Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
569 views
in Technique[技术] by (71.8m points)

knockout.js - Extending a generic parameter in Typescript

I want to create a utility function which creates a checklist by adding an isChecked knockout observable property to each item in an array. This function should look like this:

createCheckList<T>(allEntities: T[], selected: T[]) : CheckListItem<T> {
    ...
}

I am returning a CheckListItem<T> because this interface should extend T to add the isChecked property. However, typescript will not allow me to do this:

interface CheckListItem<T> extends T {
    isChecked: KnockoutObservable<boolean>;
}

Gives me the error:

An interface may only extend another class or interface.

Is there any way to do this?

See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Reply

0 votes
by (71.8m points)

Since TypeScript 1.6 you can use intersection types:

type CheckListItem<T> = T & {
  isChecked: KnockoutObservable<boolean>;
};

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
OGeek|极客中国-欢迎来到极客的世界,一个免费开放的程序员编程交流平台!开放,进步,分享!让技术改变生活,让极客改变未来! Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

...