i have this number 1234567890
this is how i want to display it
1234 567 890
im trying this
console.log('1234567899'.replace(/(\d)(?=(\d{3})+$)/g, '$1 '));
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which angular version are you using?Abhishek Mani– Abhishek Mani2018-11-23 06:33:04 +00:00Commented Nov 23, 2018 at 6:33
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Can you explain the logic you're looking for? Do you want to omit the first comma when there is only one digit in the first section, or for two as well, or for all 3, or what?CertainPerformance– CertainPerformance2018-11-23 06:33:58 +00:00Commented Nov 23, 2018 at 6:33
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4 Answers
One option would be to have an optional group that starts at the beginning of the string and (greedily) matches the number of the leading-digits-without-commas you want to have. Then, instead of replacing with just \1, replace with \1\2 (the optional group plus the second captured digits):
const format = str => str.replace(
/(^(?:\d{1,2}))?(\d{1,3})(?=(?:\d{3})+$)/g,
// ^^^ change these to change the number of unbroken leading digits
'$1$2 '
);
console.log(format('1234567899'));
console.log(format('01234567899'));
console.log(format('101234567899'));The above snippet's optional group begins with \d{1,2}, which means that there will be between 3 and 5 leading digits, unbroken by commas. To change that quantity, just change the number of repetitions.
The leading group (^(?:\d{1,2}))? means: optionally, the beginning of the string, followed by one or two digits.
answered Nov 23, 2018 at 6:53
CertainPerformance
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Comments
- Divide your string into 2 parts.
- String contains space after 4 digits
- String contains space after 3 digits.
- Add space after n digits
Try with this
var str = "1234567890";
var temp = str.substring(0,4); // get first 4 digits
//Add space after 3 digits. You can use same logic to add space after 4 digits as well
var z = [...str.substring(4)].map((d, i) => i % 3 == 0 ? ' '+d : d).join('').trim();
//Concatenate both strings
var result = temp + ' ' + z;
//Display result
console.log(temp);
console.log(z);
console.log(result);
Comments
Try this regex which ensures the numbers will be grouped in bunches of three except the first digits which can be grouped from four to five.
Match /(^\d{4}|\d{3})(?=(\d{3})*$)/g and replace with $1
Here is some sample javascript code,
console.log('1234567899' + ' --> ' + '1234567899'.replace(/(^\d{4}|\d{3})(?=(\d{3})*$)/g, '$1 '));
console.log('12345678991' + ' --> ' + '12345678991'.replace(/(^\d{4}|\d{3})(?=(\d{3})*$)/g, '$1 '));
console.log('123456789912' + ' --> ' + '123456789912'.replace(/(^\d{4}|\d{3})(?=(\d{3})*$)/g, '$1 '));
console.log('1234567899123' + ' --> ' + '1234567899123'.replace(/(^\d{4}|\d{3})(?=(\d{3})*$)/g, '$1 '));
answered Nov 23, 2018 at 7:38
Pushpesh Kumar Rajwanshi
18.4k2 gold badges22 silver badges39 bronze badges
Comments
You can do with this way.
let number = '1234567890';
let result = number.replace(/(\d{4})(\d{3})(\d{3})/, "$1 $2 $3");
console.warn(result);
