We Keep Coding

problem with JS comparison logic if/and/or

so here is the scenario
if ( ( a >= b) && ( (c < d) && (d != '') ) )
cannot get this logic on this to work correctly
so if d = '' it would cause that to be false. which would mean that the whole thing would equate to false. Problem is I need it to trigger when a >= b but also needs to include the and for c < d but only if d != '', in other words ignore the c < d part if d = '', otherwise used the c < d part to prevent a >= b from triggering.
hope this is making sense. I am trying to avoid doing and if/else or switch.

One comment would be if you are not absolutely avoid using if/else or switch then you could define funcitons use them and use thiss functions in your statement above.
Apart from that, changing the order of some expressions might do what you want. c < d would not be evaluated if d = '' is not true. a >= b would not be evaluated if the expresion before second or, ||, is true or d == '', empty string, is true.
if ((d != '' && c < d) || (d != '' && c >= d && a >= b))
which is;
if (d != '' && (c < d || a >= b))

thanks for your suggestions, I found a workaround though
I just put
if (d == '') { d = Infinity; }
then
if ( (a >= b) && (c < d) )
and that solved it
thanks for everyone's help

What does the first operand on var __decorator = (this && this.__decorate) || function() {...} try to do?

When TypeScript decorator (class-decorator) gets transpiled to JavaScript it yields this:
var __decorate = (this && this.__decorate) || function (decorators, target, key, desc) {
var c = arguments.length, r = c < 3 ? target : desc === null ? desc = Object.getOwnPropertyDescriptor(target, key) : desc, d;
if (typeof Reflect === "object" && typeof Reflect.decorate === "function") r = Reflect.decorate(decorators, target, key, desc);
else for (var i = decorators.length - 1; i >= 0; i--) if (d = decorators[i]) r = (c < 3 ? d(r) : c > 3 ? d(target, key, r) : d(target, key)) || r;
return c > 3 && r && Object.defineProperty(target, key, r), r;
};
Now, what gets assigned to var __decorate variable when the (this && this.__decorate) returns true?
Also, if it gets abandoned how does it affect the whole code?

JS does not nessesarily return true or false from a logical statement. A || B will return A if it is truthy or B if A is falsy. Similarly A && B will return A if it is falsy and B otherwise.
Both || and && are also so-called short-circuit operators where the second operand is not even executed if it is not needed in order to determine the result. So (A && A.B) is often used to get the value of A.B while guarding against A being undefined, or A = B || C used to set A to B if it is defined, or to declare the definition C if it is not.
See, for example, MDN for more details.
(this && this._decorate) looks like it is checking if _decorate is already defined on this. First it is checking if this is defined (otherwise an error would be thrown) then checking the value. If this._decorate is already defined then the function declaration will be skipped (as the existing this._decorate function will be truthy as all functions are).
The only JS operator which is guaranteed to return an actual boolean value is the not operator !.

Conditional logical operator using four variables

I have a function that returns a condition using four variables.
(payload, variables) => {
return payload.newMessage.lenderId === variables.lenderId && payload.newMessage.user.id === variables.authId
}
I want to be able to say C === D is only if C and D exists. What would be the optimal expression for this? So A === B is a sufficient condition if C and D doesn't exist, but if C and D exist, A === B and C === D both have to be met.

A and B must always be equal, and one of the following must be true:
There is no C
There is no D
C and D are equal
(A === B) && (!C || !D || C === D)
Beyond this construction, you should know what you mean by "exists" - is it enough that they not be undefined? Is it any truthy value? Etc.

function customCheck(a,b,c,d) {
const abComparison = a === b;
if (c && d) {
return abComparison && (c === d);
}
return abComparison;
}
From your description seems something like above?

how can you use ! and && instead of ||?

How can you write an or function without using || operator? ! and && are permissible.
var output = or(true, false);
console.log(output); // --> true;

There are two basic tautologies that you just have to combine. For any a, b:
De Morgan's law: !(a && b) if and only if !a || !b
Double negation: !!a if and only if a
Thus a || b if and only if !(!a && !b) so
function or(a, b) {
return !(!a && !b);
}
That being sad this is only true if a, b are boolean. Note that || operator in JavaScript has side effect:
> var a = 'foo';
> var b = 'bar';
> a || b
'foo'
but
> !(!a && !b)
true
I doubt this can be achieved solely with ! and &&. However this can be implemented without ||:
function or(a, b) {
if (a) {
return a;
}
return b;
}
This still is not 100% the same as ||. That's because a || b evaluates b only if a is false. In particular for functions f() || g() the call g() will be evaluated only when f() is false. So this is another side effect of || which I don't think can be emulated by a function at all.

When is or true?
Whenever either A is true or B is true -- but that answer includes the word "or".
So let's put it another way: whenever it is not the case that both A is false and B is false.

!a && !b = !(a || b)
!a || !b = !(a && b)
!(a || b) = (a || b) else
so to answer your question
!a && !b else // this is the or
but that is a horrible thing to do. Use positive logic and all the tools available to make your code readable.

Are you asking something like this?
var a;
var b = 3;
console.log(a || b);
console.log(!a && b);

Just because I wanted to prove to myself the original question is entirely possible ;) and I didn't use a |
function or(a, b){
var c = console.log;
console.log = function(input) {
c.apply(console, [input]);
window.output = !window.output;
}
return !(!a && !b);
}
// OP Question
var output = or(true, false);
console.log(output); // --> true;
console.log(output); // --> false;

What is the order of precedence for boolean operators in JS?

e.pageX = e.clientX + (html && html.scrollLeft || body && body.scrollLeft || 0) - (html.clientLeft || 0)
how would this expression evaluate?

You can find a precedence table for Javascript operators on MDN: https://developer.mozilla.org/en/JavaScript/Reference/Operators/Operator_Precedence. It is a great source for Javascript documentation in general.
Anyway, && has higher precedence than || so a && b || c || d is equivalent to (((a && b) || c) || d). This is similar to many other language with a C-inspired syntax.