Modify this code to remove the "year" part - javascript

I have this code and I have to remove the year part of it, it's a formatter for date values in a form, it has to be "DD/MM" but it is "MM/DD/YYYY" can someone help me?
<input type="text" name="datada" id="dtDataDa" placeholder="" value="%dtDataDa%"/>
THIS IS THE CODE I HAVE TO MODIFY
var date = document.getElementById('dtDataDa');
function checkValue(str, max) {
if (str.charAt(0) !== '0' || str == '00') {
var num = parseInt(str);
if (isNaN(num) || num <= 0 || num > max) num = 1;
str = num > parseInt(max.toString().charAt(0)) && num.toString().length == 1 ? '0' + num : num.toString();
};
return str;
};
date.addEventListener('input', function (e) {
this.type = 'text';
var input = this.value;
if (/\D\/$/.test(input)) input = input.substr(0, input.length - 3);
var values = input.split('/').map(function (v) {
return v.replace(/\D/g, '')
});
if (values[0]) values[0] = checkValue(values[0], 12);
if (values[1]) values[1] = checkValue(values[1], 31);
var output = values.map(function (v, i) {
return v.length == 2 && i < 2 ? v + ' / ' : v;
});
this.value = output.join('').substr(0, 14);
});
date.addEventListener('blur', function (e) {
this.type = 'text';
var input = this.value;
var values = input.split('/').map(function (v, i) {
return v.replace(/\D/g, '')
});
var output = '';
if (values.length == 3) {
var year = values[2].length !== 4 ? parseInt(values[2]) + 2000 : parseInt(values[2]);
var month = parseInt(values[0]) - 1;
var day = parseInt(values[1]);
var d = new Date(year, month, day);
if (!isNaN(d)) {
document.getElementById('result').innerText = d.toString();
var dates = [d.getMonth() + 1, d.getDate(), d.getFullYear()];
output = dates.map(function (v) {
v = v.toString();
return v.length == 1 ? '0' + v : v;
}).join(' / ');
};
};
this.value = output;
});
It makes the form insert only real digits, for the month (1s part) numbers until 12, for the days (2nd part) numbers until 31 and for the year (3rd part) there is no rule.. I would like to have only the month and days part of it so that an example would look like "06/18"
Here you can see it https://codepen.io/user23xx/pen/VdarOL?editors=1010
Thank you for your time.


It's actually quite simple to modify from the code you've shared.
Look for the comments at addEventListener
date.addEventListener('input', function (e) {
this.type = 'text';
var input = this.value;
if (/\D\/$/.test(input)) input = input.substr(0, input.length - 3);
var values = input.split('/').map(function (v) {
return v.replace(/\D/g, '')
});
if (values[0]) values[0] = checkValue(values[0], 31); //Put 31 instead of 12
if (values[1]) values[1] = checkValue(values[1], 12); //Put 12 instead of 31
var output = values.map(function (v, i) {
return v.length == 2 && i < 2 ? v + ' / ' : v;
});
this.value = output.join('').substr(0, 7); //Update to 7 instead of 14 because full length you want is 7
});
Next we want to update the blur method as below
if (!isNaN(d)) {
document.getElementById('result').innerText = d.toString();
var dates = [d.getDate(),d.getMonth() + 1];//Remove the year part
output = dates.map(function (v) {
v = v.toString();
return v.length == 1 ? '0' + v : v;
}).join(' / ');
};
UPDATES:
date.addEventListener('blur', function (e) {
this.type = 'text';
var input = this.value;
var values = input.split('/').map(function (v, i) {
return v.replace(/\D/g, '')
});
var output = '';
if (values.length == 2) {
var dates = [values[0], values[1]];
output = dates.map(function (v) {
v = v.toString();
return v.length == 1 ? '0' + v : v;
}).join(' / ');
};
this.value = output;
});

Related

Convert row and column range to rows and columns array

I want to convert the ranges of below fashion to rows and column notations
The example input and the expected output is given below
Input is A1 => Then output is [0,1,0,1]
B1 = [0,1,1,2]
A2 = [1,2,0,1]
C2 = [1,2,2,3]
C12 = [11,12,2,3]
A1:B4 = [0,4,0,2]
C1:D11 = [0,11,2,4]
F32:H43 = [31,43,5,8]
C:F = [null,null,2,6]
30:38 = [29,38,null,null]
76:79 = [75,79,null,null]
AA1:AD15 = [0,15,26,30]
Z2:B1 = [null,null,null,null] or undefined //any invalid range
2:B1 = [null,null,null,null] or undefined //any invalid range
0:0 = [null,null,null,null] or undefined //any invalid range
4-2 = [null,null,null,null] or undefined //any invalid range
6-7d7 = [null,null,null,null] or undefined //any invalid range
The general formula
[startrow-1,endrow,startcol-1,endcol]
30:38 = [startrow-1,endrow,null,null]
76:79 = [startrow-1,endrow,null,null]
C:F = [null,null,2,6]
A1
can also be written as
A1:A1 = [0,1,0,1]
C12
can also be written as
C12:C12 = [11,12,2,3]
My code works for these
A10:B10 //already works
A:B10 //not working so far
A10:B //not working so far
A:B //not working so far
10:10 //not working so far
<!DOCTYPE html>
<html>
<body>
<script>
function fromA1Notation(cell) {
var i, l, chr,
sum = 0,
A = "A".charCodeAt(0),
radix = "Z".charCodeAt(0) - A + 1;
if (typeof cell !== 'string' || !/^[A-Z]+$/.test(cell)) {
throw new Error("Expected column label");
}
for (i = 0, l = cell.length; i < l; i++) {
chr = cell.charCodeAt(i);
sum = sum * radix + chr - A + 1
}
return sum;
}
var input = "A1:B20";
if (input.length > 0 && input.match(/[A-Z]+[0-9]+:[A-Z]+[0-9]+/i) != null) {
var matched = input.match("([A-Za-z]+)([0-9]+):([A-Za-z]+)([0-9]+)");
console.log(JSON.stringify(matched))
if (matched != null) {
a1range = {
a1not: input,
c1: (fromA1Notation(matched[1].toUpperCase()) - 1),
r1: (matched[2] - 1),
c2: fromA1Notation(matched[3].toUpperCase()),
r2: matched[4]
};
if (a1range.c1 >= a1range.c2 || a1range.c1 >= a1range.c2) {
a1range = undefined;
}
console.log(a1range)
}
}
console.log(a1range)
</script>
</body>
</html>

Here is a solution that:
turns 'A1' pattern into 'A1:A1'
matches col, num, ':', col, num
converts col letters into 1-based index
does invalid range check
fixed the start row and start col index to be zero based
function A2N (str) {
return str.split('').reduce((acc, char, idx) => {
return acc += char.charCodeAt(0) - 65 + (idx * 26);
}, 1);
}
[ 'A1', 'B1', 'A2', 'C:F', 'A10:B10', 'A:B10', 'A10:B', 'A:B', '10:10', 'AA1', 'AAA1', 'A', '1', 'B', '20', 'Z9:A1', '#'
].forEach(str => {
let parts = str
.replace(/^(\w+)$/, '$1:$1') // turn 'A1' into 'A1:A1'
.match(/^([A-Z]*)([0-9]*)(?::([A-Z]*)([0-9]*))?$/);
let result = [ null, null, null, null ];
if(parts) {
result = [
parts[2] ? Number(parts[2]) : null,
parts[4] ? Number(parts[4]) : null,
parts[1] ? A2N(parts[1]) : null,
parts[3] ? A2N(parts[3]) : null
];
if(result[0] && result[1] && result[0] > result[1]) {
// invalid range
result[0] = null;
result[1] = null;
}
if(result[2] && result[3] && result[2] > result[3]) {
// invalid range
result[2] = null;
result[3] = null;
}
if(result[0]) {
// zero-based start row
result[0]--;
}
if(result[2]) {
// zero-based start col
result[2]--;
}
}
console.log(str, '==>', result);
});

Get amount of seconds from a formatted time string

How to convert a time string like 1m15s to 75s, 75 or 75000 ideally using momentjs.
I attempted to parse that string using new Date('1m1s') but it gives Invalid Date.
I don't want to resort to regex:
const second = (function () {
const countdownStep = '1h1m1s'.match(
/(?:(?<h>\d{0,2})h)?(?:(?<m>\d{0,2})m)?(?:(?<s>\d{0,2})s)?/i
);
return (
(countdownStep.groups.h
? parseInt(countdownStep.groups.h) * 3600
: 0) +
(countdownStep.groups.m
? parseInt(countdownStep.groups.m) * 60
: 0) +
(countdownStep.groups.s ? parseInt(countdownStep.groups.s) : 0)
);
})();

You can use the duration interface of momentjs:
let s = '1m15s';
// convert to duration format and pass to momentjs
let secs = moment.duration('PT' + s.toUpperCase()).as("seconds");
console.log("seconds: ", secs);
<script src="https://cdnjs.cloudflare.com/ajax/libs/moment.js/2.29.1/moment.min.js"></script>
Without library, it could be:
const unit = { s: 1, m: 60, h: 60*60 };
let s = '1m15s';
let secs = s.toLowerCase().match(/\d+./g)
.reduce((acc, p) => acc + parseInt(p) * unit[p.at(-1)], 0);
console.log("seconds: ", secs);

Another approach using plain js:
const getSecondsFromString = (str) => {
const hourIndex = str.indexOf("h")
const minuteIndex = str.indexOf("m")
const secondIndex = str.indexOf("s")
let hours = 0
let minutes = 0
let seconds = 0
if (hourIndex !== -1) {
hours = Number(str.substring(0, hourIndex))
}
if (minuteIndex !== -1) {
if (hourIndex !== -1) {
minutes = Number(str.substring(hourIndex + 1, minuteIndex))
} else {
minutes = Number(str.substring(0, minuteIndex))
}
}
if (secondIndex !== -1) {
if (minuteIndex !== -1) {
seconds = Number(str.substring(minuteIndex + 1, secondIndex))
} else if (hourIndex !== -1) {
seconds = Number(str.substring(hourIndex + 1, secondIndex))
} else {
seconds = Number(str.substring(0, secondIndex))
}
}
return hours * 3600 + minutes * 60 + seconds
}

Another approach without substring/replace, in O(n);
const getSecondsFromTimeString = (timeString) => {
let numberBeforeNextChar = 0;
let seconds = 0;
const symbolToSecondMap = {
"s" : 1,
"m" : 60,
"h" : 60*60,
"d" : 60*60*24
};
for (let i = 0; i < timeString.length; i++) {
let char = timeString.charAt(i);
if((+ char) <= 9 && (+ char) >= 0 ){
numberBeforeNextChar = (numberBeforeNextChar * 10) + parseInt(char);
continue;
}
if(char.toLowerCase() == char.toUpperCase()){
throw "Non alphanumeric symbol encountered";
}
if(symbolToSecondMap[char] === undefined){
throw "Invalid date alphabet encountered";
}
seconds = seconds + (numberBeforeNextChar * symbolToSecondMap[char]);
numberBeforeNextChar = 0;
}
return seconds;
}
console.log(getSecondsFromTimeString('1s'))
console.log(getSecondsFromTimeString('10s'))
console.log(getSecondsFromTimeString('1s4m10d3h'))
console.log(getSecondsFromTimeString('1s4m03h1h'))
console.log(getSecondsFromTimeString('10d'))

I tried this, is it helpful to you
let d = '1H1s';
d = d.toLowerCase();
let sec = 0;
if(d.indexOf('h') > -1) {
if (d.indexOf('m') == -1) {
d = d.substring(0, d.indexOf('h') + 1) +"0m"+d.substring(d.indexOf('h') + 1);
}
}
let newDs = d.replace('h',':').replace('m',':').replace('s','').split(':');
newDs.forEach((v, i) => sec += Math.pow(60, (newDs.length - i - 1)) * v);
console.log(sec);

24-hours time automatic regex

This is my code:
let input = document.getElementById('f2f11c3');
input.addEventListener('input', addcommas, false);
function addcommas() {
var v = document.getElementById('f2f11c3');
var t = v.value.replace(/\D/g, '');
var i, temp = '';
for (i = t.length; i >= 0; i -= 2) {
if (i == t.length) {
temp = t.substring(i - 2, i);
} else {
if (t.substring(i - 2, i) != "")
temp = t.substring(i - 2, i) + ':' + temp;
}
if (i < 0) {
temp = t.substring(0, i + 2) + ':' + temp;
break;
}
}
v.value = temp;
}
<input type="text" value="" maxlength="8" id="f2f11c3" />
But entered numbers can be greater than 24. Can you fully adapt to the time format?
Example: 23:59:59 and one more 00:00:00

Short answer
Take the string
Split that
Use modulo operator to get the result
Example
var time="24:70:65"; // Suppose we have this time
var parts=time.split(":");// Split the time with :
var part1=(+parts[0] % 24)
var part2=(+parts[1] % 60)
var part3=(+parts[2] % 60)
// Now to get answer in two digits you can
part1 = ("0" + part1).slice(-2);
part2 = ("0" + part2).slice(-2);
part3 = ("0" + part3).slice(-2);
console.log(`${part1}:${part2}:${part3}`)
Hope this will work

Auto slash(/) for date input using Javascript MM/YY

I'm working on an input form that includes a credit card expiration date for only Visa and Mastercard. These two restrict the expiration date to MM/YY. I'm using the following script elsewhere on the form to automatically introduce slashes as the user types in their birth date:
$(function() {
var date = document.getElementById('ccexp');
function checkValue(str, max) {
if (str.charAt(0) !== '0' || str == '00') {
var num = parseInt(str);
if (isNaN(num) || num <= 0 || num > max) num = 1;
str = num > parseInt(max.toString().charAt(0)) && num.toString().length == 1 ? '0' + num : num.toString();
}
return str;
}
date.addEventListener('keydown', function(e) {
this.type = 'text';
var input = this.value;
var key = e.keyCode || e.charCode;
if (key == 8 || key == 46) // checks if backspace or delete is being pressed
return false;
if (/\D\/$/.test(input)) input = input.substr(0, input.length - 1);
var values = input.split('/').map(function(v) {
return v.replace(/\D/g, '')
});
if (values[0]) values[0] = checkValue(values[0], 12); // validates month 1-12
if (values[1]) values[1] = checkValue(values[1], 31); // validates day 1-31
var output = values.map(function(v, i) {
return v.length == 2 && i < 2 ? v + '/' : v;
});
this.value = output.join('').substr(0, 10);
});
});
(The requirements on this project are for users to type in dates, rather than use a date picker). I'd like to use a version of this script to look for MM/YY, but I really don't know how to parse this well enough myself.

Displaying a number in Indian format using Javascript

I have the following code to display in Indian numbering system.
var x=125465778;
var res= x.toString().replace(/\B(?=(\d{3})+(?!\d))/g, ",");
Am getting this output :125,465,778.
I need output like this: 12,54,65,778.
Please help me to sort out this problem .

i'm late but i guess this will help :)
you can use Number.prototype.toLocaleString()
Syntax
numObj.toLocaleString([locales [, options]])
var number = 123456.789;
// India uses thousands/lakh/crore separators
document.getElementById('result').innerHTML = number.toLocaleString('en-IN');
// → 1,23,456.789
document.getElementById('result1').innerHTML = number.toLocaleString('en-IN', {
maximumFractionDigits: 2,
style: 'currency',
currency: 'INR'
});
// → ₹1,23,456.79
<div id="result"></div>
<div id="result1"></div>

For Integers:
var x=12345678;
x=x.toString();
var lastThree = x.substring(x.length-3);
var otherNumbers = x.substring(0,x.length-3);
if(otherNumbers != '')
lastThree = ',' + lastThree;
var res = otherNumbers.replace(/\B(?=(\d{2})+(?!\d))/g, ",") + lastThree;
alert(res);
Live Demo
For float:
var x=12345652457.557;
x=x.toString();
var afterPoint = '';
if(x.indexOf('.') > 0)
afterPoint = x.substring(x.indexOf('.'),x.length);
x = Math.floor(x);
x=x.toString();
var lastThree = x.substring(x.length-3);
var otherNumbers = x.substring(0,x.length-3);
if(otherNumbers != '')
lastThree = ',' + lastThree;
var res = otherNumbers.replace(/\B(?=(\d{2})+(?!\d))/g, ",") + lastThree + afterPoint;
alert(res);
Live Demo

Simple way to do,
1. Direct Method using LocalString()
(1000.03).toLocaleString()
(1000.03).toLocaleString('en-IN') # number followed by method
2. using Intl - Internationalization API
The Intl object is the namespace for the ECMAScript Internationalization API, which provides language sensitive string comparison, number formatting, and date and time formatting.
eg: Intl.NumberFormat('en-IN').format(1000)
3. Using Custom Function:
function numberWithCommas(x) {
return x.toString().split('.')[0].length > 3 ? x.toString().substring(0,x.toString().split('.')[0].length-3).replace(/\B(?=(\d{2})+(?!\d))/g, ",") + "," + x.toString().substring(x.toString().split('.')[0].length-3): x.toString();
}
console.log("0 in indian format", numberWithCommas(0));
console.log("10 in indian format", numberWithCommas(10));
console.log("1000.15 in indian format", numberWithCommas(1000.15));
console.log("15123.32 in indian format", numberWithCommas(15123.32));
if your input is 10000.5,
numberWithCommas(10000.5)
You will get output like this, 10,000.5

For integers only no additional manipulations needed.
This will match every digit from the end, having 1 or more double digits pattern after, and replace it with itself + ",":
"125465778".replace(/(\d)(?=(\d\d)+$)/g, "$1,");
-> "1,25,46,57,78"
But since we want to have 3 in the end, let's state this explicitly by adding extra "\d" before match end of input:
"125465778".replace(/(\d)(?=(\d\d)+\d$)/g, "$1,");
-> "12,54,65,778"

Given a number to below function, it returns formatted number in Indian format of digit grouping.
ex: input: 12345678567545.122343
output: 1,23,45,67,85,67,545.122343
function formatNumber(num) {
input = num;
var n1, n2;
num = num + '' || '';
// works for integer and floating as well
n1 = num.split('.');
n2 = n1[1] || null;
n1 = n1[0].replace(/(\d)(?=(\d\d)+\d$)/g, "$1,");
num = n2 ? n1 + '.' + n2 : n1;
console.log("Input:",input)
console.log("Output:",num)
return num;
}
formatNumber(prompt("Enter Number",1234567))
https://jsfiddle.net/scLtnug8/1/

I am little late in the game.
But here is the implicit way to do this.
var number = 3493423.34;
console.log(new Intl.NumberFormat('en-IN', { style: "currency", currency: "INR" }).format(number));
if you dont want currency symbol, use it like this
console.log(new Intl.NumberFormat('en-IN').format(number));

The easiest way is just to use Globalize plugin (read more about it here and here):
var value = 125465778;
var formattedValue = Globalize.format(value, 'n');

Try like below, I have found a number formatter Plugin here : Java script number Formatter
By using that i have done the below code, It works fine, Try this, It will help you..
SCRIPT :
<script src="format.20110630-1100.min.js" type="text/javascript"></script>
<script>
var FullData = format( "#,##0.####", 125465778)
var n=FullData.split(",");
var part1 ="";
for(i=0;i<n.length-1;i++)
part1 +=n[i];
var part2 = n[n.length-1]
alert(format( "#0,#0.####", part1) + "," + part2);
</script>
Inputs :
1) 125465778
2) 1234567.89
Outputs :
1) 12,54,65,778
2) 12,34,567.89

Simply use https://osrec.github.io/currencyFormatter.js/
Then all you need is:
OSREC.CurrencyFormatter.format(2534234, { currency: 'INR' });
// Returns ₹ 25,34,234.00

These will format the value in the respective systems.
$(this).replace(/\B(?=(?:\d{3})+(?!\d))/g, ',');
For US number system (millions & billions)
$(this).replace(/\B(?=(?:(\d\d)+(\d)(?!\d))+(?!\d))/g, ',');
For Indian number system (lakhs & crores)

This function can handle float value properly just addition to another answer
function convertNumber(num) {
var n1, n2;
num = num + '' || '';
n1 = num.split('.');
n2 = n1[1] || null;
n1 = n1[0].replace(/(\d)(?=(\d\d)+\d$)/g, "$1,");
num = n2 ? n1 + '.' + n2 : n1;
n1 = num.split('.');
n2 = (n1[1]) || null;
if (n2 !== null) {
if (n2.length <= 1) {
n2 = n2 + '0';
} else {
n2 = n2.substring(0, 2);
}
}
num = n2 ? n1[0] + '.' + n2 : n1[0];
return num;
}
this function will convert all function to float as it is
function formatAndConvertToFloatFormat(num) {
var n1, n2;
num = num + '' || '';
n1 = num.split('.');
if (n1[1] != null){
if (n1[1] <= 9) {
n2 = n1[1]+'0';
} else {
n2 = n1[1]
}
} else {
n2 = '00';
}
n1 = n1[0].replace(/(\d)(?=(\d\d)+\d$)/g, "$1,");
return n1 + '.' + n2;
}

Improvised Slopen's approach above, Works for both int and floats.
function getIndianFormat(str) {
str = str.split(".");
return str[0].replace(/(\d)(?=(\d\d)+\d$)/g, "$1,") + (str[1] ? ("."+str[1]): "");
}
console.log(getIndianFormat("43983434")); //4,39,83,434
console.log(getIndianFormat("1432434.474")); //14,32,434.474

Based on Nidhinkumar's question i have checked the above answers and
while handling negative numbers the output won't be correct for eg: -300 it should display as -300 but the above answers will display it as -,300 which is not good so i have tried with the below code which works even during the negative cases.
var negative = input < 0;
var str = negative ? String(-input) : String(input);
var arr = [];
var i = str.indexOf('.');
if (i === -1) {
i = str.length;
} else {
for (var j = str.length - 1; j > i; j--) {
arr.push(str[j]);
}
arr.push('.');
}
i--;
for (var n = 0; i >= 0; i--, n++) {
if (n > 2 && (n % 2 === 1)) {
arr.push(',');
}
arr.push(str[i]);
}
if (negative) {
arr.push('-');
}
return arr.reverse().join('');

Improvising #slopen's answer with decimal support and test cases.
Usage: numberToIndianFormat(555555.12) === "5,55,555.12"
utils.ts
export function numberToIndianFormat(x: number): string {
if (isNaN(x)) {
return "NaN"
} else {
let string = x.toString();
let numbers = string.split(".");
numbers[0] = integerToIndianFormat(parseInt(numbers[0]))
return numbers.join(".");
}
}
function integerToIndianFormat(x: number): string {
if (isNaN(x)) {
return "NaN"
} else {
let integer = x.toString();
if (integer.length > 3) {
return integer.replace(/(\d)(?=(\d\d)+\d$)/g, "$1,");
} else {
return integer;
}
}
}
utils.spec.ts
describe('numberToIndianFormat', () => {
it('nan should output NaN', () => {
expect(numberToIndianFormat(Number.NaN)).toEqual("NaN")
});
describe('pure integer', () => {
it('should leave zero untouched', () => {
expect(numberToIndianFormat(0)).toEqual("0")
});
it('should leave simple numbers untouched', () => {
expect(numberToIndianFormat(10)).toEqual("10")
});
it('should add comma at thousand place', () => {
expect(numberToIndianFormat(5555)).toEqual("5,555")
});
it('should add comma at lakh place', () => {
expect(numberToIndianFormat(555555)).toEqual("5,55,555")
});
it('should add comma at crore place', () => {
expect(numberToIndianFormat(55555555)).toEqual("5,55,55,555")
});
});
describe('with fraction', () => {
it('should leave zero untouched', () => {
expect(numberToIndianFormat(0.12)).toEqual("0.12")
});
it('should leave simple numbers untouched', () => {
expect(numberToIndianFormat(10.12)).toEqual("10.12")
});
it('should add comma at thousand place', () => {
expect(numberToIndianFormat(5555.12)).toEqual("5,555.12")
});
it('should add comma at lakh place', () => {
expect(numberToIndianFormat(555555.12)).toEqual("5,55,555.12")
});
it('should add comma at crore place', () => {
expect(numberToIndianFormat(55555555.12)).toEqual("5,55,55,555.12")
});
});
})

Indian money format function
function indian_money_format(amt)
{
amt=amt.toString();
var lastThree = amt.substring(amt.length-3);
var otherNumbers = amt.substring(0,amt.length-3);
if(otherNumbers != '')
lastThree = ',' + lastThree;
var result = otherNumbers.replace(/\B(?=(\d{2})+(?!\d))/g, ",") + lastThree;
alert(result)
return result;
}
indian_money_format(prompt("Entry amount",123456))

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