I want to suppress before two zeros and include zeros before values with two digits.
How to do the logic kindly help me on this.
This can be done in one line in C#:
string stringValue = "056";
string newStringValue = int.Parse(stringValue).ToString("000");
int.Parse() will automatically strip the leading zeroes and give you an int, and the "000" format in .ToString will ensure the resulting string is always three digits and pad with leading zeroes if necessary.
In javascript (or any other language), just add enough leading zeros to ensure you have a length of 3, and then take the rightmost 3 characters:
const n = '56';
const padded = n.padStart(3,'0').slice(-3) ;
// padded = '056'
It's not much different in C#, except that you need an explicit intermediate (because you need that string's length):
string n = '56';
string t = n.PadLeft(3,'0') ;
string padded = t.SubString( t.Length-3 ) ;
Javascript solution:
/^([0]{0,})/ regex replaces all leading contiguous zeros. Then we just pad it to length 3 with an initial zero.
const numbers = ['56', '056', '0056', '00056'];
const leadingZero = (nums) => {
return nums.map((num) => {
if (!num?.length) throw new Error();
return removeContigiousLeadingZeros(num);
});
};
const removeContigiousLeadingZeros = (num) => {
return num.replace(/^([0]{0,})/, '0')
};
console.log(leadingZero(numbers));
Related
example:
new Intl.NumberFormat('de-DE', { minimumFractionDigits: 2, maximumFractionDigits: 2}).format('54546464646464654650000000000000065464645')
"54.546.464.646.464.660.000.000.000.000.000.000.000.000,00"
This is because when I format it, it places the last numbers in 0 and so you have no precision in the numbers that you show for example in amounts of money.
How could I solve it?
You may be entering a number too big. Perhaps a string solution could be used here:
function format(val, decimals) {
// You cannot use variables in regex syntax (/re/) so you must create a string
// then create the regex using RegExp constructor
const reStr = `(\\d{${decimals}})$`
const decimalRe = new RegExp(reStr);
return val
.replace(decimalRe, '\,$1') // Adds "," after last N numbers
.replace(/\B(?=(\d{3})+(?!\d))/g, '.'); // Adds "." every 3 numbers after ","
}
const res = format("54546464646464654650000000000000065464645", 2)
console.log(res)
The question became a bit long, but it explains the expected behaviour.
let regex = undefined;
const format = (string) => string.replace(regex, '');
format('0')
//0
format('00')
//0
format('02')
//2
format('-03')
//-3
format('023.2323')
//23.23
format('00023.2.3.2.3')
//23.23
In the above example you can see the expected results in comments.
To summarize. I'm looking for a regex not for test, for replace which formats a string:
removes 0s from the beginning if it's followed by any numbers
allows decimal digits, but just 2
allows negative numbers
allows decimal points, but just one (followed by min 1, max 2 decimal digits)
The last one is a bit difficult to handle as the user can't enter period at the same time, I'll have two formatter functions, one will be the input in the input field, and one for the closest valid value at the moment (for example '2.' will show '2.' in the input field, but the handler will receive the value '2').
If not big favour, I'd like to see explanation of the solution, why it works, and what's the purpose of which part.
Right now I'm having string.replace(/[^\d]+(\.\[^\d{1,2}])+|^0+(?!$)/g, ''), but it doesn't fulfill all the requirements.
You may use this code:
const arr = ['0', '00', '02', '-03', '023.2323', '00023.2.3.2.3', '-23.2.3.2.3']
var narr = []
// to remove leading zeroes
const re1 = /^([+-]?)0+?(?=\d)/
// to remove multiple decimals
const re2 = /^([+-]?\d*\.\d+)\.(\d+).*/
arr.forEach( el => {
el = el.replace(re1, '$1').replace(re2, '$1$2')
if (el.indexOf('.') >= 0)
el = Number(el).toFixed(2)
narr.push(el)
})
console.log(narr)
//=> ["0", "0", "2", "-3", "23.23", "23.23"]
If you aren't bound to the String#replace method, you can try this regex:
/^([+-])?0*(?=\d+$|\d+\.)(\d+)(?:\.(\d{1,2}))?$/
Inspect on regex101.com
It collects the parts of the number into capturing groups, as follows:
Sign: the sign of the number, +, - or undefined
Integer: the integer part of the number, without leading zeros
Decimal: the decimal part of the number, undefined if absent
This regex won't match if more then 2 decimal places present. To strip it instead, use this:
/^([+-])?0*(?=\d+$|\d+\.)(\d+)(?:\.(\d{1,2})\d*)?$/
Inspect on regex101.com
To format a number using one of the above, you can use something like:
let regex = /^([+-])?0*(?=\d+$|\d+\.)(\d+)(?:\.(\d{1,2}))?$/
const format = string => {
try{
const [, sign, integer, decimal = ''] = string.match(regex)
return `${(sign !== '-' ? '' : '-')}${integer}${(decimal && `.${decimal}`)}`
}catch(e){
//Invalid format, do something
return
}
}
console.log(format('0'))
//0
console.log(format('00'))
//0
console.log(format('02'))
//2
console.log(format('-03'))
//-3
console.log(format('023.23'))
//23.23
console.log(format('023.2323'))
//undefined (invalid format)
console.log(format('00023.2.3.2.3'))
//undefined (invalid format)
//Using the 2nd regex
regex = /^([+-])?0*(?=\d+$|\d+\.)(\d+)(?:\.(\d{1,2})\d*)?$/
console.log(format('0'))
//0
console.log(format('00'))
//0
console.log(format('02'))
//2
console.log(format('-03'))
//-3
console.log(format('023.23'))
//23.23
console.log(format('023.2323'))
//23.23
console.log(format('00023.2.3.2.3'))
//undefined (invalid format)
Another option is to use pattern with 3 capturing groups. In the replacement, use all 3 groups "$1$2$3"
If the string after the replacement is empty, return a single zero.
If the string is not empty, concat group 1, group 2 and group 3 where for group 3, remove all the dots except for the first one to keep it for the decimal and take the first 3 characters (which is the dot and 2 digits)
^([-+]?)0*([1-9]\d*)((?:\.\d+)*)|0+$
In parts
^ Start of string
( Capture group 1
[-+]? Match an optional + or -
) Close group
0* Match 0+ times a zero
( Capture group 2
[1-9]\d* Match a digit 1-9 followed by optional digits 0-9
) Close group
( Capture group 3
(?:\.\d+)* Repeat 0+ times matching a dot and a digit
) Close group
| Or
0+ Match 1+ times a zero
$ End of string
Regex demo
const strings = ['0', '00', '02', '-03', '023.2323', '00023.2.3.2.3', '-23.2.3.2.3', '00001234', '+0000100005.0001']
let pattern = /^([-+]?)0*([1-9]\d*)((?:\.\d+)*)|0+$/;
let format = s => {
s = s.replace(pattern, "$1$2$3");
return s === "" ? '0' : s.replace(pattern, (_, g1, g2, g3) =>
g1 + g2 + g3.replace(/(?!^)\./g, '').substring(0, 3)
);
};
strings.forEach(s => console.log(format(s)));
For example, I got that long nubmer 1517778188788. How can i get first 6 digits from that number, like 151777 and trim another digits?
Just convert a number to string and then slice it and convert it back to Number.
const a = 1517778188788;
const str_a = a.toString();
const result = Number(str_a.slice(0, 6));
new String(your_number).substring(0,6)
(basically converting it to a string and substringing it). Don't forget to parse it back afterwards
Applicable only when you want to strip last 7 digits, and the numbers have constant length (13 in this case). Still leaving you with first 6 ones though.
const nr = 1517778188788;
const result = Math.floor(nr / 10000000)
Try this:
var num = 1517778188788; // long number
var str = num.toString(); //convert number to string
var result = str.substring(0,6) // cut six first character
result = parseInt(result); // convert it to a number
here is a working fiddle
I need to parse string that contains two numbers that may be in three cases :
"646.60 25.10" => [646.60 25.10]
"1 395.86 13.50" => [1395.86, 13.50]
"13.50 1 783.69" => [13.50, 1 783.69]
In a simple case it's enough use 'number'.join(' ') but in the some cases there is thousand separator like in second and third ones.
So how could I parse there numbers for all cases?
EDIT: All numbers have a decimal separator in the last segment of a number.
var string1 = "646.60 25.10";// => [646.60 25.10]
var string2 = "1 395.86 13,50";// => [1395.86, 13,50]
var string3 = "13.50 1 783.69"; // => [13.50, 1 783.69]
function getArray(s) {
var index = s.indexOf(" ", s.indexOf("."));
return [s.substring(0,index), s.substring(index+1) ];
}
console.log(getArray(string1));
console.log(getArray(string2));
console.log(getArray(string3));
Assuming every number ends with a dot-digit-digit (in the comments you said they do), you can use that to target the right place to split with aregex.
That way, it is robust and general for any number of numbers (although you specified you have only two) and for any number of digits in the numbers, as long as it ends with the digit-dot-digit-digit:
str1 = "4 435.89 1 333 456.90 7.54";
function splitToNumbers(str){
arr = str.replace(/\s+/g,'').split(/(\d+.\d\d)/);
//now clear the empty strings however you like
arr.shift();
arr.pop();
arr = arr.join('&').split(/&+/);
return arr;
}
console.log(splitToNumbers(str1));
//4435.89,1333456.90,7.54
If I have a string like "something12" or "something102", how would I use a regex in javascript to return just the number parts?
Regular expressions:
var numberPattern = /\d+/g;
'something102asdfkj1948948'.match( numberPattern )
This would return an Array with two elements inside, '102' and '1948948'. Operate as you wish. If it doesn't match any it will return null.
To concatenate them:
'something102asdfkj1948948'.match( numberPattern ).join('')
Assuming you're not dealing with complex decimals, this should suffice I suppose.
You could also strip all the non-digit characters (\D or [^0-9]):
let word_With_Numbers = 'abc123c def4567hij89'
let word_Without_Numbers = word_With_Numbers.replace(/\D/g, '');
console.log(word_Without_Numbers)
For number with decimal fraction and minus sign, I use this snippet:
const NUMERIC_REGEXP = /[-]{0,1}[\d]*[.]{0,1}[\d]+/g;
const numbers = '2.2px 3.1px 4px -7.6px obj.key'.match(NUMERIC_REGEXP)
console.log(numbers); // ["2.2", "3.1", "4", "-7.6"]
Update: - 7/9/2018
Found a tool which allows you to edit regular expression visually: JavaScript Regular Expression Parser & Visualizer.
Update:
Here's another one with which you can even debugger regexp: Online regex tester and debugger.
Update:
Another one: RegExr.
Update:
Regexper and Regex Pal.
If you want only digits:
var value = '675-805-714';
var numberPattern = /\d+/g;
value = value.match( numberPattern ).join([]);
alert(value);
//Show: 675805714
Now you get the digits joined
I guess you want to get number(s) from the string. In which case, you can use the following:
// Returns an array of numbers located in the string
function get_numbers(input) {
return input.match(/[0-9]+/g);
}
var first_test = get_numbers('something102');
var second_test = get_numbers('something102or12');
var third_test = get_numbers('no numbers here!');
alert(first_test); // [102]
alert(second_test); // [102,12]
alert(third_test); // null
IMO the #3 answer at this time by Chen Dachao is the right way to go if you want to capture any kind of number, but the regular expression can be shortened from:
/[-]{0,1}[\d]*[\.]{0,1}[\d]+/g
to:
/-?\d*\.?\d+/g
For example, this code:
"lin-grad.ient(217deg,rgba(255, 0, 0, -0.8), rgba(-255,0,0,0) 70.71%)".match(/-?\d*\.?\d+/g)
generates this array:
["217","255","0","0","-0.8","-255","0","0","0","70.71"]
I've butchered an MDN linear gradient example so that it fully tests the regexp and doesn't need to scroll here. I think I've included all the possibilities in terms of negative numbers, decimals, unit suffixes like deg and %, inconsistent comma and space usage, and the extra dot/period and hyphen/dash characters within the text "lin-grad.ient". Please let me know if I'm missing something. The only thing I can see that it does not handle is a badly formed decimal number like "0..8".
If you really want an array of numbers, you can convert the entire array in the same line of code:
array = whatever.match(/-?\d*\.?\d+/g).map(Number);
My particular code, which is parsing CSS functions, doesn't need to worry about the non-numeric use of the dot/period character, so the regular expression can be even simpler:
/-?[\d\.]+/g
var result = input.match(/\d+/g).join([])
Using split and regex :
var str = "fooBar0123".split(/(\d+)/);
console.log(str[0]); // fooBar
console.log(str[1]); // 0123
The answers given don't actually match your question, which implied a trailing number. Also, remember that you're getting a string back; if you actually need a number, cast the result:
item=item.replace('^.*\D(\d*)$', '$1');
if (!/^\d+$/.test(item)) throw 'parse error: number not found';
item=Number(item);
If you're dealing with numeric item ids on a web page, your code could also usefully accept an Element, extracting the number from its id (or its first parent with an id); if you've an Event handy, you can likely get the Element from that, too.
As per #Syntle's answer, if you have only non numeric characters you'll get an Uncaught TypeError: Cannot read property 'join' of null.
This will prevent errors if no matches are found and return an empty string:
('something'.match( /\d+/g )||[]).join('')
Here is the solution to convert the string to valid plain or decimal numbers using Regex:
//something123.777.321something to 123.777321
const str = 'something123.777.321something';
let initialValue = str.replace(/[^0-9.]+/, '');
//initialValue = '123.777.321';
//characterCount just count the characters in a given string
if (characterCount(intitialValue, '.') > 1) {
const splitedValue = intitialValue.split('.');
//splittedValue = ['123','777','321'];
intitialValue = splitedValue.shift() + '.' + splitedValue.join('');
//result i.e. initialValue = '123.777321'
}
If you want dot/comma separated numbers also, then:
\d*\.?\d*
or
[0-9]*\.?[0-9]*
You can use https://regex101.com/ to test your regexes.
Everything that other solutions have, but with a little validation
// value = '675-805-714'
const validateNumberInput = (value) => {
let numberPattern = /\d+/g
let numbers = value.match(numberPattern)
if (numbers === null) {
return 0
}
return parseInt(numbers.join([]))
}
// 675805714
One liner
I you do not care about decimal numbers and only need the digits, I think this one liner is rather elegant:
/**
* #param {String} str
* #returns {String} - All digits from the given `str`
*/
const getDigitsInString = (str) => str.replace(/[^\d]*/g, '');
console.log([
'?,!_:/42\`"^',
'A 0 B 1 C 2 D 3 E',
' 4 twenty 20 ',
'1413/12/11',
'16:20:42:01'
].map((str) => getDigitsInString(str)));
Simple explanation:
\d matches any digit from 0 to 9
[^n] matches anything that is not n
* matches 0 times or more the predecessor
( It is an attempt to match a whole block of non-digits all at once )
g at the end, indicates that the regex is global to the entire string and that we will not stop at the first occurrence but match every occurrence within it
Together those rules match anything but digits, which we replace by an empty strings. Thus, resulting in a string containing digits only.