i'm having a serious issue here. I have a function that pads a user's input with zeros. For example if i enter 88 it will normalize it to:
00000088. My function is this:
export default length => (value) => {
const noLeadingZeros = value.toString().replace(/(0+)/, '');
if (noLeadingZeros.length === 0) {
return value;
}
return padLeft(noLeadingZeros, length);
};
with padleft is:
export default (num, len) => (
len > num.toString().length ? '0'.repeat(len - num.toString().length) + num
: num
);
My problem is that if i entered something like this:
80112345 it convert it to 08112345. Any ideas?
Using slice:
let str = '00000000' + 88;
let resp = str.slice(-8, str.length)
console.log(resp) // 00000088
In your replace, you're replacing all the zeros in the number not just those on the left side, and even if there are zeros on the left side, why remove them if you're just going to add them back. You could use a for loop that pads the string with a zero n times (where n is the number of digits that the string needs to have length 8), or (thanks to a comment by #AndrewBone), you can use the .repeat() function that does this for you:
function padLeft(value, len) {
return '0'.repeat(String(value).length < len ? len - String(value).length : 0) + value;
}
console.log(padLeft("", 8));
console.log(padLeft("88", 8));
console.log(padLeft("00088", 8));
console.log(padLeft("12312388", 8));
console.log(padLeft("00000000", 8));
this looks wrong :
const noLeadingZeros = value.toString().replace(/(0+)/, '');
you are deleting every zeros out of your number... even those inside !
You can use this regex instead, instead of for /(0+)/ in your code :
/\b(0+)/
explanation : the \b ensures the zeros are at the beginning of a word
or this
/^(0+)/
explanation : the ^ ensure this is the beginning of the string
Just use a RegEx to assert that the number is a valid number.
/0+/
Then get the number of digits in the number:
('' + num).length;
Then put the whole thing together
var paddedNum ='';
for (var i=0;i<8-len;i++) {
paddedNum += "0";
}
paddedNum += num;
Related
i have a function to split string into 2 part, front and back. Then reverse it to back and front. Here is my code
function reverseString(string) {
let splitString = ""
let firstString = ""
for(i = 0; i <= string.length/2 - 1; i++) {
firstString += string[i]
}
for(i = string.length/2; i <= string.length; i++) {
splitString += string[i]
}
return splitString + firstString
}
Sorry for bad explanation, this is test case and expected result (first one is expected result, the second one is my result)
console.log(reverseString("aaabccc")); // "cccbaaa" "undefinedundefinedundefinedundefinedaaa"
console.log(reverseString("aab")); // "baa" "undefinedundefineda"
console.log(reverseString("aaaacccc")); // "ccccaaaa" "ccccundefinedaaa"
console.log(reverseString("abcdefghabcdef")); // "habcdefabcdefg" "habcdefundefinedabcdefg"
could you help me, whats wrong with it. Thank you
You could try another approach and use the slice function
function reverseString(string)
{
if (string.length < 2) { return string; }
let stringHalfLength = string.length / 2;
let isLengthOdd = stringHalfLength % 1 !== 0;
if (isLengthOdd) {
return string.slice(Math.ceil(stringHalfLength), string.length + 1) + string[Math.floor(stringHalfLength)] + string.slice(0, Math.floor(stringHalfLength));
}
return string.slice(stringHalfLength, string.length + 1) + string.slice(0, stringHalfLength);
}
console.log(reverseString("aaabccc") === "cccbaaa");
console.log(reverseString("aab") === "baa");
console.log(reverseString("aaaacccc") === "ccccaaaa");
console.log(reverseString("abcdefghabcdef") === "habcdefabcdefg");
A more efficient way to reverse the string would be to split the string, then use the built-in reverse javascript function (which reverses the elements of the split string), and then re-join the elements using the join function.. No need to re-invent the wheel?
You can concatenate the functions in shorthand (.split.reverse.join etc...) so your function would look something like this:
function reverseString(string) {
return string.split("").reverse().join("");
}
Try it out!
function reverseString(string) {
return string.split("").reverse().join("");
}
console.log(reverseString("hello"));
console.log(reverseString("aaabbbccc"));
If there's a particular reason you're opting not to use the in-built functions (i.e. if I've missed something?) , feel free to comment.
The short version of what you need:
function reverseString(string) {
const splitPosition = Math.ceil(string.length / 2);
return string.substring(splitPosition) + string.substring(0, splitPosition);
}
The key to your question is the middle element. To accomplish that, you probably want to use Math.floor that round under.
console.log(reverseString("aaabccc")); // "cccbaaa"
console.log(reverseString("abcdefghabcdef")); // "habcdefabcdefg"
function reverseString (str) {
if (str.length<2) {
return str
}
var half = Math.floor(str.length / 2);
return (str.slice(-half) + (str.length%2?str[half]:"") + str.slice(0,half));
}
reverseString('')
> ""
reverseString('1')
> "1"
reverseString('12')
> "21"
reverseString('123')
> "321"
reverseString('1234')
> "3412"
reverseString('12345')
> "45312"
reverseString("aaabccc")
> "cccbaaa"
reverseString("abcdefghabcdef")
> "habcdefabcdefg"
So basically your problem is not to grab 2 parts of the string and rearrange, it is to grab 3 parts.
1 part: str.slice(0,half)
2 part: str.length%2 ? str[half] : ""
3 part: str.slice(-half)
The second part is empty if the string length is even and the middle character if is odd.
So the code version in long self explanatory code:
function reverseString (str) {
if (str.length<2) {
return str
}
var half = Math.floor(str.length / 2);
var firstPart = str.slice(0,half);
var midlePart = str.length % 2 ? str[half] : ""; // we could expand also this
var endPart = str.slice(-half);
return endPart + midlePart + firstPart;
}
And also, notice the precondition, so I don't have to deal with the easy cases.
Also, in your code, you got undefined because you access in the last loop to:
string[string.length] you need to change <= by <
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))
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
i want int from string in javascript how i can get them from
test1 , stsfdf233, fdfk323,
are anyone show me the method to get the integer from this string.
it is a rule that int is always in the back of the string.
how i can get the int who was at last in my string
var s = 'abc123';
var number = s.match(/\d+$/);
number = parseInt(number, 10);
The first step is a simple regular expression - \d+$ will match the digits near the end.
On the next step, we use parseInt on the string we've matched before, to get a proper number.
You can use a regex to extract the numbers in the string via String#match, and convert each of them to a number via parseInt:
var str, matches, index, num;
str = "test123and456";
matches = str.match(/\d+/g);
for (index = 0; index < matches.length; ++index) {
num = parseInt(matches[index], 10);
display("Digit series #" + index + " converts to " + num);
}
Live Example
If the numbers really occur only at the ends of the strings or you just want to convert the first set of digits you find, you can simplify a bit:
var str, matches, num;
str = "test123";
matches = str.match(/\d+/);
if (matches) {
num = parseInt(matches[0], 10);
display("Found match, converts to: " + num);
}
else {
display("No digits found");
}
Live example
If you want to ignore digits that aren't at the end, add $ to the end of the regex:
matches = str.match(/\d+$/);
Live example
var str = "stsfdf233";
var num = parseInt(str.replace(/\D/g, ''), 10);
var match = "stsfdf233".match(/\d+$/);
var result = 0; // default value
if(match != null) {
result = parseInt(match[0], 10);
}
Yet another alternative, this time without any replace or Regular Expression, just one simple loop:
function ExtractInteger(sValue)
{
var sDigits = "";
for (var i = sValue.length - 1; i >= 0; i--)
{
var c = sValue.charAt(i);
if (c < "0" || c > "9")
break;
sDigits = c + sDigits;
}
return (sDigits.length > 0) ? parseInt(sDigits, 10) : NaN;
}
Usage example:
var s = "stsfdf233";
var n = ExtractInteger(s);
alert(n);
This might help you
var str = 'abc123';
var number = str.match(/\d/g).join("");
Use my extension to String class :
String.prototype.toInt=function(){
return parseInt(this.replace(/\D/g, ''),10);
}
Then :
"ddfdsf121iu".toInt();
Will return an integer : 121
First positive or negative number:
"foo-22bar11".match(/-?\d+/); // -22
javascript:alert('stsfdf233'.match(/\d+$/)[0])
Global.parseInt with radix is overkill here, regexp extracted decimal digits already and rigth trimmed string
I am trying to increment a number by a given value each second and retain the formatting using JavaScript or JQuery
I am struggling to do it.
Say I have a number like so:
1412015
the number which this can be incremented by each second is variable it could be anything beween 0.1 and 2.
Is it possible, if the value which it has to be incremented by each second is 0.54 to incremenet the number and have the following output:
1,412,016
1,412,017
1,412,018
Thanks
Eef
I'm not quite sure I understand your incrementation case and what you want to show.
However, I decided to chime in on a solution to format a number.
I've got two versions of a number format routine, one which parses an array, and one which formats with a regular expression. I'll admit they aren't the easiest to read, but I had fun coming up with the approach.
I've tried to describe the lines with comments in case you're curious
Array parsing version:
function formatNum(num) {
//Convert a formatted number to a normal number and split off any
//decimal places if they exist
var parts = String( num ).replace(/[^\d.]-/g,'').split('.');
//turn the string into a character array and reverse
var arr = parts[0].split('').reverse();
//initialize the return value
var str = '';
//As long as the array still has data to process (arr.length is
//anything but 0)
//Use a for loop so that it keeps count of the characters for me
for( var i = 0; arr.length; i++ ) {
//every 4th character that isn't a minus sign add a comma before
//we add the character
if( i && i%3 == 0 && arr[0] != '-' ) {
str = ',' + str ;
}
//add the character to the result
str = arr.shift() + str ;
}
//return the final result appending the previously split decimal place
//if necessary
return str + ( parts[1] ? '.'+parts[1] : '' );
}
Regular Expression version:
function formatNum(num) {
//Turn a formatted number into a normal number and separate the
//decimal places
var parts = String( num ).replace(/[^\d.]-/g,'').split('.');
//reverse the string
var str = parts[0].split('').reverse().join('');
//initialize the return value
var retVal = '';
//This gets complicated. As long as the previous result of the regular
//expression replace is NOT the same as the current replacement,
//keep replacing and adding commas.
while( retVal != (str = str.replace(/(\d{3})(\d{1,3})/,'$1,$2')) ) {
retVal = str;
}
//If there were decimal points return them back with the reversed string
if( parts[1] ) {
return retVal.split('').reverse().join('') + '.' + parts[1];
}
//return the reversed string
return retVal.split('').reverse().join('');
}
Assuming you want to output a formatted number every second incremented by 0.54 you could use an interval to do your incrementation and outputting.
Super Short Firefox with Firebug only example:
var num = 1412015;
setInterval(function(){
//Your 0.54 value... why? I don't know... but I'll run with it.
num += 0.54;
console.log( formatNum( num ) );
},1000);
You can see it all in action here: http://jsbin.com/opoze
To increment a value on every second use this structure:
var number = 0; // put your initial value here
function incrementNumber () {
number += 1; // you can increment by anything you like here
}
// this will run incrementNumber() every second (interval is in ms)
setInterval(incrementNumber, 1000);
This will format numbers for you:
function formatNumber(num) {
num = String(num);
if (num.length <= 3) {
return num;
} else {
var last3nums = num.substring(num.length - 3, num.length);
var remindingPart = num.substring(0, num.length - 3);
return formatNumber(remindingPart) + ',' + last3nums;
}
}
function rounded_inc(x, n) {
return x + Math.ceil(n);
}
var x = 1412015;
x = rounded_inc(x, 0.54);