Dealing with optional properties can be quite tedious. Consider the object test1 in TypeScript:
interface Test {
a?: { b?: { c?: { d?: string } } };
}
const test1: Test = { a: { b: { c: { d: 'e' } } } };
Handling the absence of each property is crucial to avoid errors when trying to access the value 'e'. For example:
let result;
const test2: Test = { a: { b: {} } };
Directly accessing the property d will throw an error if c is undefined, as it does not have the property d:
result = test2.a.b.c.d // TypeError: undefined is not an object
Manually checking each property is necessary:
let result;
if (test2 && test2.a && test2.a.b && test2.a.b.c && test2.a.b.c.d !== undefined) {
result = test2.a.b.c.d;
}
What is the most efficient and best practice approach to tackle this common issue?
Using a try/catch block is an option, but it may not be the shortest solution. Passing test2.a.b.c.d as a function argument to handle the error also seems problematic, as the error would occur before the function is executed.