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regex to extract numbers starting from second symbol

Sorry for one more to the tons of regexp questions but I can't find anything similar to my needs. I want to output the string which can contain number or letter 'A' as the first symbol and numbers only on other positions. Input is any string, for example:
---INPUT--- -OUTPUT-
A123asdf456 -> A123456
0qw#$56-398 -> 056398
B12376B6f90 -> 12376690
12A12345BCt -> 1212345
What I tried is replace(/[^A\d]/g, '') (I use JS), which almost does the job except the case when there's A in the middle of the string. I tried to use ^ anchor but then the pattern doesn't match other numbers in the string. Not sure what is easier - extract matching characters or remove unmatching.

I think you can do it like this using a negative lookahead and then replace with an empty string.
In an non capturing group (?:, use a negative lookahad (?! to assert that what follows is not the beginning of the string followed by ^A or a digit \d. If that is the case, match any character .
(?:(?!^A|\d).)+
var pattern = /(?:(?!^A|\d).)+/g;
var strings = [
"A123asdf456",
"0qw#$56-398",
"B12376B6f90",
"12A12345BCt"
];
for (var i = 0; i < strings.length; i++) {
console.log(strings[i] + " ==> " + strings[i].replace(pattern, ""));
}

You can match and capture desired and undesired characters within two different sides of an alternation, then replace those undesired with nothing:
^(A)|\D
JS code:
var inputStrings = [
"A-123asdf456",
"A123asdf456",
"0qw#$56-398",
"B12376B6f90",
"12A12345BCt"
];
console.log(
inputStrings.map(v => v.replace(/^(A)|\D/g, "$1"))
);

You can use the following regex : /(^A)?\d+/g
var arr = ['A123asdf456','0qw#$56-398','B12376B6f90','12A12345BCt', 'A-123asdf456'],
result = arr.map(s => s.match(/(^A|\d)/g).join(''));
console.log(result);

Need help finding a plus sign using javascript regex

I am using the code below to find a match for a plus sign but it keeps returning false. I am not sure what I am doing wrong. Any help will be really appreciated it. Thanks!
var str = '+2443';
var result = /d\+1/.test(str);
console.log(result); // true

var str = '+2443';
var result = /\+/.test(str);
console.log(result); // true

Your /d\+1/ regex matches the first occurrence of a d+1 substring in any string.
To check if a string contains a +, you do not need a regex. Use indexOf:
var str = '+2443';
if (~str.indexOf("+")) {
console.log("Found a `+`");
} else {
console.log("A `+` is not found");
}
A regex will be more appropriate when you need to match a + in some context. For example, to check if the string starts with a plus, and then only contains digits, you would use
var str = '+2443';
var rx = /^\+\d+$/;
console.log(rx.test(str));
where ^ assets the position at the end of the string, \+ matches a literal +, \d+ matches 1+ digits and the $ anchor asserts the position at the end of the string.

Extract word between '=' and '('

I have the following string
234234=AWORDHERE('sdf.'aa')
where I need to extract AWORDHERE.
Sometimes there can be space in between.
234234= AWORDHERE('sdf.'aa')
Can I do this with a regular expression?
Or should I do it manually by finding indexes?
The datasets are huge, so it's important to do it as fast as possible.

Try this regex:
\d+=\s?(\w+)\(
Check Demo
in Javascript it would like that:
var myString = "234234=AWORDHERE('sdf.'aa')";// or 234234= AWORDHERE('sdf.'aa')
var myRegexp = /\d+=\s?(\w+)\(/g;
var match = myRegexp.exec(myString);
console.log(match[1]); // AWORDHERE

You could do this at least three ways. You need to benchmark to see what's fastest.
Substring w/ indexes
function extract(from) {
var ixEq = from.indexOf("=");
var ixParen = from.indexOf("(");
return from.substring(ixEq + 1, ixParen);
}
.
Splits
function extract(from) {
var spEq = from.split("=");
var spParen = spEq[1].split("(");
return spParen[0];
}
Regex (demo)
Here is some sample regex you could use
/[^=]+=([^(]+).*/g
This says
[^=]+ - One or more character which is not an =
= - The = itself
( - creates a matching group so you can access your match in code
[^(]+ - One or more character which is not a (
) - closes the matching group
.* - Matches the rest of the line
the /g on the end tells it to perform the match on all lines.

Using look around you can search for string preceded by = and followed by ( as following.
Regex: (?<==)[A-Z ]+(?=\()
Explanation:
(?<==) checks if [A-Z ] is preceded by an =.
[A-Z ]+ matches your pattern.
(?=\() checks if matched pattern is followed by a (.
Regex101 Demo

var str = "234234= AWORDHERE('sdf.'aa')";
var regexp = /.*=\s+(\w+)\(.*\)/g;
var match = regexp.exec(str);
alert( match[1] );

I made my solution for this just a little more general than you asked for, but I don't think it takes much more time to execute. I didn't measure. If you need greater efficiency than this provides, comment and I or someone else can help you with that.
Here's what I did, using the command prompt of node:
> var s = "234234= AWORDHERE('sdf.'aa')"
undefined
> var a = s.match(/(\w+)=\s*(\w+)\s*\(.*/)
undefined
> a
[ '234234= AWORDHERE(\'sdf.\'aa\')',
'234234',
'AWORDHERE',
index: 0,
input: '234234= AWORDHERE(\'sdf.\'aa\')' ]
>
As you can see, this matches the number before the = in a[1], and it matches the AWORDHERE name as you requested in a[2]. This will work with any number (including zero) spaces before and/or after the =.

What's the exact regex to match the proper string?

My string has [1212,1212],[1212,11212],...
I'd like to extract each value into an array for example I'd want 1212,1212 as one pair and evaluate a series of steps.
Tried /[[0-9],[0-9]]/ but It wasn't doing the task as I wanted. Basically I'm a noob in Regex, could someone please help.
Thanks in advance

You need some modifications for your regular expression for it to work correctly:
/\[[0-9]+,[0-9]+\]/g
You need to escape square brackets [ because they have special meaning.
[0-9] matches only one digits, you need the + quantifier to match one or more digits and thus [0-9]+.
Use the global modifier g to extract all matches.
Then you can extract all the values into an array like this:
var input = "[1212,1212],[1212,11212]";
var pattern = /\[[0-9]+,[0-9]+\]/g;
var result = [];
var currentMatch;
while((currentMatch = pattern.exec(input)) != null) {
result.push(currentMatch.toString());
}
result;
Or if you don't need to find the matches successively one at a time, then you can use String.match() as #Matthew Mcveigh did:
var input = "[1212,1212],[1212,11212]";
var result = input.match(/\[[0-9]+,[0-9]+\]/g);

It seems like you just need to match one or more digits before and after a comma, so you could do the following:
"[1212,1212],[1212,11212]".match(/\d+,\d+/g)
Which will give you the array: ["1212,1212", "1212,11212"]
To extract the pairs:
var result = "[1212,1212],[1212,11212]".match(/\d+,\d+/g);
for (var i = 0; i < result.length; i++) {
var pair = result[i].match(/\d+/g),
left = pair[0], right = pair[1];
alert("left: " + left + ", right: " + right);
}

You need to escape the literal brackets that you want to match. You can also use \d to match "any digit", which makes it tidier. Also, you're only matching one digit. You need to match "one or more" (+ quantifier)
/\[\d+,\d+\]/g
That g modifier finds all matches in the string, otherwise only the first one is found.

How can I remove all characters up to and including the 3rd slash in a string?

I'm having trouble with removing all characters up to and including the 3 third slash in JavaScript. This is my string:
http://blablab/test
The result should be:
test
Does anybody know the correct solution?

To get the last item in a path, you can split the string on / and then pop():
var url = "http://blablab/test";
alert(url.split("/").pop());
//-> "test"
To specify an individual part of a path, split on / and use bracket notation to access the item:
var url = "http://blablab/test/page.php";
alert(url.split("/")[3]);
//-> "test"
Or, if you want everything after the third slash, split(), slice() and join():
var url = "http://blablab/test/page.php";
alert(url.split("/").slice(3).join("/"));
//-> "test/page.php"

var string = 'http://blablab/test'
string = string.replace(/[\s\S]*\//,'').replace(/[\s\S]*\//,'').replace(/[\s\S]*\//,'')
alert(string)
This is a regular expression. I will explain below
The regex is /[\s\S]*\//
/ is the start of the regex
Where [\s\S] means whitespace or non whitespace (anything), not to be confused with . which does not match line breaks (. is the same as [^\r\n]).
* means that we match anywhere from zero to unlimited number of [\s\S]
\/ Means match a slash character
The last / is the end of the regex

var str = "http://blablab/test";
var index = 0;
for(var i = 0; i < 3; i++){
index = str.indexOf("/",index)+1;
}
str = str.substr(index);
To make it a one liner you could make the following:
str = str.substr(str.indexOf("/",str.indexOf("/",str.indexOf("/")+1)+1)+1);

You can use split to split the string in parts and use slice to return all parts after the third slice.
var str = "http://blablab/test",
arr = str.split("/");
arr = arr.slice(3);
console.log(arr.join("/")); // "test"
// A longer string:
var str = "http://blablab/test/test"; // "test/test";

You could use a regular expression like this one:
'http://blablab/test'.match(/^(?:[^/]*\/){3}(.*)$/);
// -> ['http://blablab/test', 'test]
A string’s match method gives you either an array (of the whole match, in this case the whole input, and of any capture groups (and we want the first capture group)), or null. So, for general use you need to pull out the 1th element of the array, or null if a match wasn’t found:
var input = 'http://blablab/test',
re = /^(?:[^/]*\/){3}(.*)$/,
match = input.match(re),
result = match && match[1]; // With this input, result contains "test"

let str = "http://blablab/test";
let data = new URL(str).pathname.split("/").pop();
console.log(data);