"07999323384"
How do I convert the number above to number, but once I use any numeric method it takes out the first zero
You use Number("07999323384"); except it will take out the first 0, its not possible to have a number beginning with 0 (except the number 0) because js removes leading 0s.
There is nothing you can do about this however that shouldnt matter because there should be no situation where you need it as a number type but also need the first 0
Create a copy in a seperate variable to do calculations on (which will drop the leading zero) and use the original variable to maintain the leading zero, it's either that or pad out the number after using your numeric methods presuming it's a fixed length as per this: How can I pad a value with leading zeros?
Related
I am trying to retain leading zeroes in an integer/number within JavaScript.
In this example I have an "ItemName" key which has either a string or an integer as its value.
I only want to have integers to be the value, but when I try converting any string value to a number, it will remove the leading zeroes.
I need to retain the zeroes so that the data sent will reliably be 6 digits, where in the example it would only be 4 digits.
The value is intended to be an identifier for a given Item.
{"Item": [{"ItemName":"007730","BusinessUnit":1}] }
{"Item": [{"ItemName":917730,"BusinessUnit":1}] }
This is for my work and we use ES6, but if there's an option outside of that, I'm certainly open to it.
Thanks in advance!
You can't have a number with leading zeroes in Javascript, because, as Randy Casburn said, they don't have any value. You have to convert it to a string and use String.padStart() to pad the string with zeroes. parseInt will work with leading zeroes. For example:
(294).toString().padStart(6, "0") --> "000294"
parseInt("000294") --> 294
I am parsing a string of multiple numbers between 1 and 10 with the eventual goal of adding them to a set.
There will be multiple concatenated numbers after a text identifier such as {text}12345678910.
I am currently using match(/\d/g) to grab the numbers but it separates 1 and 0 in 10. I then look for 0 in my String Array, see if there's a 1 in the element before it, turn it into a 10 and delete the other entry. Not very elegant.
How can I clean up my matching code? I definitely don't need to use regex for this, but it makes grabbing the numbers fairly easy.
You could just match with this regex:
/10|\d/g
(instead of the one you use currently, not additionally)
Regex is executed left-to-right, so first it finds any occurrences of 10, and then of other digits (so using, for example /\d|10/g or even /\d|(10)/g won't work either).
How can I extract and get just the numeric value after the hyphen in a string?
Here is the input string:
var x = "-2147467259"
After some processing.... return:
alert(2147467259)
How do I accomplish this?
You could replace away the hyphen:
alert(+x.replace("-", ""));
And yes, the + is important. It converts a string to a number; so you're removing the hypen by replacing it with nothing, and then essentially casting the result of that operation into a number. This operation will also work if no hyphen is present.
You could also use substr to achieve this:
alert(+x.substr(1));
You could also use parseInt to convert the string to a number (which will end up negative if a hyphen is persent), and then find its absolute value:
alert(Math.abs(parseInt(x, 10));
As Bergi notes, if you can be sure that the first character in the string is always a hyphen, you can simple return its negative, which will by default cast the value into a number and then perform the negative operation on it:
alert(-x);
You could also check to see if the number is negative or positive via a tertiary operator and then perform the respective operation on it to ensure that it is a positive Number:
x = x >= 0 ? +x : -x;
This may be cheaper in terms of performance than using Math.abs, but the difference will be minuscule either way.
As you can see, there really are a variety of ways to achieve this. I'd recommend reading up on JavaScript string functions and number manipulation in general, as well as examining JavaScript's Math object to get a feel for what tools are available to you when you go to solve a problem.
How about:
Math.abs(parseInt("-2147467259"))
Or
"-2147467259".replace('-','')
or
"-2147467259".replace(/\-/,'')
#1 option is converting the string to numbers. The #2 approach is removing all - from the string and the #3 option even though it will not be necessary on this example uses Regular Expression but I wanted to show the possibility of using RegEx in replace situations.
If you need a number as the final value #1 is your choice if you need strings #2 is your choice.
I have the problem that when i round a number to 2 decimals the parseFloat() function removes .00 from the number. I have tried
var num = parseFloat(Math.round(19 * 100) / 100).toFixed(2);
The return: num="19.00"
The return i need: num = 19.00
I know 19 = 19.00, but i am using a service that always require two decimals .00
The function returns a string with the right value. When i parse it to float the .00 is removed.
You cannot get 19.00 as float, only as string, because numbers always remove trailing zeros.
Maybe you can show us a bit more code to get an idea, there you need these trailing zeros?
Numbers do and can not hold information about their representation. They are only a numerical value.
When you display a number using window.alert, console.log or similar, you are not looking at a number, but at a string. Those display functions convert numbers to strings before displaying them. Number.toFixed also converts numbers into strings, with the difference being that it rounds them to two decimal places, so you end up with another representation of the same number.
What I am trying to say is that to display a number, you cannot get around converting it to a string. Whether you do it explicitly or the display function does it for you. When you send the number to the service that you are using, you are probably also sending a string (JSON, XML, etc. are always strings once you send them). If you need the value of the number for calculations, use it, then convert it in the end. No matter how, you have to do it in the end if you want those 0's at the end.
I have a button where in the code behind I add a onclick and I pass a unique ID which will be passed to the js function. The id starts with a 0.
It wasn't working and eventually I figured out that the number, id, it was passing was wrong...
Ie. see this: js fiddle
It works with a ' at the start and end of the number. Just wondering why 013 turns to 11. I did some googling and couldn't find anything...
Cheers
Robin
Edit:
Thanks guys. Yep understand now.
As in this case the 0 at the start has a meaning, here the recipient ID in a mailing list, I will use '013' instead of just 013, i.e. a string. I can then split the values in js as each of the 3 values represents a different id which will always be only 1 character long, i.e. 0-9.
A numeric literal that starts with a 0 is treated as an octal number. So 13 from base 8 is 11 in base 10...
Octal numeric literals have been deprecated, but still work if you are not in strict mode.
(You didn't ask, but) A numeric literal that starts with 0x is treated as hexadecimal.
More info at MDN.
In your demo the parameter is called id, which implies you don't need to do numerical operations on it - if so, just put it in quotes and use it as a string.
If you need to be able to pass a leading zero but still have the number treated as base 10 to do numerical operations on it you can enclose it in quotes to pass it as a string and then convert the string to a number in a way that forces base 10, e.g.:
something('013');
function something(id){
alert(+id); // use unary plus operator to convert
// OR
alert(parseInt(id,10)); // use parseInt() to convert
}
Demo: http://jsfiddle.net/XYa6U/5/
013 is octal, not decimal, it's equal 11 in decimal
You should note that 013 starts with a 0. In Javascript, this causes the number to be considered octal. In general you'll want to use the decimal, and hexadecimal number systems. Occasionally though, octal numbers are useful, as this question shows.
I hope this helps! :)
If the first digit of a number is a zero, parseInt interprets the number as an octal.
You can specify a base of ten like this:
parseInt(numberString, 10)
You could also remove such zeros with a regex like this (the result will be a string):
numberString.replace(/^0+/g, '');