Develop Reference

Adding a whitespace in front of the first number in a string

I need a whitespace to be added in front of the first number of a string, unless there is one already and unless the number is the first character in the string.
So I wrote this JS code:
document.getElementById('billing:street1').addEventListener('blur', function() {
var value = document.getElementById('billing:street1').value;
var array = value.match(/\d{1,}/g);
if (array !== null) {
var number = array[0];
var index = value.indexOf(number);
if(index !== 0){
var street = value.substring(0, index);
var housenumber = value.substring(index);
if (street[street.length - 1] !== ' ') {
document.getElementById('billing:street1').value = street + ' ' + housenumber;
}
}
}
});
Fiddle
It works fine, but I feel like this can probably be done in a smarter, more compact way.
Also, JQuery suggestions welcome, I am just not very familiar with it.

Try this one :
const addSpace = (str) => {
return str.replace(/(\D)(\d)/, "$1 $2")
}
console.log(addSpace("12345")) // "12345"
console.log(addSpace("city12345")) // "city 12345"
console.log(addSpace("city")) // "city"
(\D) captures a non-digit
(\d) captures a digit
so (\D)(\d) means : non-digit followed by a digit
that we replace with "$1 $2" = captured1 + space + captured2

You can do it by using only regular expressions. For example:
var s = "abcde45";
if(!s.match(/\s\d/)){
s = s.replace(/(\d)/, ' $1');
}
console.log(s); // "abcde 45"
UPD : Of course, if you have string with a wrong syntax(e.g no numbers inside), that code wouldn't work.

adding a space to every space in a string, then cycling back around until length is met

I have the following while loop as part of my text justify function. The idea is that I have text strings (str) that need to be justified (spaces added to existing spaces in between words) to equal to a given length (len)
The catch is I can only add one space to an existing space at a time before I iterate over to the next space in the string and add another space there. If that's it for all spaces in the string and it's still not at the required length, I cycle back over to the original space (now two spaces) and add another. Then it goes to the next space between words and so on and so on. The idea is that any spaces between words in the string should not have a differential of more than one space (i.e. Lorem---ipsum--dolor--sit, not Lorem----ipsum--dolor-sit)
From my research, I decided that using a substring method off the original string to add that first extra space, then I will increment the index and move to the next space in the string and repeat the add. Here's my code:
var indexOf = str.indexOf(" ", 0);
if ( indexOf > -1 ) {
while ( indexOf > -1 && str.length < len ) {
//using a regexp to find a space before a character
var space = /\s(?=\b)/.exec(str);
str = str.substring(0, indexOf + 1) + " " + str.substring(indexOf + 1);
//go to next space in string
indexOf = str.indexOf(space, indexOf + 2);
if ( indexOf === -1 ) {
//loops back to beginning of string
indexOf = str.indexOf(space, 0);
}
}
}
finalResults.push(str);
This code works most of the time, but I noticed that there are instances where the cycle of spacing is not correct. For example, it generates the following string:
sit----amet,--blandit
when the correct iteration would be
sit---amet,---blandit
Any assistance in making this code properly iterate over every space (to add one space) in the string once, then cycling back around to the beginning of the string to start over until the desired length is achieved would be most appreciated.

I think it's more efficient to compute the number spaces required in the beginning.
var s = "today is a friday";
var totalLength = 40;
var tokens = s.split(/\s+/);
var noSpaceLength = s.replace(/\s+/g,'').length;
var minSpace = Math.floor((totalLength - noSpaceLength)/(tokens.length-1));
var remainder = (totalLength - noSpaceLength) % (tokens.length-1);
var out = tokens[0];
for (var i = 1; i < tokens.length; i++) {
var spaces = (i <= remainder ? minSpace+1 : minSpace);
out += "-".repeat(spaces) + tokens[i];
}
$('#out').text(out);
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<div id="out"></div>

This solution
splits the string (s) into words in an array (a)
finds the number of spaces to be added between all words (add)
finds the remainder of spaces to be added between first words (rem)
then sticks the words with add spaces + one if rem is not exhausted
Code
var s = "Caballo sin Nombre"; // assume one space between words
var len = 21; // desired length
var need = len - s.length;
var a = s.split(/ /); // split s
// need>0 and at least two words
if (need > 0 && a.length>1) {
var add = Math.floor(need / (a.length-1)) + 1; // all spaces need that (+existing)
var rem = need % (a.length-1); // remainder
var sp = '';
while (add-- > 0) sp += ' ';
// replace
var i,res = ''; // result
for (i=0 ; i<a.length-1 ; i++) {
res += a[i] + sp;
if (rem-- > 0) res += ' '; // remainder
}
res += a[i];
s = res;
}
console.log("'" + s + "' is " + s.length + " chars long.");

This function adds the spaces using a global replace, carefully limiting the text size.
function expand (txt, colwidth) {
txt = txt.replace (/\s\s+/, ' '); // Ensure no multiple spaces in txt
for (var spaces = ' ', // Spaces to check for
limit = colwidth - txt.length; // number of additional spaces required
limit > 0; // do while limit is positive
spaces += ' ') // add 1 to spaces to search for
txt = txt.replace (RegExp (spaces, 'g'),
function (r) {
// If limit > 0 then add a space else do not.
return limit > 0 && --limit ? r + ' ' : r
});
return txt;
}
for (var w = 21; w--;) console.log (expand ('this is a test.', w));
Shows this on console:
this is a test.
this is a test.
this is a test.
this is a test.
14 this is a test.

Replace the last and first integers of a string

I have a string like this : var input = "/first_part/5/another_part/3/last_part"
I want to replace the last occurence of integers (3 in my string), then the first occurence (5).
I tried this: input .replace(/\d+/, 3); which replace all occurences. But how to only target the last / first one.
Thanks in advance.

This will replace the first and last single digit in the input string with 3
input.replace(/^(.*?)\d(.*)\d(.*)$/, "$13$23$3");
Here's a reFiddle link that demos this: http://refiddle.com/1am9
More readable:
var replacement = '3';
input.replace(/^(.*?)\d(.*)\d(.*)$/, "$1" + replacement + "$2" + replacement + "$3");
or input.replace(/^(.*?)\d(.*)\d(.*)$/, ["$1", "$2", "$3"].join(replacement)); if that's your thing.

You can use this negative lookahead based regex:
var input = "/first_part/5/another_part/3/last_part";
// replace first number
var r = input.replace(/\d+/, '9').replace(/\d+(?=\D*$)/, '7');
//=> /first_part/9/another_part/7/last_part
Here \d+(?=\D*$) means match 1 or more digits that are followed by all non-digits till end of line.

Here is a pretty rigid approach to your problem, you might want to adapt it to your needs, but it shows one way you can get things done.
// input string
var string = "/first_part/5/another_part/3/last_part";
//match all the parts of the string
var m = string.match(/^(\D+)(\d+)+(\D+)(\d+)(.+)/);
// ["/first_part/5/another_part/3/last_part", "/first_part/", "5", "/another_part/", "3", "/last_part"]
// single out your numbers
var n1 = parseInt(m[2], 10);
var n2 = parseInt(m[4], 10);
// do any operations you want on them
n1 *= 2;
n2 *= 2;
// put the string back together
var output = m[1] + n1 + m[3] + n2 + m[5];
// /first_part/10/another_part/6/last_part

Remove/ truncate leading zeros by javascript/jquery

Suggest solution for removing or truncating leading zeros from number(any string) by javascript,jquery.

You can use a regular expression that matches zeroes at the beginning of the string:
s = s.replace(/^0+/, '');

I would use the Number() function:
var str = "00001";
str = Number(str).toString();
>> "1"
Or I would multiply my string by 1
var str = "00000000002346301625363";
str = (str * 1).toString();
>> "2346301625363"

Maybe a little late, but I want to add my 2 cents.
if your string ALWAYS represents a number, with possible leading zeros, you can simply cast the string to a number by using the '+' operator.
e.g.
x= "00005";
alert(typeof x); //"string"
alert(x);// "00005"
x = +x ; //or x= +"00005"; //do NOT confuse with x+=x, which will only concatenate the value
alert(typeof x); //number , voila!
alert(x); // 5 (as number)
if your string doesn't represent a number and you only need to remove the 0's use the other solutions, but if you only need them as number, this is the shortest way.
and FYI you can do the opposite, force numbers to act as strings if you concatenate an empty string to them, like:
x = 5;
alert(typeof x); //number
x = x+"";
alert(typeof x); //string
hope it helps somebody

Since you said "any string", I'm assuming this is a string you want to handle, too.
"00012 34 0000432 0035"
So, regex is the way to go:
var trimmed = s.replace(/\b0+/g, "");
And this will prevent loss of a "000000" value.
var trimmed = s.replace(/\b(0(?!\b))+/g, "")
You can see a working example here

parseInt(value) or parseFloat(value)
This will work nicely.

I got this solution for truncating leading zeros(number or any string) in javascript:
<script language="JavaScript" type="text/javascript">
<!--
function trimNumber(s) {
while (s.substr(0,1) == '0' && s.length>1) { s = s.substr(1,9999); }
return s;
}
var s1 = '00123';
var s2 = '000assa';
var s3 = 'assa34300';
var s4 = 'ssa';
var s5 = '121212000';
alert(s1 + '=' + trimNumber(s1));
alert(s2 + '=' + trimNumber(s2));
alert(s3 + '=' + trimNumber(s3));
alert(s4 + '=' + trimNumber(s4));
alert(s5 + '=' + trimNumber(s5));
// end hiding contents -->
</script>

Simply try to multiply by one as following:
"00123" * 1; // Get as number
"00123" * 1 + ""; // Get as string

1. The most explicit is to use parseInt():
parseInt(number, 10)
2. Another way is to use the + unary operator:
+number
3. You can also go the regular expression route, like this:
number.replace(/^0+/, '')

Try this,
function ltrim(str, chars) {
chars = chars || "\\s";
return str.replace(new RegExp("^[" + chars + "]+", "g"), "");
}
var str =ltrim("01545878","0");
More here

You should use the "radix" parameter of the "parseInt" function :
https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/parseInt?redirectlocale=en-US&redirectslug=JavaScript%2FReference%2FGlobal_Objects%2FparseInt
parseInt('015', 10) => 15
if you don't use it, some javascript engine might use it as an octal
parseInt('015') => 0

If number is int use
"" + parseInt(str)
If the number is float use
"" + parseFloat(str)

const number = '0000007457841';
console.log(+number) //7457841;
OR number.replace(/^0+/, '')

Regex solution from Guffa, but leaving at least one character
"123".replace(/^0*(.+)/, '$1'); // 123
"012".replace(/^0*(.+)/, '$1'); // 12
"000".replace(/^0*(.+)/, '$1'); // 0

I wanted to remove all leading zeros for every sequence of digits in a string and to return 0 if the digit value equals to zero.
And I ended up doing so:
str = str.replace(/(0{1,}\d+)/, "removeLeadingZeros('$1')")
function removeLeadingZeros(string) {
if (string.length == 1) return string
if (string == 0) return 0
string = string.replace(/^0{1,}/, '');
return string
}

One another way without regex:
function trimLeadingZerosSubstr(str) {
var xLastChr = str.length - 1, xChrIdx = 0;
while (str[xChrIdx] === "0" && xChrIdx < xLastChr) {
xChrIdx++;
}
return xChrIdx > 0 ? str.substr(xChrIdx) : str;
}
With short string it will be more faster than regex (jsperf)

const input = '0093';
const match = input.match(/^(0+)(\d+)$/);
const result = match && match[2] || input;

Use "Math.abs"
eg: Math.abs(003) = 3;
console.log(Math.abs(003))

Adding extra zeros in front of a number using jQuery?

I have file that are uploaded which are formatted like so
MR 1
MR 2
MR 100
MR 200
MR 300
ETC.
What i need to do is add extra two 00s before anything before MR 10 and add one extra 0 before MR10-99
So files are formatted
MR 001
MR 010
MR 076
ETC.
Any help would be great!

Assuming you have those values stored in some strings, try this:
function pad (str, max) {
str = str.toString();
return str.length < max ? pad("0" + str, max) : str;
}
pad("3", 3); // => "003"
pad("123", 3); // => "123"
pad("1234", 3); // => "1234"
var test = "MR 2";
var parts = test.split(" ");
parts[1] = pad(parts[1], 3);
parts.join(" "); // => "MR 002"

I have a potential solution which I guess is relevent, I posted about it here:
https://www.facebook.com/antimatterstudios/posts/10150752380719364
basically, you want a minimum length of 2 or 3, you can adjust how many 0's you put in this piece of code
var d = new Date();
var h = ("0"+d.getHours()).slice(-2);
var m = ("0"+d.getMinutes()).slice(-2);
var s = ("0"+d.getSeconds()).slice(-2);
I knew I would always get a single integer as a minimum (cause hour 1, hour 2) etc, but if you can't be sure of getting anything but an empty string, you can just do "000"+d.getHours() to make sure you get the minimum.
then you want 3 numbers? just use -3 instead of -2 in my code, I'm just writing this because I wanted to construct a 24 hour clock in a super easy fashion.

Note: see Update 2 if you are using latest ECMAScript...
Here a solution I liked for its simplicity from an answer to a similar question:
var n = 123
String('00000' + n).slice(-5); // returns 00123
('00000' + n).slice(-5); // returns 00123
UPDATE
As #RWC suggested you can wrap this of course nicely in a generic function like this:
function leftPad(value, length) {
return ('0'.repeat(length) + value).slice(-length);
}
leftPad(123, 5); // returns 00123
And for those who don't like the slice:
function leftPad(value, length) {
value = String(value);
length = length - value.length;
return ('0'.repeat(length) + value)
}
But if performance matters I recommend reading through the linked answer before choosing one of the solutions suggested.
UPDATE 2
In ES6 the String class now comes with a inbuilt padStart method which adds leading characters to a string. Check MDN here for reference on String.prototype.padStart(). And there is also a padEnd method for ending characters.
So with ES6 it became as simple as:
var n = '123';
n.padStart(5, '0'); // returns 00123
Note: #Sahbi is right, make sure you have a string otherwise calling padStart will throw a type error.
So in case the variable is or could be a number you should cast it to a string first:
String(n).padStart(5, '0');

function addLeadingZeros (n, length)
{
var str = (n > 0 ? n : -n) + "";
var zeros = "";
for (var i = length - str.length; i > 0; i--)
zeros += "0";
zeros += str;
return n >= 0 ? zeros : "-" + zeros;
}
//addLeadingZeros (1, 3) = "001"
//addLeadingZeros (12, 3) = "012"
//addLeadingZeros (123, 3) = "123"

This is the function that I generally use in my code to prepend zeros to a number or string.
The inputs are the string or number (str), and the desired length of the output (len).
var PrependZeros = function (str, len) {
if(typeof str === 'number' || Number(str)){
str = str.toString();
return (len - str.length > 0) ? new Array(len + 1 - str.length).join('0') + str: str;
}
else{
for(var i = 0,spl = str.split(' '); i < spl.length; spl[i] = (Number(spl[i])&& spl[i].length < len)?PrependZeros(spl[i],len):spl[i],str = (i == spl.length -1)?spl.join(' '):str,i++);
return str;
}
};
Examples:
PrependZeros('MR 3',3); // MR 003
PrependZeros('MR 23',3); // MR 023
PrependZeros('MR 123',3); // MR 123
PrependZeros('foo bar 23',3); // foo bar 023

If you split on the space, you can add leading zeros using a simple function like:
function addZeros(n) {
return (n < 10)? '00' + n : (n < 100)? '0' + n : '' + n;
}
So you can test the length of the string and if it's less than 6, split on the space, add zeros to the number, then join it back together.
Or as a regular expression:
function addZeros(s) {
return s.replace(/ (\d$)/,' 00$1').replace(/ (\d\d)$/,' 0$1');
}
I'm sure someone can do it with one replace, not two.
Edit - examples
alert(addZeros('MR 3')); // MR 003
alert(addZeros('MR 23')); // MR 023
alert(addZeros('MR 123')); // MR 123
alert(addZeros('foo bar 23')); // foo bar 023
It will put one or two zeros infront of a number at the end of a string with a space in front of it. It doesn't care what bit before the space is.

Just for a laugh do it the long nasty way....:
(NOTE: ive not used this, and i would not advise using this.!)
function pad(str, new_length) {
('00000000000000000000000000000000000000000000000000' + str).
substr((50 + str.toString().length) - new_length, new_length)
}

I needed something like this myself the other day, Pud instead of always a 0, I wanted to be able to tell it what I wanted padded ing the front. Here's what I came up with for code:
function lpad(n, e, d) {
var o = ''; if(typeof(d) === 'undefined'){ d='0'; } if(typeof(e) === 'undefined'){ e=2; }
if(n.length < e){ for(var r=0; r < e - n.length; r++){ o += d; } o += n; } else { o=n; }
return o; }
Where n is what you want padded, e is the power you want it padded to (number of characters long it should be), and d is what you want it to be padded with. Seems to work well for what I needed it for, but it would fail if "d" was more than one character long is some cases.

var str = "43215";
console.log("Before : \n string :"+str+"\n Length :"+str.length);
var max = 9;
while(str.length < max ){
str = "0" + str;
}
console.log("After : \n string :"+str+"\n Length :"+str.length);
It worked for me !
To increase the zeroes, update the 'max' variable
Working Fiddle URL : Adding extra zeros in front of a number using jQuery?:

str could be a number or a string.
formatting("hi",3);
function formatting(str,len)
{
return ("000000"+str).slice(-len);
}
Add more zeros if needs large digits

In simple terms we can written as follows,
for(var i=1;i<=31;i++)
i=(i<10) ? '0'+i : i;
//Because most of the time we need this for day, month or amount matters.

Know this is an old post, but here's another short, effective way:
edit: dur. if num isn't string, you'd add:
len -= String(num).length;
else, it's all good
function addLeadingZeros(sNum, len) {
len -= sNum.length;
while (len--) sNum = '0' + sNum;
return sNum;
}

Try following, which will convert convert single and double digit numbers to 3 digit numbers by prefixing zeros.
var base_number = 2;
var zero_prefixed_string = ("000" + base_number).slice(-3);

By adding 100 to the number, then run a substring function from index 1 to the last position in right.
var dt = new Date();
var month = (100 + dt.getMonth()+1).toString().substr(1, 2);
var day = (100 + dt.getDate()).toString().substr(1, 2);
console.log(month,day);
you will got this result from the date of 2020-11-3
11,03
I hope the answer is useful