading backslash when it should be a literal, except in brackets where it does not have any special meaning.
Characters that do need to be escaped in character classes are e.g. the escape character itself (backslash), the expression delimimiter (slash), the closing bracket and the range operator (dash/minus), the latter except at the end of the character class as in my code example.
A character class [] matches one character. A quantifier, e.g. [abc]{2} would match "aa", "bb", but "ab" as well.
You can use a backreference to a match in parentheses:
/(abc)\1
Here the \1 refers to the first parentheses (abc). The entire expression would match "abcabc".
To clarify again: We could use a quantifier on the backreference:
/([+\/*-])\1{9}/g
This matches exactly 10 equal characters out of the class, the subpattern itself and 9 backreferences more.
/.../g finds all occurrences due to the modifier global (g).
test-case on regextester.com

Define your pattern like this:
var patt1 = /\+\+|--|\/\/|\*\*/;
Now it should do what you want.
More info about regular expressions: https://developer.mozilla.org/en-US/docs/Web/JavaScript/Guide/Regular_Expressions

You can use:
/\+\+|--|\/\/|\*\*/
as your expression.
Here I have escaped the special characters by using a backslash before each (\).
I've also used .test(str) on the regular expression as all you need is a boolean (true/false) result.
See working example below:
var str = document.getElementById('screen').innerHTML;
var res = str.substring(0, str.length);
var patt1 = /\+\+|--|\/\/|\*\*/;
var result = patt1.test(res);
if (result) {
alert("you cant do this :l");
document.getElementById('screen').innerHTML = '';
}
<div id="screen">
This is some++ text
</div>

Try this:-
As
n+:- Matches any string that contains at least one n
n* Matches any string that contains zero or more occurrences of n
We need to use backslash before this special characters.
var str = document.getElementById('screen').innerHTML;
var res = str.substring(0, str.length);
var patt1 = /\+\+|--|\/\/|\*\*/;
var result = str.match(patt1);
if (result)
{
alert("you cant do this :l");
document.getElementById('screen').innerHTML='';
}
<div id="screen">2121++</div>

Replace all “?” by “&” except first

I’d would to replace all “?” by “&” except the first one by javascript. I found some regular expressions but they didn’t work.
I have something like:
home/?a=1
home/?a=1?b=2
home/?a=1?b=2?c=3
And I would like:
home/?a=1
home/?a=1&b=2
home/?a=1&b=2&c=3
Someone know how to I can do it?
Thanks!

I don't think it's possible with regex but you can split the string and then join it back together, manually replacing the first occurance:
var split = 'home/?a=1?b=2'.split('?'); // [ 'home/', 'a=1', 'b=2' ]
var replaced = split[0] + '?' + split.slice(1).join('&') // 'home/?a=1&b=2'
console.log(replaced);

You could match from the start of the string not a question mark using a negated character class [^?]+ followed by matching a question mark and capture that in the first capturing group. In the second capturing group capture the rest of the string.
Use replace and pass a function as the second parameter where you return the first capturing group followed by the second capturing group where all the question marks are replaced by &
let strings = [
"home/?a=1",
"home/?a=1?b=2",
"home/?a=1?b=2?c=3"
];
strings.forEach((str) => {
let result = str.replace(/(^[^?]+\?)(.*)/, function(match, group1, group2) {
return group1 + group2.replace(/\?/g, '&')
});
console.log(result);
});

You can split it by "?" and then rewrap the array:
var string = "home/?a=1?b=2";
var str = string.split('?');
var new = str[0] + '?'; // text before first '?' and first '?'
for( var x = 1; x < str.length; x++ ) {
new = new + str[x];
if( x != ( str.length - 1 ) ) new = new + '&'; //to avoid place a '&' after the string
}

You can use /([^\/])\?/ as pattern in regex that match any ? character that isn't after / character.
var str = str.replace(/([^\/])\?/g, "$1&");
var str = "home/?a=1\nhome/?a=1?b=2\nhome/?a=1?b=2?c=3\n".replace(/([^\/])\?/g, "$1&");
console.log(str);

Javascript regex to add to every character of a string a backslash

So as the title says I'd like to add to every character of a string a backslash, whether the string has special characters or not. The string should not be considered 'safe'
eg:
let str = 'dj%^3&something';
str = str.replace(x, y);
// str = '\d\j\%\^\3\&\s\o\m\e\t\h\i\n\g'

You could capture every character in the string with (.) and use \\$1 as replacement, I'm not an expert but basically \\ will render to \ and $1 will render to whatever (.) captures.
HIH
EDIT
please refer to Wiktor Stribiżew's comment for an alternative which will require less coding. Changes as follows:
str = str.replace(/(.)/g, '\\$1'); for str = str.replace(/./g, '\\$&');
Also, for future reference I strongly advice you to visit regexr.com when it comes to regular expressions, it's helped ME a lot
let str = 'dj%^3&something';
str = str.replace(/(.)/g, '\\$1');
console.log(str);

If you just want to display a string safely, you should just do:
let str = 'dj%^3&something';
let node = document.createTextNode(str);
let dest = document.querySelector('.whatever');
dest.appendChild(node);
And then you are guaranteed that it will be treated as text, and won't be able to execute a script or anything.
For example: https://jsfiddle.net/s6udj03L/1/

You can split the string to an array, add a \ to each element to the array, then joint the array back to the string that you wanted.
var str = 'dj%^3&something';
var split = str.split(""); // split string into array
for (var i = 0; i < split.length; i++) {
split[i] = '\\' + split[i]; // add backslash to each element in array
}
var joint = split.join('') // joint array to string
console.log(joint);

If you don't care about creating a new string and don't really have to use a regex, why not just iterate over the existing one and place a \ before each char. Notice to you have to put \\ to escape the first \.
To consider it safe, you have to encode it somehow. You could replace typical 'non-safe' characters like in the encode() function below. Notice how & get's replaced by &
let str = 'dj%^3&something';
let out = "";
for(var i = 0; i < str.length; i++) {
out += ("\\" + str[i]);
}
console.log(out);
console.log(encode(out));
function encode(string) {
return String(string).replace(/&/g, '&').replace(/</g, '<').replace(/>/g, '>').replace(/"/g, '"');
}

Replace last character of string using JavaScript

I have a very small query. I tried using concat, charAt, slice and whatnot but I didn't get how to do it.
Here is my string:
var str1 = "Notion,Data,Identity,"
I want to replace the last , with a . it should look like this.
var str1 = "Notion,Data,Identity."
Can someone let me know how to achieve this?

You can do it with regex easily,
var str1 = "Notion,Data,Identity,".replace(/.$/,".")
.$ will match any character at the end of a string.

You can remove the last N characters of a string by using .slice(0, -N), and concatenate the new ending with +.
var str1 = "Notion,Data,Identity,";
var str2 = str1.slice(0, -1) + '.';
console.log(str2);
Notion,Data,Identity.
Negative arguments to slice represents offsets from the end of the string, instead of the beginning, so in this case we're asking for the slice of the string from the beginning to one-character-from-the-end.

This isn't elegant but it's reusable.
term(str, char)
str: string needing proper termination
char: character to terminate string with
var str1 = "Notion,Data,Identity,";
function term(str, char) {
var xStr = str.substring(0, str.length - 1);
return xStr + char;
}
console.log(term(str1,'.'))

You can use simple regular expression
var str1 = "Notion,Data,Identity,"
str1.replace(/,$/,".")

Getting each 'word' after every underscore in a string in Javascript using regex

I'm wanting to extract each block of alphanumeric characters that come after underscores in a Javascript string. I currently have it working using a combination of string methods and regex like so:
var string = "ignore_firstMatch_match2_thirdMatch";
var firstValGone = string.substr(string.indexOf('_'));
// returns "_firstMatch_match2_thirdMatch"
var noUnderscore = firstValGone.match(/[^_]+/g);
// returns ["firstMatch", "match2" , "thirdMatch"]
I'm wondering if there's a way to do it purely using regex? Best I've managed is:
var string = "ignore_firstMatch_match2_thirdMatch";
var matchTry = string.match(/_[^_]+/g);
// returns ["_firstMatch", "_match2", "_thirdMatch"]
but that returns the preceding underscore too. Given you can't use lookbehinds in JS I don't know how to match the characters after, but exclude the underscore itself. Is this possible?

You can use a capture group (_([^_]+)) and use RegExp#exec in a loop while pushing the captured values into an array:
var re = /_([^_]+)/g;
var str = 'ignore_firstMatch_match2_thirdMatch';
var res = [];
while ((m = re.exec(str)) !== null) {
res.push(m[1]);
}
document.body.innerHTML = "<pre>" + JSON.stringify(res, 0, 4) + "</pre>";
Note that using a string#match() with a regex defined with a global modifier /g will lose all the captured texts, that's why you cannot just use str.match(/_([^_]+)/g).

Since lookbehind is not supported in JS the only way I can think of is using a group like this.
Regex: _([^_]+) and capture group using \1 or $1.
Regex101 Demo
var myString = "ignore_firstMatch_match2_thirdMatch";
var myRegexp = /_([^_]+)/g;
match = myRegexp.exec(myString);
while (match != null) {
document.getElementById("match").innerHTML += "<br>" + match[0];
match = myRegexp.exec(myString);
}
<div id="match">
</div>
An alternate way using lookahead would be something like this.
But it takes long in JS. Killed my page thrice. Would make a good ReDoS exploit
Regex: (?=_([A-Za-z0-9]+)) and capture groups using \1 or $1.
Regex101 Demo

Why do you assume you need regex? a simple split will do the job:
string str = "ignore_firstMatch_match2_thirdMatch";
IEnumerable<string> matches = str.Split('_').Skip(1);