Develop Reference

Counting numbers after decimal point in JavaScript

I have a problem in JavaScript. Is it possible to check how many numbers are after the decimal point? I tried to do it using a.toString().split(".")[1]), but if there is no decimal point in the number, there is an error. What should I do if I want the system to do nothing if there is no decimal point?

You're on the right track. You can also .includes('.') to test if it contains a decimal along with .length to return the length of the decimal portion.
function decimalCount (number) {
// Convert to String
const numberAsString = number.toString();
// String Contains Decimal
if (numberAsString.includes('.')) {
return numberAsString.split('.')[1].length;
}
// String Does Not Contain Decimal
return 0;
}
console.log(decimalCount(1.123456789)) // 9
console.log(decimalCount(123456789)) // 0

Convert to a string, split on “.”, then when there is no “.” to split on, assume it’s empty string '' (the part you’re missing), then get said string’s length:
function numDigitsAfterDecimal(x) {
var afterDecimalStr = x.toString().split('.')[1] || ''
return afterDecimalStr.length
}
console.log(numDigitsAfterDecimal(1.23456))
console.log(numDigitsAfterDecimal(0))

You could check if no dot is available, then return zero, otherwise return the delta of the lenght and index with an adjustment.
function getDigits(v) {
var s = v.toString(),
i = s.indexOf('.') + 1;
return i && s.length - i;
}
console.log(getDigits(0));
console.log(getDigits(0.002));
console.log(getDigits(7.654321));
console.log(getDigits(1234567890.654321));

The condition you need is:
number.split('.')[1].length
It checks if there are any numbers after the dot which separates the number from its decimal part.
I'm not sure if you are able to use split on numbers though. If not, parse it to a string.

You first need to convert the decimal number to string and then get the count of character after decimal point,
var a = 10.4578;
var str = a.toString();
if(str){
var val = str.split('.');
if(val && val.length == 2){
alert('Length of number after decimal point is ', val[1].length);
} else {
alert('Not a decimal number');
}
}
The output is 4

parse integer with javascript using parseInt and a radix

I am doing an online test and it asks me to write basic javascript code.
It asks me to parse a numberic string and convert it to a number of a different base. It needs me to return -1 if for whatever reason the conversion cannot be done.
I have written this:
function convert(strNumber, radix) {
var result = parseInt(strNumber, radix);
if(isNaN(result))
{return -1;}
return result;
}
Then it runs my code through various tests and all pass. Except one.
Apparently convert("ASD", 15) should be invalid according to the test and it expects it to be -1.
But Javascript happily converts it to number 10
I tried various things such as to add a try{}catch{} block and other things, but javascript never complains about converting "ASD" to base 15.
Is the test wrong, or is parseInt wrong?
By the way strNumber can be any base under 36.
So for instance:
convert("Z", 36) is 35

As I stated in the comment, parseInt will convert up to the point where it fails. So "A" is valid in that radix and "S" is not. So you would need to add a check.
var nums = "0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ".substr(0, radix)
var re = new RegExp("^[" + nums + "]+$","i")
if (!re.test(strNumber)) {
return -1
}

parseInt is behaving normally and is converting the letter A into 10 in base 15 (similar to how hex uses A for the number 10). The S and D are discarded, as parseInt accepts this type of malformed input.
From the parseInt documentation:
If parseInt encounters a character that is not a numeral in the specified radix, it ignores it and all succeeding characters and returns the integer value parsed up to that point.

As per official documentation the parseInt function behaves as following
For radices above 10, the letters of the alphabet indicate numerals
greater than 9. For example, for hexadecimal numbers (base 16), A
through F are used.
and
If parseInt encounters a character that is not a numeral in the
specified radix, it ignores it and all succeeding characters and
returns the integer value parsed up to that point.
Thus to prevent invalid arguments from being parsed they have to be validated first
function convert(strNumber, radix) {
if (isValidRadix(radix) && isValidInteger(strNumber, radix))
return parseInt(strNumber, radix);
return -1;
}
function isValidInteger(str, radix) {
var letters = ['0','1','2','3','4','5','6','7','8','9','A','B','C','D','E','F'].slice(0,radix);
str = str.toUpperCase();
for (var i=0; i<str.length; i++) {
var s = str.charAt(i);
if (letters.indexOf(s) == -1) return false;
}
return true;
}
function isValidRadix(radix) {
// 16 up to HEX system
return radix > 0 && radix <= 16;
}
console.log(convert("ASD", 15));
console.log(parseInt("ASD", 15));
console.log(convert("AAA", 15));

Evaluating numbers that include thousand separators

I understand that if the parseFloat function encounters any character other than numeric characters (0-9+-. and exponents) it just evaluates the number up to that character, discarding anything else.
I'm having a problem where I need to be able to validate numbers with thousand separators like so:
var number = "10,000.01"; //passes
var numberWithoutThousand = "10000.01"; //fails
//i.e:
if(parseFloat(number) <= 10000) {
return true;
}
//passess
problem is the above code returns true when technically that number is larger than 10,000.
What's the best way to get around this? I've considered stripping out the comma before testing the number, but not sure that is a good strategy.

You don't have numbers, you have strings, so just removing the comma is the way to go
number = number.replace(/\,/g, '');

Your "stripping the comma" strategy seems good to me.
if ( parseFloat( number.replace(",","") ) ) { etc(); }

As has been suggested, to have , in the number it must be a string and so do a search and replace. If you are having to do this on a regular basis then make yourself a reusable function.
Javascript
function myParseFloat(value) {
if (typeof value === 'string') {
value = value.replace(/,/g, '');
}
return parseFloat(value);
}
var number1 = "10,000.01",
number2 = "10000.01",
number3 = 10000.01;
console.log(myParseFloat(number1), myParseFloat(number2), myParseFloat(number3));
Output
10000.01 10000.01 10000.01
On jsFiddle

Javascript string/integer comparisons

I store some parameters client-side in HTML and then need to compare them as integers. Unfortunately I have come across a serious bug that I cannot explain. The bug seems to be that my JS reads parameters as strings rather than integers, causing my integer comparisons to fail.
I have generated a small example of the error, which I also can't explain. The following returns 'true' when run:
console.log("2" > "10")

Parse the string into an integer using parseInt:
javascript:alert(parseInt("2", 10)>parseInt("10", 10))

Checking that strings are integers is separate to comparing if one is greater or lesser than another. You should always compare number with number and string with string as the algorithm for dealing with mixed types not easy to remember.
'00100' < '1' // true
as they are both strings so only the first zero of '00100' is compared to '1' and because it's charCode is lower, it evaluates as lower.
However:
'00100' < 1 // false
as the RHS is a number, the LHS is converted to number before the comparision.
A simple integer check is:
function isInt(n) {
return /^[+-]?\d+$/.test(n);
}
It doesn't matter if n is a number or integer, it will be converted to a string before the test.
If you really care about performance, then:
var isInt = (function() {
var re = /^[+-]?\d+$/;
return function(n) {
return re.test(n);
}
}());
Noting that numbers like 1.0 will return false. If you want to count such numbers as integers too, then:
var isInt = (function() {
var re = /^[+-]?\d+$/;
var re2 = /\.0+$/;
return function(n) {
return re.test((''+ n).replace(re2,''));
}
}());
Once that test is passed, converting to number for comparison can use a number of methods. I don't like parseInt() because it will truncate floats to make them look like ints, so all the following will be "equal":
parseInt(2.9) == parseInt('002',10) == parseInt('2wewe')
and so on.
Once numbers are tested as integers, you can use the unary + operator to convert them to numbers in the comparision:
if (isInt(a) && isInt(b)) {
if (+a < +b) {
// a and b are integers and a is less than b
}
}
Other methods are:
Number(a); // liked by some because it's clear what is happening
a * 1 // Not really obvious but it works, I don't like it

Comparing Numbers to String Equivalents Without Using parseInt
console.log(Number('2') > Number('10'));
console.log( ('2'/1) > ('10'/1) );
var item = { id: 998 }, id = '998';
var isEqual = (item.id.toString() === id.toString());
isEqual;

use parseInt and compare like below:
javascript:alert(parseInt("2")>parseInt("10"))

Always remember when we compare two strings.
the comparison happens on chacracter basis.
so '2' > '12' is true because the comparison will happen as
'2' > '1' and in alphabetical way '2' is always greater than '1' as unicode.
SO it will comeout true.
I hope this helps.

You can use Number() function also since it converts the object argument to a number that represents the object's value.
Eg: javascript:alert( Number("2") > Number("10"))

+ operator will coerce the string to a number.
console.log( +"2" > +"10" )

The answer is simple. Just divide string by 1.
Examples:
"2" > "10" - true
but
"2"/1 > "10"/1 - false
Also you can check if string value really is number:
!isNaN("1"/1) - true (number)
!isNaN("1a"/1) - false (string)
!isNaN("01"/1) - true (number)
!isNaN(" 1"/1) - true (number)
!isNaN(" 1abc"/1) - false (string)
But
!isNaN(""/1) - true (but string)
Solution
number !== "" && !isNaN(number/1)

The alert() wants to display a string, so it will interpret "2">"10" as a string.
Use the following:
var greater = parseInt("2") > parseInt("10");
alert("Is greater than? " + greater);
var less = parseInt("2") < parseInt("10");
alert("Is less than? " + less);

Regular expression for DN addresses

How to write a regular expression in javascript that must follow the conditions
All segment in the DN address should follow the sequence cn=<name>,ou=<name>,o=<bic8>,o=swift
All segments should be separated with ,.
The DN address should have maximum of 100 characters.
No space is allowed.
Minimum of 2 and maximum of 10 segments are allowed in a DN address.
The <name> part must contain minimum of 2 characters and maximum 20 alphanumeric characters. The characters should be in lower case. Only one special character is allowed to be used i.e. -(Hypen).
The DN address will have maximum 2 numbers. The <name> part can contain maximum of 2 numerical digits.
Thanks in advance

I think .split() is a lot easier to use in this case.
First split the entire string on the ,'s and then split every separate segment of the resulting array on the ='s.
Especially on a well defined spec as this, split is more then enough to handle it.

Untested code follows, don't blame me if it blows up your computer:
var parseDn(str)
var m = /^cn=(.*?),ou=(.*?),o=(.*?),o=swift$/.exec(str);
if (!m) { return null; } // (a) and (b).
if (s.length > 100) { return null; } // (c).
if (/\s/.exec(s)) { return null; } // (d).
var x = {cn:m[1], ou:m[2], o:m[3]};
var isValidName = function(s) { return (/^[a-z-]{2,20}$/).exec(s); }
if (!isValidName(x.cn) || !isValidName(x.ou) || !isValidName(x.o)) {
return null; // (f).
}
var countNumbers = function(s) { return s.replace(/\D/g, "").length; }
if (countNumbers(x.cn)>2 || countNumbers(x.ou)>2 || countNumbers(x.o)>2) {
return null; // (g).
}
return x; // => {"cn":"name", "ou":"name", "o":"bic8"}
}
Note that (e) and a few of the points regarding "segments" are completely unchecked since the description is vague. But this should get you started...