Develop Reference

How can I distinguish 0 and empty string in javascript?

Say
a=""; // empty String
b=0;
then
a==b; // returns true
What Test could I build to return true only if I compare two empty strings or two zero's?

Use the strict comparison operator, ===. This will not use JavaScript's default type coercion, so you will get the correct result.
"" === 0; // false

use === instead of == for checking undifined and zero or false compare

What's wrong with if (json.RowKey != json.NewRowKey)

I am using lint and for the following:
if (json.RowKey != json.NewRowKey)
It gives me a message:
Expected '!==' and instead saw '!='.
Can someone explain what this means?

== will attempt to convert the types of the two operands to be the same before comparing them. Thus "2" == 2 is true.
=== will not attempt to convert the types of the two operands so if they are not the same type, they will never be ===. Thus "2" === 2 is false.
It is better to use === and !== as your default choice and only use == or != when you explicitly want to allow type conversion. If you are not expecting the two operands to be different types, then === and !== is more likely to do what you are expecting.
This avoids coding mistakes when things are not really equal and your code doesn't intend for them to be equal.
Some odd things that happen with == (and do not happen with ===) that you may not expect and can lead to trouble. All of these evaluate to true:
null == undefined
false == '0'
0 == '0'
0 == ''
There's more detail written here: Which equals operator (== vs ===) should be used in JavaScript comparisons?

!== is not equal (neither value or type)
Eg:
var x = 5; x !== "5" returns true
var x = 5; x !== 5 returns false

It's a warning that tells you that you should consider using the !== operator instead of !=.
On the surface they both do the same thing - compare whether the two items are not equal. However, !== makes sure that both items are of the same type first (that is, it won't try to convert, say, a string to a number if one item being compared is a string and the other is a number), so it's generally safer to use and helps you avoid comparison errors.
For more details, see here: Which equals operator (== vs ===) should be used in JavaScript comparisons?

Yes,
if("1" == 1) //Returns true because it will convert the second 1 into a string to match
but
"1" !== 1 //Returns false because the first 1 is a string and the second 1 is an integer.
We are talking about variable types; with two equal signs, javascript (and php) will typecast the variable for you (convert the integer 1 into a string 1 to make the match and be true), but with the second statement, it will not.
1 === 1 //True
1 === "1" //False
1 == "1" //True
1 == 1 //True
Hope my answer makes sense to you.

In JavaScript, why is "0" equal to false, but when tested by 'if' it is not false by itself?

The following shows that "0" is false in Javascript:
>>> "0" == false
true
>>> false == "0"
true
So why does the following print "ha"?
>>> if ("0") console.log("ha")
ha

Tables displaying the issue:
and ==
Moral of the story use ===
table generation credit: https://github.com/dorey/JavaScript-Equality-Table

The reason is because when you explicitly do "0" == false, both sides are being converted to numbers, and then the comparison is performed.
When you do: if ("0") console.log("ha"), the string value is being tested. Any non-empty string is true, while an empty string is false.
Equal (==)
If the two operands are not of the same type, JavaScript converts the operands then applies strict comparison. If either operand is a number or a boolean, the operands are converted to numbers if possible; else if either operand is a string, the other operand is converted to a string if possible. If both operands are objects, then JavaScript compares internal references which are equal when operands refer to the same object in memory.
(From Comparison Operators in Mozilla Developer Network)

It's according to spec.
12.5 The if Statement
.....
2. If ToBoolean(GetValue(exprRef)) is true, then
a. Return the result of evaluating the first Statement.
3. Else,
....
ToBoolean, according to the spec, is
The abstract operation ToBoolean converts its argument to a value of type Boolean according to Table 11:
And that table says this about strings:
The result is false if the argument is the empty String (its length is zero);
otherwise the result is true
Now, to explain why "0" == false you should read the equality operator, which states it gets its value from the abstract operation GetValue(lref) matches the same for the right-side.
Which describes this relevant part as:
if IsPropertyReference(V), then
a. If HasPrimitiveBase(V) is false, then let get be the [[Get]] internal method of base, otherwise let get
be the special [[Get]] internal method defined below.
b. Return the result of calling the get internal method using base as its this value, and passing
GetReferencedName(V) for the argument
Or in other words, a string has a primitive base, which calls back the internal get method and ends up looking false.
If you want to evaluate things using the GetValue operation use ==, if you want to evaluate using the ToBoolean, use === (also known as the "strict" equality operator)

It's PHP where the string "0" is falsy (false-when-used-in-boolean-context). In JavaScript, all non-empty strings are truthy.
The trick is that == against a boolean doesn't evaluate in a boolean context, it converts to number, and in the case of strings that's done by parsing as decimal. So you get Number 0 instead of the truthiness boolean true.
This is a really poor bit of language design and it's one of the reasons we try not to use the unfortunate == operator. Use === instead.

// I usually do this:
x = "0" ;
if (!!+x) console.log('I am true');
else console.log('I am false');
// Essentially converting string to integer and then boolean.

Your quotes around the 0 make it a string, which is evaluated as true.
Remove the quotes and it should work.
if (0) console.log("ha")

It is all because of the ECMA specs ... "0" == false because of the rules specified here http://ecma262-5.com/ELS5_HTML.htm#Section_11.9.3 ...And if ('0') evaluates to true because of the rules specified here http://ecma262-5.com/ELS5_HTML.htm#Section_12.5

== Equality operator evaluates the arguments after converting them to numbers.
So string zero "0" is converted to Number data type and boolean false is converted to Number 0.
So
"0" == false // true
Same applies to `
false == "0" //true
=== Strict equality check evaluates the arguments with the original data type
"0" === false // false, because "0" is a string and false is boolean
Same applies to
false === "0" // false
In
if("0") console.log("ha");
The String "0" is not comparing with any arguments, and string is a true value until or unless it is compared with any arguments.
It is exactly like
if(true) console.log("ha");
But
if (0) console.log("ha"); // empty console line, because 0 is false
`

This is because JavaScript uses type coercion in Boolean contexts and your code
if ("0")
will be coerced to true in boolean contexts.
There are other truthy values in Javascript which will be coerced to true in boolean contexts, and thus execute the if block are:-
if (true)
if ({})
if ([])
if (42)
if ("0")
if ("false")
if (new Date())
if (-42)
if (12n)
if (3.14)
if (-3.14)
if (Infinity)
if (-Infinity)

This is the reason why you should whenever possible use strict equality === or strict inequality !==
"100" == 100
true because this only checks value, not the data type
"100" === 100
false this checks value and data type

The "if" expression tests for truthiness, while the double-equal tests for type-independent equivalency. A string is always truthy, as others here have pointed out. If the double-equal were testing both of its operands for truthiness and then comparing the results, then you'd get the outcome you were intuitively assuming, i.e. ("0" == true) === true. As Doug Crockford says in his excellent JavaScript: the Good Parts, "the rules by which [== coerces the types of its operands] are complicated and unmemorable.... The lack of transitivity is alarming." It suffices to say that one of the operands is type-coerced to match the other, and that "0" ends up being interpreted as a numeric zero, which is in turn equivalent to false when coerced to boolean (or false is equivalent to zero when coerced to a number).

if (x)
coerces x using JavaScript's internal toBoolean (http://es5.github.com/#x9.2)
x == false
coerces both sides using internal toNumber coercion (http://es5.github.com/#x9.3) or toPrimitive for objects (http://es5.github.com/#x9.1)
For full details see http://javascriptweblog.wordpress.com/2011/02/07/truth-equality-and-javascript/

I have same issue, I found a working solution as below:
The reason is
if (0) means false, if (-1, or any other number than 0) means true. following value are not truthy, null, undefined, 0, ""empty string, false, NaN
never use number type like id as
if (id) {}
for id type with possible value 0, we can not use if (id) {}, because if (0) will means false, invalid, which we want it means valid as true id number.
So for id type, we must use following:
if ((Id !== undefined) && (Id !== null) && (Id !== "")){
} else {
}
for other string type, we can use if (string) {}, because null, undefined, empty string all will evaluate at false, which is correct.
if (string_type_variable) { }

In JS "==" sign does not check the type of variable. Therefore, "0" = 0 = false (in JS 0 = false) and will return true in this case, but if you use "===" the result will be false.
When you use "if", it will be "false" in the following case:
[0, false, '', null, undefined, NaN] // null = undefined, 0 = false
So
if("0") = if( ("0" !== 0) && ("0" !== false) && ("0" !== "") && ("0" !== null) && ("0" !== undefined) && ("0" !== NaN) )
= if(true && true && true && true && true && true)
= if(true)

I came here from search looking for a solution to evaluating "0" as a boolean.
The technicalities are explained above so I won't go into it but I found a quick type cast solves it.
So if anyone else like me is looking to evaluate a string 1 or 0 as boolean much like PHP. Then you could do some of the above or you could use parseInt() like:
x = "0";
if(parseInt(x))
//false

Why does isNaN(" ") (string with spaces) equal false?

In JavaScript, why does isNaN(" ") evaluate to false, but isNaN(" x") evaluate to true?
I’m performing numerical operations on a text input field, and I’m checking if the field is null, "", or NaN. When someone types a handful of spaces into the field, my validation fails on all three, and I’m confused as to why it gets past the isNaN check.

JavaScript interprets an empty string as a 0, which then fails the isNAN test. You can use parseInt on the string first which won't convert the empty string to 0. The result should then fail isNAN.

You may find this surprising or maybe not, but here is some test code to show you the wackyness of the JavaScript engine.
document.write(isNaN("")) // false
document.write(isNaN(" ")) // false
document.write(isNaN(0)) // false
document.write(isNaN(null)) // false
document.write(isNaN(false)) // false
document.write("" == false) // true
document.write("" == 0) // true
document.write(" " == 0) // true
document.write(" " == false) // true
document.write(0 == false) // true
document.write(" " == "") // false
so this means that
" " == 0 == false
and
"" == 0 == false
but
"" != " "
Have fun :)

To understand it better, please open Ecma-Script spec pdf on page 43 "ToNumber Applied to the String Type"
if a string has a numerical syntax, which can contain any number of white-space characters, it can be converted to Number type. Empty string evaluates to 0. Also the string 'Infinity' should give
isNaN('Infinity'); // false

Try using:
alert(isNaN(parseInt(" ")));
Or
alert(isNaN(parseFloat(" ")));

From MDN reason for the issue you are facing
When the argument to the isNaN function is not of type Number, the value is first coerced to a Number. The resulting value is then tested to determine whether it is NaN.
You may want to check the following comprehensive answer which covers the NaN comparison for equality as well.
How to test if a JavaScript variable is NaN

I think it's because of Javascript's typing: ' ' is converted to zero, whereas 'x' isn't:
alert(' ' * 1); // 0
alert('x' * 1); // NaN

The Not-Entirely-Correct Answer
Antonio Haley's highly upvoted and accepted answer here makes a wrong assumption that this process goes through JavaScript's parseInt function:
You can use parseInt on the string ... The result should then fail isNAN.
We can easily disprove this statement with the string "123abc":
parseInt("123abc") // 123 (a number...
isNaN("123abc") // true ...which is not a number)
With this, we can see that JavaScript's parseInt function returns "123abc" as the number 123, yet its isNaN function tells us that "123abc" isn't a number.
The Correct Answer
ECMAScript-262 defines how the isNaN check works in section 18.2.3.
18.2.3 isNaN (Number)
The isNaN function is the %isNaN% intrinsic object. When the isNaN function is called with one argument number, the following steps are taken:
Let num be ? ToNumber(number).
If num is NaN, return true.
Otherwise, return false.
The ToNumber function it references is also defined in ECMAScript-262's section 7.1.3. In here, we get told how JavaScript handles Strings which are passed in to this function.
The first example given in the question is a string containing nothing but white space characters. This section states that:
A StringNumericLiteral that is empty or contains only white space is converted to +0.
The " " example string is therefore converted to +0, which is a number.
The same section also states:
If the grammar cannot interpret the String as an expansion of StringNumericLiteral, then the result of ToNumber is NaN.
Without quoting all of the checks contained within that section, the " x" example given in the question falls into the above condition as it cannot be interpreted as a StringNumericLiteral. " x" is therefore converted to NaN.

If you would like to implement an accurate isNumber function, here is one way to do it from Javascript: The Good Parts by Douglas Crockford [page 105]
var isNumber = function isNumber(value) {
return typeof value === 'number' &&
isFinite(value);
}

The function isNaN("") performs a String to Number type coercion
ECMAScript 3-5 defines the following return values for the typeof operator:
undefined
object (null, objects, arrays)
boolean
number
string
function
Better to wrap our test in a function body:
function isNumber (s) {
return typeof s == 'number'? true
: typeof s == 'string'? (s.trim() === ''? false : !isNaN(s))
: (typeof s).match(/object|function/)? false
: !isNaN(s)
}
This function is not intented to test variable type, instead it tests the coerced value. For instance, booleans and strings are coerced to numbers, so perhaps you may want to call this function as isNumberCoerced()
if there's no need to test for types other than string and number, then the following snippet might be used as part of some condition:
if (!isNaN(s) && s.toString().trim()!='') // 's' can be boolean, number or string
alert("s is a number")

NaN !== "not a number"
NaN is a value of Number Type
this is a definition of isNaN() in ECMAScript
1. Let num be ToNumber(number).
2. ReturnIfAbrupt(num).
3. If num is NaN, return true.
4. Otherwise, return false.
Try to convert any value to Number.
Number(" ") // 0
Number("x") // NaN
Number(null) // 0
If you want to determine if the value is NaN, you should try to convert it to a Number value firstly.

I suggest you to use the following function if you really want a proper check if it is an integer:
function isInteger(s)
{
return Math.ceil(s) == Math.floor(s);
}

That isNaN(" ") is false is part of the confusing behavior of the isNaN global function due to its coercion of non-numbers to a numeric type.
From MDN:
Since the very earliest versions of the isNaN function specification, its behavior for non-numeric arguments has been confusing. When the argument to the isNaN function is not of type Number, the value is first coerced to a Number. The resulting value is then tested to determine whether it is NaN. Thus for non-numbers that when coerced to numeric type result in a valid non-NaN numeric value (notably the empty string and boolean primitives, which when coerced give numeric values zero or one), the "false" returned value may be unexpected; the empty string, for example, is surely "not a number."
Note also that with ECMAScript 6, there is also now the Number.isNaN method, which according to MDN:
In comparison to the global isNaN() function, Number.isNaN() doesn't suffer the problem of forcefully converting the parameter to a number. This means it is now safe to pass values that would normally convert to NaN, but aren't actually the same value as NaN. This also means that only values of the type number, that are also NaN, return true.
Unfortunately:
Even the ECMAScript 6 Number.isNaN method has its own issues, as outlined in the blog post - Fixing the ugly JavaScript and ES6 NaN problem.

The isNaN function expects a Number as its argument, so arguments of any other type (in your case a string) will be converted to Number before the actual function logic is performed. (Be aware that NaN is also a value of type Number!)
Btw. this is common for all built-in functions - if they expect an argument of a certain type, the actual argument will be converted using the standard conversion functions. There are standard conversions between all the basic types (bool, string, number, object, date, null, undefined.)
The standard conversion for String to Number can be invoked explicit with Number(). So we can see that:
Number(" ") evaluates to 0
Number(" x") evaluates to NaN
Given this, the result of the isNaN function is completely logical!
The real question is why the standard String-to-Number conversion works like it does. The string-to-number conversion is really intended to convert numeric strings like "123" or "17.5e4" to the equivalent numbers. The conversion first skips initial whitespace (so " 123" is valid) and then tries to parse the rests as a number. If it is not parseable as a number ("x" isn't) then the result is NaN. But there is the explicit special rule that a string which is empty or only whitespace is converted to 0. So this explains the conversion.
Reference: http://www.ecma-international.org/ecma-262/5.1/#sec-9.3.1

I'm not sure why, but to get around the problem you could always trim whitespace before checking. You probably want to do that anyway.

I wrote this quick little function to help solve this problem.
function isNumber(val) {
return (val != undefined && val != null && val.toString().length > 0 && val.toString().match(/[^0-9\.\-]/g) == null);
};
It simply checks for any characters that aren't numeric (0-9), that aren't '-' or '.', and that aren't undefined, null or empty and returns true if there's no matches. :)

As other explained the isNaN function will coerce the empty string into a number before validating it, thus changing an empty string into 0 (which is a valid number).
However, I found that the parseInt function will return NaN when trying to parse an empty string or a string with only spaces. As such the following combination seems to be working well:
if ( isNaN(string) || isNaN(parseInt(string)) ) console.log('Not a number!');
This check will work for positive numbers, negative numbers and numbers with a decimal point, so I believe it covers all common numerical cases.

This function seemed to work in my tests
function isNumber(s) {
if (s === "" || s === null) {
return false;
} else {
var number = parseInt(s);
if (number == 'NaN') {
return false;
} else {
return true;
}
}
}

What about
function isNumberRegex(value) {
var pattern = /^[-+]?\d*\.?\d*$/i;
var match = value.match(pattern);
return value.length > 0 && match != null;
}

The JavaScript built-in isNaN function, is - as should be expected by default - a "Dynamic Type Operator".
Therefore all values which (during the DTC process) may yield a simple true | false such as "", " ", " 000", cannot be NaN.
Meaning that the argument supplied will first undergo a conversion as in:
function isNaNDemo(arg){
var x = new Number(arg).valueOf();
return x != x;
}
Explanation:
In the top line of the function body, we are (first) trying to successfully convert the argument into a number object. And (second), using the dot operator we are - for our own convenience - immediately stripping off, the primitive value of the created object.
In the second line, we are taking the value obtained in the previous step, and the advantage of the fact that NaN is not equal to anything in the universe, not even to itself, e.g.: NaN == NaN >> false to finally compare it (for inequality) with itself.
This way the function return will yield true only when, and only if, the supplied argument-return, is a failed attempt of conversion to a number object, i.e., a not-a-number number; e.g., NaN.
isNaNstatic( )
However, for a Static Type Operator - if needed and when needed - we can write a far simpler function such as:
function isNaNstatic(x){
return x != x;
}
And avoid the DTC altogether so that if the argument is not explicitly a NaN number, it will return false. Wherefore, testing against the following:
isNaNStatic(" x"); // will return false because it's still a string.
However:
isNaNStatic(1/"x"); // will of course return true. as will for instance isNaNStatic(NaN); >> true.
But contrary to isNaN, the isNaNStatic("NaN"); >> false because it (the argument) is an ordinary string.
p.s.:
The static version of isNaN can be very useful in modern coding scenarios. And it may very well be one of the main reasons I took my time for posting this.
Regards.

isNAN(<argument>) is a function to tell whether given argument is illegal number.
isNaN typecasts the arguments into Number type. If you want to check if argument is Numeric or not? Please use $.isNumeric() function in jQuery.
That is, isNaN(foo) is equivalent to isNaN(Number(foo))
It accepts any strings having all numerals as numbers for obvious reasons. For ex.
isNaN(123) //false
isNaN(-1.23) //false
isNaN(5-2) //false
isNaN(0) //false
isNaN('123') //false
isNaN('Hello') //true
isNaN('2005/12/12') //true
isNaN('') //false
isNaN(true) //false
isNaN(undefined) //true
isNaN('NaN') //true
isNaN(NaN) //true
isNaN(0 / 0) //true

I use this
function isNotANumeric(val) {
if(val.trim && val.trim() == "") {
return true;
} else {
return isNaN(parseFloat(val * 1));
}
}
alert(isNotANumeric("100")); // false
alert(isNotANumeric("1a")); // true
alert(isNotANumeric("")); // true
alert(isNotANumeric(" ")); // true

When checking if certain string value with whitespace or " "is isNaN maybe try to perform string validation, example :
// value = "123 "
if (value.match(/\s/) || isNaN(value)) {
// do something
}

I find it convenient to have a method specific to the Number class (since other functions that do conversions like parseInt have different outputs for some of these values) and use prototypal inheritance.
Object.assign(Number.prototype, {
isNumericallyValid(num) {
if (
num === null
|| typeof num === 'boolean'
|| num === ''
|| Number.isNaN(Number(num))
) {
return false;
}
return true;
}
});

I use the following.
x=(isNaN(parseFloat(x)))? 0.00 : parseFloat(x);

let isNotNumber = val => isNaN(val) || (val.trim && val.trim() === '');
console.log(isNotNumber(' '));
console.log(isNotNumber('1'));
console.log(isNotNumber('123x'));
console.log(isNotNumber('x123'));
console.log(isNotNumber('0'));
console.log(isNotNumber(3));
console.log(isNotNumber(' x'));
console.log(isNotNumber('1.23'));
console.log(isNotNumber('1.23.1.3'));
if(!isNotNumber(3)){
console.log('This is a number');
}

Implied string comparison, 0=='', but 1=='1'

I was debugging something and discovered some strangeness in JavaScript:
alert(1=='') ==> false
alert(0=='') ==> true
alert(-1=='') ==> false
It would make sense that an implied string comparison that 0 should = '0'. This is true for all non-zero values, but why not for zero?

According to the Mozilla documentation on Javascript Comparison Operators
If the two operands are not of the same type, JavaScript converts the
operands then applies strict
comparison. If either operand is a
number or a boolean, the operands are
converted to numbers; if either
operand is a string, the other one is
converted to a string
What's actually happening is that the strings are being converted to numbers.
For example:
1 == '1' becomes 1 == Number('1') becomes 1 == 1: true
Then try this one:
1 == '1.' becomes 1 == Number('1.') becomes 1 == 1: true
If they were becoming strings, then you'd get '1' == '1.', which would be false.
It just so happens that Number('') == 0, therefore 0 == '' is true

When javascript does implicit type conversions, the empty string literal will match the 0 integer. Do your comparison like this and you'll get your expected result:
alert(1==='') ==> false
alert(0==='') ==> false
alert(-1==='') ==> false

ECMA-262, 3rd edition, 11.9.3 regarding x == y, step 16:
If Type(x) is Number and Type(y) is String, return the result of the comparison x == ToNumber(y).
The empty string '' gets converted to 0 before the comparison.

This is just one of the truly hideous mangles that went into the JavaScript compromise. '' and 0 are both uninitialized values (equal to boolean false) and, therefore, equal.
To protect yourself from weird bugs like this, it's better to always use the === operator.

Javascript, like PHP, is weakly typed*. So when you compare 0 to '', the JS engine converts them to a similar datatype. Since 0 and '' both equate to boolean(false), "false == false" is true.
*Weakly typed languages do not require variables to be any specific data type, so you can set one variable as a string, change it to int, float, and back to string without the processor throwing errors.

In many languages, the empty string can be coerced to false.
Beware of doing comparisons with == instead of ===:
alert('' == '0'); //false
alert(0 == ''); // true
alert(0 =='0'); // true
== is not transitive.