let deletedQuestionsDto = { examId: this.examId, questionId: q.id } as DeletedQuestionsDto;

您可以使用Class设置默认值。

没有类构造函数:

interface IModal {
  content: string;
  form: string;
  href: string;
  isPopup: boolean;
};

class Modal implements IModal {
  content = "";
  form = "";
  href: string;  // will not be added to object
  isPopup = true;
}

const myModal = new Modal();
console.log(myModal); // output: {content: "", form: "", isPopup: true}

使用类构造函数

interface IModal {
  content: string;
  form: string;
  href: string;
  isPopup: boolean;
}

class Modal implements IModal {
  constructor() {
    this.content = "";
    this.form = "";
    this.isPopup = true;
  }

  content: string;

  form: string;

  href: string; // not part of constructor so will not be added to object

  isPopup: boolean;
}

const myModal = new Modal();
console.log(myModal); // output: {content: "", form: "", isPopup: true}

let deletedQuestionsDto = { examId: this.examId, questionId: q.id } as DeletedQuestionsDto;

下面是另一种方法:

你可以像这样简单地创建一个ESLint友好的对象

const modal: IModal = {} as IModal;

或者基于接口的默认实例，如果有默认值的话

const defaultModal: IModal = {
    content: "",
    form: "",
    href: "",
    $form: {} as JQuery,
    $message: {} as JQuery,
    $modal: {} as JQuery,
    $submits: {} as JQuery
};

然后通过重写一些属性来改变默认实例

const confirmationModal: IModal = {
    ...defaultModal,     // all properties/values from defaultModal
    form: "confirmForm"  // override form only
}

由于问题包括TypeScript的使用，替换

var modal: IModal;

by

let modal: IModal = {} as IModal;

应该回答这个问题。

因为我没有在上面找到一个相等的答案，所以我的答案是不同的。 我做的事:

modal: IModal = <IModal>{}