interface Human {
firstName: string;
lastName: string;
}
let human1: Human = {};
human1.firstName = "John"
human1.lastName = "Doe"
Upon declaring human1, an error pops up:
Type '{}' is missing the following properties from type Humaninterface Human {
firstName: string;
lastName: string;
}
let human1: Human = {};
human1.firstName = "John"
human1.lastName = "Doe"
Upon declaring human1, an error pops up:
Type '{}' is missing the following properties from type HumanHere is an improved approach:
let person1: Person = {name: '', surname: ''};
If you specifically need an empty object, you can do it like this:
let person1: Person = {} as Person;
Update following a comment:
Take a look at the unpredictableFunction:
const unpredictableFunction = (): string|number|string[] => {
return Math.random() > 0.5 ? 'string' : Math.random() > 0.5 ? 9999 : ['1', '2', '3']
};
This function may return a number, a string, or an array of strings.
const person: Person = {name: '', surname: ''};
person.name = unpredictableFunction (); // this illustrates your point
In such a scenario, you will encounter
Type 'string | number | string[]' is not assignable to type 'string'.Solutions include:
Review your code and ensure only strings are assigned to Person properties,
Alternatively, modify the interface to accommodate different value types:
interface Person {
name: string | number | string[];
surname: string;
}
If you have created an interface with two mandatory properties, when creating an object with the Person interface type, you need to specify these properties immediately like so:
let person: Person = {
name: '',
surname: ''
}
However, if you consider these properties to be optional rather than mandatory, you can modify your interface as follows:
interface Person {
name?: string;
surname?: string;
}
By using the ? symbol, you indicate that the property is optional. The following code should now function correctly:
let person: Person = {};
Enhanced functionality in Typescript 2.0
let person1 ! : Person;
The exclamation mark "!" signifies the Non-null assertion operator
Referencing the official documentation
A new ! post-fix expression operator may be used to assert that its operand is non-null and non-undefined in contexts where the type checker is unable to conclude that fact. Specifically, the operation x! produces a value of the type of x with null and undefined excluded. Similar to type assertions of the forms x and x as T, the ! non-null assertion operator is simply removed in the emitted JavaScript code.
Simply include the following code in your TypeScript file:
} | {};
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