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RegEx change function name and parameter of string

I'm awful with RegEx to begin with. Anyway I tried my best and think I got pretty far, but I'm not exactly there yet...
What I have:
A javascript source file that I need to process in Node.js. Can look like that:
var str = "require(test < 123)\n\nrequire(test2 !== test)\n\nfunction(dontReplaceThisParam) {\n console.log(dontReplaceThisParam)\n}";
What I came up with:
console.log(str.replace(/\(\s*([^)].+?)\s*\)/g, 'Debug$&, \'error_1\''))
Theres a few problems:
I want that the string error gets inside the paranthesis so it acts as a second parameter.
All function calls, or I think even everything with paranthesis will be replaced. But only function calls to "require(xxx)" should be touched.
Also, the error codes should somehow increment if possible...
So a string like "require(test == 123)" should convert to "requireDebug(test == 123, 'error_N')" but only calls to "require"...
What currently gets outputted by my code:
requireDebug(test < 123), 'error_1'
requireDebug(test2 !== test), 'error_1'
functionDebug(dontReplaceThisParam), 'error_1' {
console.logDebug(dontReplaceThisParam), 'error_1'
}
What I need:
requireDebug(test < 123, 'error_1')
requireDebug(test2 !== test, 'error_2')
function(dontReplaceThisParam) {
console.log(dontReplaceThisParam)
}
I know I could just do things like that manually but we're talking here about a few hundred source files. I also know that doing such things is not a very good way, but the debugger inside the require function is not working so I need to make my own debug function with an error code to locate the error. Its pretty much all I can do at the moment...
Any help is greatly appreciated!

Start the regex with require, and since you need an incrementing counter, pass a function as the second arg to replace, so that you can increment and insert the counter for each match.
var str = "require(test < 123)\n\nrequire(test2 !== test)\n\nfunction(dontReplaceThisParam) {\n console.log(dontReplaceThisParam)\n}";
var counter = 0;
console.log(str.replace(/require\(\s*([^)].+?)\s*\)/g, (s, g2) =>
`requireDebug(${g2}, \'error_${++counter}\')`
));
Other than that, your code was unaltered.

Why do I get undefined here?

function myFunc() {
var word = document.getElementById("Text1").value;
var num = parseInt(document.getElementById("Text2").value);
var numstr = num.split(",");
var wordstr = word.split("");
for (i = 0; i < word.length; i++) {
}
document.getElementById("myDiv").innerHTML += (wordstr[(numstr[i])-1]);
}
did I parseInt incorrectly? I've tried toString(), with ParseInt it doesn't do anything and without it I get 'undefined'

The parseInt() function parses a string and returns an integer.
You check your input with id "Text2" and show your HTML here to clearify the issue.

Without knowing more about your problem, it looks like you are misunderstanding how parseInt() works. Despite the misleading name, it will read your string character by character, attempting to create an integer. It will stop as soon as it finds a character that can't be part of an integer.
If you pass it "1,2,3,4" then it will read the 2 fine, but as a comma cannot be parsed as part of an integer, it will return the number 2. It doesn't make sense to call split on a number.
As others have said, you really need to give us more details for us to be able to help, but I suspect a large part of the problem is not understanding what some of these functions do.
Maybe you could explain what you're trying to achieve, then we can help you get there. Right now, your code isn't clear enough without extra information.

Format numbers to integers with SI prefix

I am drawing a chart, which gets dollar values (from 0 to millions), and I am trying to show nice ticks. I already used d3.nice to get 5 ticks that all have nice values, it's very cool. But since there's such a large variance, I am struggling to display my dollar values correctly.
I wish to do:
0-999: shows itself
1,000 - 999,999: shows 1k-999k (it's ok if 999,500 shows 1M, but not ok to show 1.00k by using d3.format('.3s'), or having 467k go to 400k by using d3.format('.1s'))
1,000,000 - 999,999,999: shows 1M-999M (also ok if it rolls over when rounding)
Prior to d3 version 4, this was easy. You could do:
.ticks((d) => {
var prefix = d3.formatPrefix(d);
return prefix(d).toFixed()+''+prefix.symbol;
})
But now, I am reading the d3 v4 docs after this fails on me, and it says:
The d3.formatPrefix method has been changed. Rather than returning an SI-prefix string, it returns an SI-prefix format function for a given specifier and reference value. For example, to format thousands:
var f = d3.formatPrefix(",.0", 1e3);
f(1e3); // "1k"
f(1e4); // "10k"
f(1e5); // "100k"
f(1e6); // "1,000k"
This seems impossible to accomplish now, then, because I want to vary the amount of significant digits, but I see no obvious way to accomplish that. Am I missing something simple?

If I understand your desired output correctly, you can use the number itself to define the value of d3.formatPrefix.
Look at this demo:
[1, 999, 1000, 999000, 1000000].forEach(function(d) {
var prefix = d3.formatPrefix(".0", d)
console.log(d + ": " + prefix(d))
})
<script src="https://d3js.org/d3.v4.min.js"></script>
PS: Using this approach, 999500 will give you 1000k, not 1M.

Okay well I suppose this was easy enough, but a little hacky. Here's the formatter function I used to make this behaviour:
(val) => {
let prefix = d3.formatPrefix('.3s', val);
let str = prefix(val);
if (val === 0) {
return '0';
}
return parseInt(str.slice(0, str.length - 2))+''+str.slice(str.length - 1);
}
I would be happy to use something better so if you have a cleaner answer, post and I'll accept.

How to reverse a string recursively

I was in an interview the other day and I was asked to write a method that reverses a string recursively.
I started writing a method that calls itself and got stuck.
Here is what I was asked, reverse a string "Obama" recursively in JavaScript.
Here is how far I got.
function reverseString(strToReverse)
{
reverseString(strToReverse);
};
And got stuck, they said NO for i loops.
Anyone got any ideas?

Look at it this way: the reversed string will start with the last letter of the original, followed by all but the last letter, reversed.
So:
function reverseString(strToReverse)
{
if (strToReverse.length <= 1)
return strToReverse;
// last char +
// 0 .. second-last-char, reversed
return strToReverse[strToReverse.length - 1] +
reverseString( strToReverse.substring(0, strToReverse.length - 1) );
}

See the solution by #MaxZoom below for a more concise version.
Note that the tail-recursive style in my own answer provides no advantage over the plain-recursive version since JS interpreters are not required to perform tail call elimination.
[Original]
Here's a tail recursive version that works by removing a character from the front of the input string and prepending it to the front of the "accumulator" string:
function reverse(s, acc) {
if (s.length <= 0) { return acc; }
return reverse(s.slice(1), s.charAt(0) + (acc||''));
}
reverse('foobar'); // => "raboof"

The simplest one:
function reverse(input) {
if (input == null || input.length < 2) return input;
return reverse(input.substring(1)) + input.charAt(0);
}
console.log(reverse('Barack Obama'));

Single line solution. And if they asked I tell them it's recursive in the native code part and not to ask any more stupid questions.
var result = "Obama".split('').reverse().join('');
Output: amabO

The real problem here is not "how to reverse a string". The real problem is, "do you understand recursion". That is what the interview question is about!
So, in order to solve the problem, you need to show you know what recursion is about, not that you can reverse the string "Obama". If all you needed to do was reverse the string "Obama", you could write return "amabO"; see?
In other words, this specific programming task is not what it's all about! The real solution is not to copy and paste the code from the answers here, but to know about recursion.
In brief,
Recursion involves calling the same function again, yes, but that's not all
In order to prevent stack overflow, you MUST ensure that the function doesn't call itself indefinitely
So there's always a condition under which the function can exit without calling itself (again)
And when it does call itself again, it should do so with parameters that make the above condition more likely.
In the case of string operations, one way to make that all happen is to make sure that it calls itself only with strings that are shorter than the one it was called with. Since strings are not of an infinite length, the function can't call itself an infinite number of times that way. So the condition can be that the string has a length of zero, in which case it's impossible to call itself with a shorter string.
If you can prove that you know all this, and can use it in a real world program, then you're on your way to passing the interview. Not if you copy and paste some source you found on the internet.
Hope this helps!

We can easily reverse a string in the recursion method using the ternary operator
function reverseString(strToReverse) {
return str.length > 1 ? reverse(str.slice(1)) + str.charAt(0) : str;
}
reverseString("America");

Not the smartest way to reverse a string, but it is recursive:
function reverse(input, output) {
output = output || '';
if (input.length > 0) {
output = output.concat(input[input.length -1]);
return reverse(input.substr(0, input.length - 1), output);
}
return output;
}
console.log(reverse('Obama'));
Here's a jsfiddle

Maybe something like this?
var base = 'Obama',
index = base.length,
result = '';
function recursive(){
if (index == 0) return;
index -= 1;
result += base[index];
recursive();
}
recursive();
alert(result);
jsfiddle:
https://jsfiddle.net/hy1d84jL/
EDIT: You can think of recursion as an infinite for..loop. Let's just use it in "controlled" way and define the bounds - 0 for minimum and the length of Obama word as the maximum. Now, let's just make it call itself whatever number of times and do what you need in order to reverse the string, which is - decrement the index variable by one and sum the character from the end. Hope it helps. Nice question.

If the function can only have the single input i would split the string into smaller and smaller pieces, and add them all together in the reverse order
function reverseString(strToReverse){
if (strToReverse.length <= 1) { return strToReverse; }
return reverseString(strToReverse.substr(1, strToReverse.length - 1) + strToReverse[0];
}

JavaScript if Statement .length not working. Why?

I have this if statement i have came up with here:
var TotalMoney=0;
var Orbs=0;
if (TotalMoney.length==2) {
Orbs+=1;
}
What this code is supposed to do is if the the "TotalMoney" value digit length equals 2,
example (the number 10 has 2 digits)
then it will add 1 "Orb" to the "Orbs" value. Currently, it does nothing. There is HTML and CSS linked to this code but i figured the problem is in this code as it works fine for everything else. Please fix it as i have been trying for hours. Thanks!
For my second question that i just found out with this code here:
var totalMoney=0;
var orbs=0;
if (totalMoney.toString().length==2) {
orbs+=1;
}
This works on getting the number value digits as 2 digits long. The problem now is that once it reaches 10, every time that number goes up (10-99) all the way up, it will add 1 orb each time. I only want it to add 1 orb only when it gets to the 2 digit number (10) and stops adding 1 orb after it reaches it. How can i achieve this? Thanks!

TotalMoney is a number, so it doesn't have a length property. You can check the length of the number by first converting to a string: TotalMoney.toString().length.

Number object in js has no length property, so TotalMoney.length return undefined.
If you want count digits you may use this:
if (TotalMoney.toString().length == 2) {
Orbs+=1;
}
But if TotalMoney will be negative, -1 for exmple, Orbs wil be incremented.
I think there are better way to find all 2-digits number:
if (TotalMoney>9 && TotalMoney<100) {
Orbs+=1;
}

TotalMoney is numeric
so to find its length use this code
TotalMoney.toString().length;
Instead of
TotalMoney.length;
so try to modify your code as below:
var TotalMoney=0;
var Orbs=0;
if (TotalMoney.toString().length==2) {
Orbs+=1;
}

Length is property of array & string.It can not be applied on other variables.
If you want to count number of digits you can do
if(TotalMoney>9)
Or you can convert it to string then check it's length
if(TotalMoney.toSting().length>2)

here are some ideas and general comments on your code.
// recommended to start with lower case. upper case such as 'TotalMoney'
// is stylistically reserved for constructors.
var totalMoney=0;
// again - changing form Orbs to orbs;
var orbs=0;
// recommended to use '===' until you are more experienced with JavaScript and
// know about the 'gotchas' that '==' might produce.
// you will be able to check the length of totalMoney only after converting it to a string.
if (totalMoney.toString().length === 2) {
orbs+=1;
}
Finally, totalMoney of 0 will not add one to orbs. But totalMoney of 10, as you mentioned, will.