In my node backend project, I've encountered a situation with my TypeScript code where a variable occasionally prints as the string 'undefined' instead of just being undefined. Is there a more elegant way to check that the variable is not equal to both 'undefined' and undefined? The current method of using the following code does not look very clean: if (xxx && xxx != 'undefined') {}
if(String(one) === 'undefined') {
// returns true for undefined and 'undefined'
}
The first approach, involving creating a string, avoids false positives for other falsey values such as null and false, but it may be slightly harder to grasp at first glance.
The second method, which uses two comparisons, is easier to understand but comes with the drawback of potential false positives for falsey values like null, 0, and false.
The third option, requiring more complex comparisons, is clearer to comprehend but results in a longer expression.
const one = undefined;
const two = 'undefined';
const three = 0;
// using the string creation method
if(String(one) === 'undefined') console.log("one is undefined");
if(String(two) === 'undefined') console.log("two is undefined");
if(String(three) === 'undefined') console.log("three is undefined");
// using two checks; easier to understand but may yield false positives
if(!one || one === 'undefined') console.log("one is undefined");
if(!two || two === 'undefined') console.log("two is undefined");
if(!three || three === 'undefined') console.log("three is undefined"); // false positive
// using two checks; easier to understand but with a longer expression
if(one === undefined || one === 'undefined') console.log("one is undefined");
if(two === undefined || two === 'undefined') console.log("two is undefined");
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