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Ternary operator inside if condition not transpiled properly using babel

let a = null, b = null;
if ((a !== null) && a.active && ((a === null) ? (a === b) : true)) {
}
After babel transpile above condition is transformed into
if (a !== null && a.active && a === null ? a === b : true) {
}
and it is always returning true because ternary operator has less precedency than comparison operators
any solution not to omit parentheses in this kind of conditions?

if statement with javascript condition operator

just have 2 question regarding JS conditional operator, is the below 2 expression valid?
1.
if(isUser && isUser === true || isGuest && isGuest === true){
//...
}
I am wondering do I have to add () to make it like and still have the same functioning:
if((isUser && isUser === true) || (isGuest && isGuest === true)){
//...
}
const items = list.orderList && list.orderList.isUser === true || list.orderList.isGuest ? list.items : [];
I am wondering do I have to add () to make it like and functioning the same as above conditional operator:
const items = list.orderList && (list.orderList.isUser === true || list.orderList.isGuest === true) ? list.items : [];

As per Operator Precedence in the MDN docs, logical AND takes precedence over logical OR. Therefore,
expression1 || expression2 && expression3
will evaluate to
expression1 || (expression2 && expression3)
Therefore,
isUser && isUser === true || isGuest && isGuest === true
naturally evaluates to
(isUser && isUser === true) || (isGuest && isGuest === true)
anyway, so you do not need parentheses..
But since, in your second example, you want to evaluate OR then AND, you do need parentheses for it to evaluate the way you require, as
list.orderList && list.orderList.isUser === true || list.orderList.isGuest
will evaluate to
(list.orderList && list.orderList.isUser === true) || list.orderList.isGuest

JavaScript short circuit evaluation error?

The two below snippets of JS code have had me confused, in my eyes both should work the same, due to short circuit evaluation. But for some reason snippet '1' causes the error (on the third line):
Cannot read property 'match' of undefined
Array 'a' holds 3 character values user entered into inputs. I want the code to return true if the char is undefined, an empty string, or a letter or number.
To be clear, this fails when a = ['a', '/'];
Snippet 1)
return typeof a[0] === 'undefined' || a[0] === '' || a[0].match(/^[a-z0-9]+$/i)
&& typeof a[1] === 'undefined' || a[1] === '' || a[1].match(/^[a-z0-9]+$/i)
&& typeof a[2] === 'undefined' || a[2] === '' || a[2].match(/^[a-z0-9]+$/i);
Snippet 2)
if (typeof a[0] === 'undefined' || a[0] === '' || a[0].match(/^[a-z0-9]+$/i)) {
if (typeof a[1] === 'undefined' || a[1] === '' || a[1].match(/^[a-z0-9]+$/i)) {
if (typeof a[2] === 'undefined' || a[2] === '' || a[2].match(/^[a-z0-9]+$/i)) {
return true;
}
return false;
}
return false;
}
return false;
Surely a[2].match should never be evaluated if a[2] is undefined due to the first conditional in the 'if'?

The answer is simple. Take a look to the order of operations .
AND binds more than OR.
In your Snippet 1 the expression is like:
a1 || b1 || (c1 && a2) || b2 || (c2 && a3) || b3 || c3
Your Snippet 2 is like:
(a1 || b1 || c1) && (a2 || b2 || c2) && (a3 || b3 || c3)

#Christoph is right but you also need to add something like !== null after match like
return (typeof a[0] === 'undefined' || a[0] === '' || a[0].match(/^[a-z0-9]+$/i) !==null ) && (typeof a[1] === 'undefined' || a[1] === '' || a[1].match(/^[a-z0-9]+$/i) !== null ) && (typeof a[2] === 'undefined' || a[2] === '' || a[2].match(/^[a-z0-9]+$/i) !== null);
You can take a look at this fiddle http://jsfiddle.net/dv360q1p/1/ which implements your question

How can I loop through two arrays comparing values in Javascript?

I have the following Javascript function where the parameters newValue and oldValue are arrays of integers and the same length. Any values in these arrays can be an integer, undefined or null:
function (newValue, oldValue) {
});
Is there some way that I could check the values in the arrays one element at a time and then do an action only if:
newValue[index] is >= 0 and < 999
oldValue[index] is >= 0 and < 999
newValue[index] is not equal to oldValue[index]
What I am not sure of is how can I handle in my checks and ignore the cases where newValue or oldValue are not null and not undefined? I know I can do a check as in if (newValue) but then this will show false when it's a 0.
Update:
I had a few quick answers so far but none are checking the right things which I listed above.

compare against null and undefined:
if (newValue[index] !== null && typeof newValue[index] !== 'undefined') {}
for OPs update:
n = newValue[index];
o = oldValue[index];
if (
n !== null && typeof n !== 'undefined' && n >= 0 && n < 999 &&
o !== null && typeof o !== 'undefined' && o >= 0 && o < 999
) {
// your code
}
for array-elements its not necessary to use typeof so n !== undefined is ok because the variable will exist.
n = newValue[index];
o = oldValue[index];
if (
n !== null && n !== undefined && n >= 0 && n < 999 &&
o !== null && o !== undefined && o >= 0 && o < 999 &&
n !== o
) {
// your code
}

This will do it:
function isEqual (newValue, oldValue) {
for (var i=0, l=newValue.length; i<l; i++) {
if (newValue[i] == null || newValue[i] < 0 || newValue[i] >= 999
|| oldValue[i] == null || oldValue[i] < 0 || oldValue[i] >= 999)
continue;
if (newVale[i] !== oldValue[i])
return false;
}
return true;
}

if (newValue != null || newValue != undefined) && (oldValue != null || oldValue != undefined)

&& Operator precendence in JavaScript

if(e.FN === ' ' && e.GN === ' ' && e.LN === ' ' && e.DB === ' '){
This condition is never evaluated at all. Is this the way to check if all the values are null.

Right now you are checking if those values are all equal to a Space. The === not only compares the values but ensures they are the same type. However, if e.FN is null both e.FN == ' ' and e.FN === ' " will always return false. I think what you want is
if(e.FN === null && e.GN === null && e.LN === null && e.DB === null)
or even better, if you don't care if they are null, undefined or 0 you could do
if(e.FN && e.GN && e.LN && e.DB)

try:
if(e.FN === null && e.GN === null && e.LN === null && e.DB === null){

To check if all the values are null you should do:
if(e.FN === null && e.GN === null && e.LN === null && e.DB === null){
if the first value is not null javascript doesn't evaluate other condition, because if the first condition is false it's not possible for all the conditions to be true.