Angular，将泛型类型传递给另一个函数。 如何输入参数？

Question

I have function like:

getSomething<T>(value: T): any{
    const temp = this.doSomething<T>(value);
...
}

doSomething<T>(value: T): any {
    return value.replace(/\s/g, '');
}

Initially the value passed has a type specific type T, I know full well that in the final function it will be a string. However, doing it this way I can't use replace function. Help.

Answer 1

You need to say that value would have string methods. Like this:

getSomething<T extends string>(value: T): any{
    const temp = this.doSomething<T>(value);
...
}

doSomething<T extends string>(value: T): any {
    return value.replace(/\s/g, '');
}

Answer 2

Think about your problem in the following way:

1. How can I be sure that provided type has "replace()" method?
2. What would happen if I provide different type eg number or null?

You have to somehow tell Typescript that you expect to get an argument that has method "replace()". You can do it by creating additional interface and mention that generic type extends that interface:

interface Replacable {
  replace: (regexp: RegExp, str: string) => string
}

function getSomething<T extends Replacable>(value: T): any{
    const temp = doSomething<T>(value);
    console.log(temp);
}

function doSomething<T  extends Replacable>(value: T): any {
    return value.replace(/\s/g, '');
}

Read more about Generic Constraints here: https://www.typescriptlang.org/docs/handbook/2/generics.html#generic-constraints