I'm working on creating a third interface that depends on the value of properties from two other interfaces, while also introducing a unique third property. I've chosen not to extend the third property from the first two interfaces as it may not be necessary for all cases, and making it optional doesn't seem like the right solution.
Here is my approach:
- Create two interfaces named A & B with similar properties but possibly different values
- Define a type that represents the union of these two interfaces
- Create a new type including the third property and combine it with the union of the two types
Here's how I have implemented this:
const A_PROP1 = 'A_PROP1';
const A_PROP2_VALUES = {
type1: 'type1',
type2: 'type2',
type3: 'type3',
};
const B_PROP1 = 'B_PROP1';
const B_PROP2_VALUES = {
type4: 'type4',
type5: 'type5',
};
interface A {
prop1: typeof A_PROP1;
prop2: keyof typeof A_PROP2_VALUES;
}
interface B {
prop1: typeof B_PROP1;
prop2: keyof typeof B_PROP2_VALUES;
}
type AB = A | B;
type ABC = AB & {
prop3: 'prop3_value';
}
However, I am encountering an issue when defining the following:
const test: ABC = {
prop3: 'prop3_value',
prop1: A_PROP1,
prop2: 'random entered value', //The suggestion for this one shows the combination of A_PROP2_VALUES and B_PROP2_VALUES
}
The warning arises in prop2 where it displays a union of A_prop2_values and B_prop2_values when I specifically want only A_prop2_values. How can I resolve this?
Edit: Following suggestions, I suspect this could be an IDE issue rather than a problem with Typescript itself (I didn't check the Typescript log). It seems that the IDE automatically combines possible values to provide guidance, leading to a type error if an incorrect value is selected.
This matter has been resolved, apologies for any confusion