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Javascript how to adding minus number and positive number

i made a simple program convert number to word. and i got problem when i want to convert minus number.
I want to find the index in the array 'satuan' so that later I add the word 'minus'
in my code i use Indonesian language btw.
js code:
// i have array like this
var satuan = ['', 'satu', 'dua', 'tiga', 'empat', 'lima', 'enam', 'tujuh', 'delapan', 'sembilan'];
// and my function to convert minus number like this
function convert_minesPuluhan(num) {
var c = Math.abs(num) + Math.abs(num);
if (num < 0 && num > -10) {
return satuan[num + c];
}
}
but when i console.log(num + c). the result is (example the num value is -1) -12.
but what I want is -1 + 1 + 1 = 1
// so i want like this
satuan[1];
how to solve this?

You are most likely getting a string for num
So, you'll have to do num = parseInt(num) first to convert it into an int datatype.
i.e.
function convert_minesPuluhan(num) {
num = parseInt(num)
var c = Math.abs(num) + Math.abs(num);
if (num < 0 && num > -10) {
return satuan[num + c];
}
}

Try this, your code is working fine. I understand that you need to add the 'minus' keyword before negative numbers.
var satuan = ['', 'satu', 'dua', 'tiga', 'empat', 'lima', 'enam', 'tujuh', 'delapan', 'sembilan'];
// and my function to convert minus number like this
function convert_minesPuluhan(num) {
var c = Math.abs(num) + Math.abs(num);
if (num < 0 && num > -10) {
console.log("minus " + satuan[num + c])
return "minus " + satuan[num + c];
}
}
convert_minesPuluhan(-5)

You need to convert the input of your function to integer
You are adding 2 strings in your function "-1" and "2" and thats why youre getting "-12" which is undefined.
let num = parseInt(num);
Try this inside your function and rest all is the same.

Wrong Convert amount

The following function works perfect, but when the amount over 1 million, the function don't work exactly.
Example:
AMOUNTPAID = 35555
The output is: 35.555,00 - work fine
But when the amount paid is for example: 1223578 (over 1 Million),
is the output the following output value: 1.223.235,00 (but it must be: 1.223.578,00) - there is a deviation of 343
Any ideas?
I call the function via HTML as follows:
<td class="tr1 td2"><p class="p2 ft4"><script type="text/javascript">document.write(ConvertBetrag('{{NETAMOUNT}}'))</script> €</P></TD>
#
Here ist the Javascript:
function Convertamount( amount ){
var number = amount;
number = Math.round(number * Math.pow(12, 2)) / Math.pow(12, 2);
number = number.toFixed(2);
number = number.toString();
var negative = false;
if (number.indexOf("-") == 0)
{
negative = true ;
number = number.replace("-","");
}
var str = number.toString();
str = str.replace(".", ",");
// number before decimal point
var intbeforedecimaln = str.length - (str.length - str.indexOf(","));
// number of delimiters
var intKTrenner = Math.floor((intbeforedecimaln - 1) / 3);
// Leading digits before the first dot
var intZiffern = (intbeforedecimaln % 3 == 0) ? 3 : (intbeforedecimaln % 3);
// Provided digits before the first thousand separator with point
strNew = str.substring(0, intZiffern);
// Auxiliary string without the previously treated digits
strHelp = str.substr(intZiffern, (str.length - intZiffern));
// Through thousands of remaining ...
for(var i=0; i<intKTrenner; i++)
{
// attach 3 digits of the nearest thousand group point to String
strNew += "." + strHelp.substring(0, 3);
// Post new auxiliary string without the 3 digits being treated
strHelp = strHelp.substr(intZiffern, (strHelp.length - intZiffern));
}
// attach a decimal
var szdecimal = str.substring(intbeforedecimaln, str.length);
if (szdecimal.length < 3 )
{
strNew += str.substring(intbeforedecimaln, str.length) + '0';
}
else
{
strNew += str.substring(intbeforedecimaln, str.length);
}
var number = strNew;
if (negative)
{
number = "- " + number ;
}
return number;
}

JavaScript's Math functions have a toLocaleString method. Why don't you just use this?
var n = (1223578.00).toLocaleString();
-> "1,223,578.00"
The locale you wish to use can be passed in as a parameter, for instance:
var n = (1223578.00).toLocaleString('de-DE');
-> "1.223.578,00"

Re-write malformed dates yyyy-M-d to yyyy-MM-dd in javascript using RegEx

So I'm rewriting dates in javacript and as familiar js spits dates like 2013-1-1 that isn't very useful always. Instead I'm looking for a routine that will form this date to the correct iso-version 2013-01-01
Today I make this by using string
var b = new Date('2013-1-1');
var result = b.getFullYear() + "-" +
(b.getMonth().toString().length == 1 ? "0" + parseInt(b.getMonth() + 1) : parseInt(b.getMonth() + 1)) + "-" +
(b.getDate().toString().length == 1 ? "0" + b.getDate() : b.getDate());
This works but it is ugly. Is there a better way to perform this using RegEx?
Please spare me of any anti-regex comments

A non-regex solution would be a generic padding function. First get your date in the non-padded version then you can split on the separator and pad it as necessary. Something like this:
var date = '2013-1-1';
var pad = function(n) {
return function(str) {
while (str.length < n) {
str = '0'+ str;
}
return str;
}
};
date = date.split(/-/g).map(pad(2)).join('-'); //=> 2013-01-01

may be this could help:
var str="2013-1-1";
var m = str.match(/^(\d{4})-(\d{1})-(\d{1})$/);
console.log([m[1], "0".concat([2]-1), "0".concat(m[3])].join('-'));

based on elclanrs suggestion I wrote an extension method
// Add 0 to single numbers
Number.prototype.padDate = function () {
// Add +1 if input is 0 (js months starts at 0)
var number = this == 0 ? 1 : this;
return number.toString().length == 1 ? "0" + number : number;
};
This allows me to build dates like this
var b = new Date('2013-1-1');
var result = b.getFullYear() + "-" + b.getMonth().padDate() + "-" + b.getDate().padDate();
Much cleaner, thanks

Javascript adding zeros to the beginning of a string (max length 4 chars)

var number = 1310;
should be left alone.
var number = 120;
should be changed to "0120";
var number = 10;
should be changed to "0010";
var number = 7;
should be changed to "0007";

In all modern browsers you can use
numberStr.padStart(4, "0");
https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/String/padStart
function zeroPad(num) {
return num.toString().padStart(4, "0");
}
var numbers = [1310, 120, 10, 7];
numbers.forEach(
function(num) {
var paddedNum = zeroPad(num);
console.log(paddedNum);
}
);

function pad_with_zeroes(number, length) {
var my_string = '' + number;
while (my_string.length < length) {
my_string = '0' + my_string;
}
return my_string;
}

try these:
('0000' + number).slice(-4);
or
(number+'').padStart(4,'0');

Here's another way. Comes from something I did that needs to be done thousands of times on a page load. It's pretty CPU efficient to hard code a string of zeroes one time, and chop as many as you need for the pad as many times as needed. I do really like the power of 10 method -- that's pretty flexible.
Anyway, this is as efficient as I could come up with:
For the original question, CHOOSE ONE of the cases...
var number = 1310;
var number = 120;
var number = 10;
var number = 7;
then
// only needs to happen once
var zeroString = "00000";
// one assignment gets the padded number
var paddedNum = zeroString.substring((number + "").length, 4) + bareNum;
//output
alert("The padded number string is: " + paddedNum);
Of course you still need to validate the input. Because this ONLY works reliably under the following conditions:
Number of zeroes in the zeroString is desired_length + 1
Number of digits in your starting number is less than or equal to your desired length
Backstory:
I have a case that needs a fixed length (14 digit) zero-padded number. I wanted to see how basic I could make this. It's run tens of thousands of times on a page load, so efficiency matters. It's not quite re-usable as-is, and it's a bit inelegant. Except that it is very very simple.
For desired n digits padded string, this method requires a string of (at least) n+1 zeroes. Index 0 is the first character in the string, which won't ever be used, so really, it could be anything.
Note also that string.substring() is different from string.substr()!
var bareNum = 42 + '';
var zeroString = "000000000000000";
var paddedNum = zeroString.substring(bareNumber.length, 14) + bareNum
This pulls zeroes from zeroString starting at the position matching the length of the string, and continues to get zeroes to the necessary length of 14. As long as that "14" in the third line is a lower integer than the number of characters in zeroString, it will work.

function pad(n, len) {
return (new Array(len + 1).join('0') + n).slice(-len);
}
might not work in old IE versions.

//to: 0 - to left, 1 - to right
String.prototype.pad = function(_char, len, to) {
if (!this || !_char || this.length >= len) {
return this;
}
to = to || 0;
var ret = this;
var max = (len - this.length)/_char.length + 1;
while (--max) {
ret = (to) ? ret + _char : _char + ret;
}
return ret;
};
Usage:
someString.pad(neededChars, neededLength)
Example:
'332'.pad('0', 6); //'000332'
'332'.pad('0', 6, 1); //'332000'

An approach I like is to add 10^N to the number, where N is the number of zeros you want. Treat the resultant number as a string and slice off the zeroth digit. Of course, you'll want to be careful if your input number might be larger than your pad length, but it's still much faster than the loop method:
// You want to pad four places:
>>> var N = Math.pow(10, 4)
>>> var number = 1310
>>> number < N ? ("" + (N + number)).slice(1) : "" + number
"1310"
>>> var number = 120
>>> number < N ? ("" + (N + number)).slice(1) : "" + number
"0120"
>>> var number = 10
>>> number < N ? ("" + (N + number)).slice(1) : "" + number
"0010"
…
etc. You can make this into a function easily enough:
/**
* Pad a number with leading zeros to "pad" places:
*
* #param number: The number to pad
* #param pad: The maximum number of leading zeros
*/
function padNumber(number, pad) {
var N = Math.pow(10, pad);
return number < N ? ("" + (N + number)).slice(1) : "" + number
}

I wrote a general function for this. It takes an input control and pad length as input.
function padLeft(input, padLength) {
var num = $("#" + input).val();
$("#" + input).val(('0'.repeat(padLength) + num).slice(-padLength));
}

With RegExp/JavaScript:
var number = 7;
number = ('0000'+number).match(/\d{4}$/);
console.log(number);
With Function/RegExp/JavaScript:
var number = 7;
function padFix(n) {
return ('0000'+n).match(/\d{4}$/);
}
console.log(padFix(number));

No loop, no functions
let n = "" + 100;
let x = ("0000000000" + n).substring(n.length);//add your amount of zeros
alert(x + "-" + x.length);

Nate as the best way I found, it's just way too long to read. So I provide you with 3 simples solutions.
1. So here's my simplification of Nate's answer.
//number = 42
"0000".substring(number.toString().length, 4) + number;
2. Here's a solution that make it more reusable by using a function that takes the number and the desired length in parameters.
function pad_with_zeroes(number, len) {
var zeroes = "0".repeat(len);
return zeroes.substring(number.toString().length, len) + number;
}
// Usage: pad_with_zeroes(42,4);
// Returns "0042"
3. Here's a third solution, extending the Number prototype.
Number.prototype.toStringMinLen = function(len) {
var zeroes = "0".repeat(len);
return zeroes.substring(self.toString().length, len) + self;
}
//Usage: tmp=42; tmp.toStringMinLen(4)

Use String.JS librairy function padLeft:
S('123').padLeft(5, '0').s --> 00123

How to output numbers with leading zeros in JavaScript? [duplicate]

This question already has answers here:
How can I pad a value with leading zeros?
(76 answers)
Closed 3 years ago.
Is there a way to prepend leading zeros to numbers so that it results in a string of fixed length? For example, 5 becomes "05" if I specify 2 places.

NOTE: Potentially outdated. ECMAScript 2017 includes String.prototype.padStart.
You'll have to convert the number to a string since numbers don't make sense with leading zeros. Something like this:
function pad(num, size) {
num = num.toString();
while (num.length < size) num = "0" + num;
return num;
}
Or, if you know you'd never be using more than X number of zeros, this might be better. This assumes you'd never want more than 10 digits.
function pad(num, size) {
var s = "000000000" + num;
return s.substr(s.length-size);
}
If you care about negative numbers you'll have to strip the - and read it.

UPDATE: Small one-liner function using the ES2017 String.prototype.padStart method:
const zeroPad = (num, places) => String(num).padStart(places, '0')
console.log(zeroPad(5, 2)); // "05"
console.log(zeroPad(5, 4)); // "0005"
console.log(zeroPad(5, 6)); // "000005"
console.log(zeroPad(1234, 2)); // "1234"
Another ES5 approach:
function zeroPad(num, places) {
var zero = places - num.toString().length + 1;
return Array(+(zero > 0 && zero)).join("0") + num;
}
zeroPad(5, 2); // "05"
zeroPad(5, 4); // "0005"
zeroPad(5, 6); // "000005"
zeroPad(1234, 2); // "1234" :)

You could extend the Number object:
Number.prototype.pad = function(size) {
var s = String(this);
while (s.length < (size || 2)) {s = "0" + s;}
return s;
}
Examples:
(9).pad(); //returns "09"
(7).pad(3); //returns "007"

From https://gist.github.com/1180489
function pad(a, b){
return(1e15 + a + '').slice(-b);
}
With comments:
function pad(
a, // the number to convert
b // number of resulting characters
){
return (
1e15 + a + // combine with large number
"" // convert to string
).slice(-b) // cut leading "1"
}

function zfill(num, len) {return (Array(len).join("0") + num).slice(-len);}

Just for fun (I had some time to kill), a more sophisticated implementation which caches the zero-string:
pad.zeros = new Array(5).join('0');
function pad(num, len) {
var str = String(num),
diff = len - str.length;
if(diff <= 0) return str;
if(diff > pad.zeros.length)
pad.zeros = new Array(diff + 1).join('0');
return pad.zeros.substr(0, diff) + str;
}
If the padding count is large and the function is called often enough, it actually outperforms the other methods...