In Perl I can repeat a character multiple times using the syntax:
$a = "a" x 10; // results in "aaaaaaaaaa"
Is there a simple way to accomplish this in Javascript? I can obviously use a function, but I was wondering if there was any built in approach, or some other clever technique.
These days, the repeat string method is implemented almost everywhere. (It is not in Internet Explorer.) So unless you need to support older browsers, you can simply write:
"a".repeat(10)
Before repeat, we used this hack:
Array(11).join("a") // create string with 10 a's: "aaaaaaaaaa"
(Note that an array of length 11 gets you only 10 "a"s, since Array.join puts the argument between the array elements.)
Simon also points out that according to this benchmark, it appears that it's faster in Safari and Chrome (but not Firefox) to repeat a character multiple times by simply appending using a for loop (although a bit less concise).
In a new ES6 harmony, you will have native way for doing this with repeat. Also ES6 right now only experimental, this feature is already available in Edge, FF, Chrome and Safari
"abc".repeat(3) // "abcabcabc"
And surely if repeat function is not available you can use old-good Array(n + 1).join("abc")
Convenient if you repeat yourself a lot:
String.prototype.repeat = String.prototype.repeat || function(n){
n= n || 1;
return Array(n+1).join(this);
}
alert( 'Are we there yet?\nNo.\n'.repeat(10) )
Array(10).fill('a').join('')
Although the most voted answer is a bit more compact, with this approach you don't have to add an extra array item.
An alternative is:
for(var word = ''; word.length < 10; word += 'a'){}
If you need to repeat multiple chars, multiply your conditional:
for(var word = ''; word.length < 10 * 3; word += 'foo'){}
NOTE: You do not have to overshoot by 1 as with word = Array(11).join('a')
The most performance-wice way is https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/String/repeat
Short version is below.
String.prototype.repeat = function(count) {
if (count < 1) return '';
var result = '', pattern = this.valueOf();
while (count > 1) {
if (count & 1) result += pattern;
count >>>= 1, pattern += pattern;
}
return result + pattern;
};
var a = "a";
console.debug(a.repeat(10));
Polyfill from Mozilla:
if (!String.prototype.repeat) {
String.prototype.repeat = function(count) {
'use strict';
if (this == null) {
throw new TypeError('can\'t convert ' + this + ' to object');
}
var str = '' + this;
count = +count;
if (count != count) {
count = 0;
}
if (count < 0) {
throw new RangeError('repeat count must be non-negative');
}
if (count == Infinity) {
throw new RangeError('repeat count must be less than infinity');
}
count = Math.floor(count);
if (str.length == 0 || count == 0) {
return '';
}
// Ensuring count is a 31-bit integer allows us to heavily optimize the
// main part. But anyway, most current (August 2014) browsers can't handle
// strings 1 << 28 chars or longer, so:
if (str.length * count >= 1 << 28) {
throw new RangeError('repeat count must not overflow maximum string size');
}
var rpt = '';
for (;;) {
if ((count & 1) == 1) {
rpt += str;
}
count >>>= 1;
if (count == 0) {
break;
}
str += str;
}
// Could we try:
// return Array(count + 1).join(this);
return rpt;
}
}
If you're not opposed to including a library in your project, lodash has a repeat function.
_.repeat('*', 3);
// → '***
https://lodash.com/docs#repeat
For all browsers
The following function will perform a lot faster than the option suggested in the accepted answer:
var repeat = function(str, count) {
var array = [];
for(var i = 0; i < count;)
array[i++] = str;
return array.join('');
}
You'd use it like this :
var repeatedString = repeat("a", 10);
To compare the performance of this function with that of the option proposed in the accepted answer, see this Fiddle and this Fiddle for benchmarks.
For moderns browsers only
In modern browsers, you can now do this using String.prototype.repeat method:
var repeatedString = "a".repeat(10);
Read more about this method on MDN.
This option is even faster. Unfortunately, it doesn't work in any version of Internet explorer. The numbers in the table specify the first browser version that fully supports the method:
In ES2015/ES6 you can use "*".repeat(n)
So just add this to your projects, and your are good to go.
String.prototype.repeat = String.prototype.repeat ||
function(n) {
if (n < 0) throw new RangeError("invalid count value");
if (n == 0) return "";
return new Array(n + 1).join(this.toString())
};
String.repeat() is supported by 96.39% of browsers as of now.
function pad(text, maxLength){
return text + "0".repeat(maxLength - text.length);
}
console.log(pad('text', 7)); //text000
/**
* Repeat a string `n`-times (recursive)
* #param {String} s - The string you want to repeat.
* #param {Number} n - The times to repeat the string.
* #param {String} d - A delimiter between each string.
*/
var repeat = function (s, n, d) {
return --n ? s + (d || "") + repeat(s, n, d) : "" + s;
};
var foo = "foo";
console.log(
"%s\n%s\n%s\n%s",
repeat(foo), // "foo"
repeat(foo, 2), // "foofoo"
repeat(foo, "2"), // "foofoo"
repeat(foo, 2, "-") // "foo-foo"
);
Just for the fun of it, here is another way by using the toFixed(), used to format floating point numbers.
By doing
(0).toFixed(2)
(0).toFixed(3)
(0).toFixed(4)
we get
0.00
0.000
0.0000
If the first two characters 0. are deleted, we can use this repeating pattern to generate any repetition.
function repeat(str, nTimes) {
return (0).toFixed(nTimes).substr(2).replaceAll('0', str);
}
console.info(repeat('3', 5));
console.info(repeat('hello ', 4));
Another interesting way to quickly repeat n character is to use idea from quick exponentiation algorithm:
var repeatString = function(string, n) {
var result = '', i;
for (i = 1; i <= n; i *= 2) {
if ((n & i) === i) {
result += string;
}
string = string + string;
}
return result;
};
For repeat a value in my projects i use repeat
For example:
var n = 6;
for (i = 0; i < n; i++) {
console.log("#".repeat(i+1))
}
but be careful because this method has been added to the ECMAScript 6 specification.
function repeatString(n, string) {
var repeat = [];
repeat.length = n + 1;
return repeat.join(string);
}
repeatString(3,'x'); // => xxx
repeatString(10,'🌹'); // => "🌹🌹🌹🌹🌹🌹🌹🌹🌹🌹"
This is how you can call a function and get the result by the helps of Array() and join()
using Typescript and arrow fun
const repeatString = (str: string, num: number) => num > 0 ?
Array(num+1).join(str) : "";
console.log(repeatString("🌷",10))
//outputs: 🌷🌷🌷🌷🌷🌷🌷🌷🌷🌷
function repeatString(str, num) {
// Array(num+1) is the string you want to repeat and the times to repeat the string
return num > 0 ? Array(num+1).join(str) : "";
}
console.log(repeatString("a",10))
// outputs: aaaaaaaaaa
console.log(repeatString("🌷",10))
//outputs: 🌷🌷🌷🌷🌷🌷🌷🌷🌷🌷
Here is what I use:
function repeat(str, num) {
var holder = [];
for(var i=0; i<num; i++) {
holder.push(str);
}
return holder.join('');
}
I realize that it's not a popular task, what if you need to repeat your string not an integer number of times?
It's possible with repeat() and slice(), here's how:
String.prototype.fracRepeat = function(n){
if(n < 0) n = 0;
var n_int = ~~n; // amount of whole times to repeat
var n_frac = n - n_int; // amount of fraction times (e.g., 0.5)
var frac_length = ~~(n_frac * this.length); // length in characters of fraction part, floored
return this.repeat(n) + this.slice(0, frac_length);
}
And below a shortened version:
String.prototype.fracRepeat = function(n){
if(n < 0) n = 0;
return this.repeat(n) + this.slice(0, ~~((n - ~~n) * this.length));
}
var s = "abcd";
console.log(s.fracRepeat(2.5))
I'm going to expand on #bonbon's answer. His method is an easy way to "append N chars to an existing string", just in case anyone needs to do that. For example since "a google" is a 1 followed by 100 zeros.
for(var google = '1'; google.length < 1 + 100; google += '0'){}
document.getElementById('el').innerText = google;
<div>This is "a google":</div>
<div id="el"></div>
NOTE: You do have to add the length of the original string to the conditional.
Lodash offers a similar functionality as the Javascript repeat() function which is not available in all browers. It is called _.repeat and available since version 3.0.0:
_.repeat('a', 10);
var stringRepeat = function(string, val) {
var newString = [];
for(var i = 0; i < val; i++) {
newString.push(string);
}
return newString.join('');
}
var repeatedString = stringRepeat("a", 1);
Can be used as a one-liner too:
function repeat(str, len) {
while (str.length < len) str += str.substr(0, len-str.length);
return str;
}
In CoffeeScript:
( 'a' for dot in [0..10]).join('')
String.prototype.repeat = function (n) { n = Math.abs(n) || 1; return Array(n + 1).join(this || ''); };
// console.log("0".repeat(3) , "0".repeat(-3))
// return: "000" "000"
I have an html template, I want to make a generalize function that itself finds all the numbers in my template and insert comma after every 3 digits. Currently I am using this function which may take some value as an input to convert it into comma separated form.
function commafy( num ) {
var str = num.toString().split('.');
if (str[0].length >= 5) {
str[0] = str[0].replace(/(\d)(?=(\d{3})+$)/g, '$1,');
}
if (str[1] && str[1].length >= 5) {
str[1] = str[1].replace(/(\d{3})/g, '$1 ');
}
return str.join('.');
}
My template contains labels, tables & input fields.
Please help me how can I do that?
If the templates know where the numbers are, just put a class for them. After the template being rendered, select and change the numbers together. Note: there'is a jquery plugin called format_currency.
If the templates does NOT know, do it after render done:
$('#TEMPLATE_CONTAINER input,#TEMPLATE_CONTAINER label,#TEMPLATE_CONTAINER td').each(function(){
FORMAT($(this));//format only if $(this).text() is a number
})
Working Demo http://jsfiddle.net/cse_tushar/P3CSF/
you can use this awesome function for number formatting
e.g
var num=123456789.12345
num.formatMoney(5) //returns 123,456,789.12345
num.formatMoney(2) //returns 123,456,789.12
function code is here :-
Number.prototype.formatMoney = function(c, d, t){
var n = this,
c = isNaN(c = Math.abs(c)) ? 2 : c,
d = d == undefined ? "." : d,
t = t == undefined ? "," : t,
s = n < 0 ? "-" : "",
i = parseInt(n = Math.abs(+n || 0).toFixed(c)) + "",
j = (j = i.length) > 3 ? j % 3 : 0;
return s + (j ? i.substr(0, j) + t : "") + i.substr(j).replace(/(\d{3})(?=\d)/g, "$1" + t) + (c ? d + Math.abs(n - i).toFixed(c).slice(2) : "");
};
console.log((123456789.12345).formatMoney(5, '.', ','));
console.log((123456789.12345).formatMoney(2));
if you want plugins :- Plugins
There are many ways:
You can wrap your money values inside a <span class='money'></span> span, and then using jQuery format those values, on DOM ready.
Using Angular JS, you can give your money values a directive, say ng-money and define that directive such that it formats the content of that span.
Third way is to send money values already commified (as you mentioned that :D ), and use them directly inside your template. Usually on server side, it's easier to format numbers and values.
I've a solution which probably works the way you want. This is only as far my understanding of the problem.
Here are few lines of code. The fiddle is also linked
var formatedNumbers=function(){
count=document.getElementsByTagName('input').length;
for(i=0;i<count;i++){
if(!isNaN(document.getElementsByTagName('input')[i].value) && document.getElementsByTagName('input')[i].value!=""){
document.getElementsByTagName('input')[i].value=formatNumber(document.getElementsByTagName('input')[i].value);
}
}
};
I am trying without much success to convert a very large hex number to decimal.
My problem is that using deciaml = parseInt(hex, 16)
gives me errors in the number when I try to convert a hex number above 14 digits.
I have no problem with this in Java, but Javascript does not seem to be accurate above 14 digits of hex.
I have tried "BigNumber" but tis gives me the same erroneous result.
I have trawled the web to the best of my ability and found web sites that will do the conversion but cannot figure out how to do the conversion longhand.
I have tried getting each character in turn and multiplying it by its factor i.e. 123456789abcdef
15 * Math.pow(16, 0) + 14 * Math.pow(16, 1).... etc but I think (being a noob) that my subroutines may not hev been all they should be because I got a completely (and I mean really different!) answer.
If it helps you guys I can post what I have written so far for you to look at but I am hoping someone has simple answer for me.
<script>
function Hex2decimal(hex){
var stringLength = hex.length;
var characterPosition = stringLength;
var character;
var hexChars = new Array();
hexChars[0] = "0";
hexChars[1] = "1";
hexChars[2] = "2";
hexChars[3] = "3";
hexChars[4] = "4";
hexChars[5] = "5";
hexChars[6] = "6";
hexChars[7] = "7";
hexChars[8] = "8";
hexChars[9] = "9";
hexChars[10] = "a";
hexChars[11] = "b";
hexChars[12] = "c";
hexChars[13] = "d";
hexChars[14] = "e";
hexChars[15] = "f";
var index = 0;
var hexChar;
var result;
// document.writeln(hex);
while (characterPosition >= 0)
{
// document.writeln(characterPosition);
character = hex.charAt(characterPosition);
while (index < hexChars.length)
{
// document.writeln(index);
document.writeln("String Character = " + character);
hexChar = hexChars[index];
document.writeln("Hex Character = " + hexChar);
if (hexChar == character)
{
result = hexChar;
document.writeln(result);
}
index++
}
// document.write(character);
characterPosition--;
}
return result;
}
</script>
Thank you.
Paul
The New 'n' Easy Way
var hex = "7FDDDDDDDDDDDDDDDDDDDDDD";
if (hex.length % 2) { hex = '0' + hex; }
var bn = BigInt('0x' + hex);
var d = bn.toString(10);
BigInts are now available in most browsers (except IE).
Earlier in this answer:
BigInts are now available in both node.js and Chrome. Firefox shouldn't be far behind.
If you need to deal with negative numbers, that requires a bit of work:
How to handle Signed JS BigInts
Essentially:
function hexToBn(hex) {
if (hex.length % 2) {
hex = '0' + hex;
}
var highbyte = parseInt(hex.slice(0, 2), 16)
var bn = BigInt('0x' + hex);
if (0x80 & highbyte) {
// You'd think `bn = ~bn;` would work... but it doesn't
// manually perform two's compliment (flip bits, add one)
// (because JS binary operators are incorrect for negatives)
bn = BigInt('0b' + bn.toString(2).split('').map(function (i) {
return '0' === i ? 1 : 0
}).join('')) + BigInt(1);
bn = -bn;
}
return bn;
}
Ok, let's try this:
function h2d(s) {
function add(x, y) {
var c = 0, r = [];
var x = x.split('').map(Number);
var y = y.split('').map(Number);
while(x.length || y.length) {
var s = (x.pop() || 0) + (y.pop() || 0) + c;
r.unshift(s < 10 ? s : s - 10);
c = s < 10 ? 0 : 1;
}
if(c) r.unshift(c);
return r.join('');
}
var dec = '0';
s.split('').forEach(function(chr) {
var n = parseInt(chr, 16);
for(var t = 8; t; t >>= 1) {
dec = add(dec, dec);
if(n & t) dec = add(dec, '1');
}
});
return dec;
}
Test:
t = 'dfae267ab6e87c62b10b476e0d70b06f8378802d21f34e7'
console.log(h2d(t))
prints
342789023478234789127089427304981273408912349586345899239
which is correct (feel free to verify).
Notice that "0x" + "ff" will be considered as 255, so convert your hex value to a string and add "0x" ahead.
function Hex2decimal(hex)
{
return ("0x" + hex) / 1;
}
If you are using the '0x' notation for your Hex String, don't forget to add s = s.slice(2) to remove the '0x' prefix.
Keep in mind that JavaScript only has a single numeric type (double), and does not provide any separate integer types. So it may not be possible for it to store exact representations of your numbers.
In order to get exact results you need to use a library for arbitrary-precision integers, such as BigInt.js. For example, the code:
var x = str2bigInt("5061756c205768697465",16,1,1);
var s = bigInt2str(x, 10);
$('#output').text(s);
Correctly converts 0x5061756c205768697465 to the expected result of 379587113978081151906917.
Here is a jsfiddle if you would like to experiment with the code listed above.
The BigInt constructor can take a hex string as argument:
/** #param hex = "a83b01cd..." */
function Hex2decimal(hex) {
return BigInt("0x" + hex).toString(10);
}
Usage:
Hex2decimal("100");
Output:
256
A rip-off from the other answer, but without the meaningless 0 padding =P
<script type="text/javascript">
function setYear()
{
var d = new Date();
var cur = document.getElementById('CYear');
var prev = document.getElementById('PYear');
var next = document.getElementById('NYear');
var optionC, optionP, optionN;
for (var i = d.getFullYear() - 10; i <= d.getFullYear() + 10; i++)
{
optionC = document.createElement("option");
optionC.setAttribute("value", i);
optionC.innerHTML = i;
cur.appendChild(optionC);
};
for (var i = d.getFullYear() - 10; i < d.getFullYear() + 10; i++)
{
optionP = document.createElement("option");
optionP.setAttribute("value", i);
optionP.innerHTML = i;
prev.appendChild(optionP);
};
for (var i = d.getFullYear() - 10; i <= d.getFullYear() + 10; i++)
{
optionN = document.createElement("option");
optionN.setAttribute("value", i);
optionN.innerHTML = i;
next.appendChild(optionN);
};
cur.value=d.getFullYear();
prev.value=d.getFullYear()-1;
next.value=d.getFullYear()+1;
}
function setYear1(obj)
{
var prev = document.getElementById('PYear');
var next = document.getElementById('NYear');
prev.value = Number(obj.value) - 1;
next.value = Number(obj.value) + 1;
}
function onlyNumbers(evt)
{
var evt = (evt) ? evt : event;
var charCode = (evt.charCode) ? evt.charCode : ((evt.keyCode) ? evt.keyCode : ((evt.which) ? evt.which : 0));
if (charCode > 31 && (charCode < 48 || charCode > 57) && charCode!=44)
{
alert("Enter numerals only in this field!");
return false;
}
return true;
}
</script>
I need to change this two function from javascript to JQuery can any one tell me how to conver. Are there any converters for converting from javascript to Jquery? Thanks in advance.
Adding jQuery to an existing project with JavaScript on it shouldn't break any existing JavaScript unless you're using a conflicting framework such as prototype, in which case you can often get away with both being present simply by setting jQuery to noConflict mode.
There's certainly no need to rewrite working JavaScript functions into jQuery if they're working fine as they are unless your primary goal is to improve efficiency / write shorter code or even if it's simply because the JavaScript has become too cumbersome to maintain.
In summary if you just want to add some new jQuery functionality to a website; why not just add it and leave the existing JavaScript as is?
https://stackoverflow.com/a/2305763/1188942
As for converters, I haven't been able to find any.
jQuery is just an extension of javascript that deals (mostly) with selecting and manipulating DOM elements. So no, there isn't going to be a converter. I'd suggest actually learning jQuery -- it's unbelievably useful.
Is there a special name for numbers in the format of 001.
For example the number 20 would be 020 and 1 would be 001. Its hard to Google around when you don`t know somethings name! Since I am already wasting your guys time does any one know a function for changing numbers to this format.
I think this is usually called "padding" the number.
Its called left zero padded numbers.
It's called padding.
Well, if you're talking about that notation within the context of certain programming languages, 020 as opposed to 20 would be Octal rather than Decimal.
Otherwise, you're referring to padding.
A quick google search revealed this nice snippet of code for Number Padding:
http://sujithcjose.blogspot.com/2007/10/zero-padding-in-java-script-to-add.html
function zeroPad(num,count)
{
var numZeropad = num + '';
while(numZeropad.length < count) {
numZeropad = "0" + numZeropad;
}
return numZeropad;
}
You can use a simple function like:
function addZ(n) {
return (n<10? '00' : n<100? '0' : '') + n;
}
Or a more robust function that pads the left hand side with as many of whatever character you like, e.g.
function padLeft(n, c, len) {
var x = ('' + n).length;
x = (x < ++len)? new Array(len - x) : [];
return x.join(c) + n
}
Try this one:
Number.prototype.toMinLengthString = function (n) {
var isNegative = this < 0;
var number = isNegative ? -1 * this : this;
for (var i = number.toString().length; i < n; i++) {
number = '0' + number;
}
return (isNegative ? '-' : '') + number;
}