Skipping the validation of a variable in a return statement with Angular and Javascript

Question

Is there a way to exclude level a.3 from the return in the completion form if a.3 is undefined?


  scope.isValid = function() {
      return a.1 && a.2 && a.3;
    };


ng-disabled="!isValid()"

Answer 1

Answer №1

Here's a solution I believe will work:

scope.checkValidity = function() {
    return (b.3 === undefined ?  b.1 && b.2  : b.1 && b.2 && b.3);
}

I recommend familiarizing yourself with ternary operators, as they can be incredibly useful once you understand how to use them effectively.

Answer 2

Here's a solution I believe will work:

scope.checkValidity = function() {
    return (b.3 === undefined ?  b.1 && b.2  : b.1 && b.2 && b.3);
}

I recommend familiarizing yourself with ternary operators, as they can be incredibly useful once you understand how to use them effectively.

Answer 3

To achieve normalization, you must simplify the expression to

(a.1 && a.2 && a.3) || (a.1 && a.2),

Alternatively, you have the option of utilizing array reduce to locate the final non-null/empty value.

[a.1, a.2, a.3].reduce(function(a,b) { return b || a; }, null);

Answer 4

To achieve normalization, you must simplify the expression to

(a.1 && a.2 && a.3) || (a.1 && a.2),

Alternatively, you have the option of utilizing array reduce to locate the final non-null/empty value.

[a.1, a.2, a.3].reduce(function(a,b) { return b || a; }, null);

Answer 5

Using the ternary operator is crucial in this task

result = (x.y === undefined)?(x.z && x.w):(x.z && x.w && x.y);

While it may be possible to achieve the same outcome without it, I find this approach to be more clear and readable.

Answer 6

Using the ternary operator is crucial in this task

result = (x.y === undefined)?(x.z && x.w):(x.z && x.w && x.y);

While it may be possible to achieve the same outcome without it, I find this approach to be more clear and readable.