angular - 在 Angular 8 中初始化嵌套对象的有效方法

Question

我有一个带有嵌套Address 类的User 类。在组件中，我想用 let user = new User() 或 let user:User 之类的东西初始化 User 类型的对象。

但是会抛出错误，因为 Angular/Typescript 希望我分配值。有没有一种有效的方法可以做到这一点，而不将所有空 "" 值分配给对象。

在类下面:

class Address {
    constructor(
        public street: string,
        public number: number,
        public suffix: string,
        public houseLetter: string,
        public city: string,
        public zipcode: string,
    ) { }
}

export class User {
    constructor(
        public id: number,
        public firstName: string,
        public lastName: string,
        public email: string,
        public address: Address,
    ) { }
}

Answer 1

您可以将这些参数标记为可选，如下所示:

export class User {
    constructor(
        public id?: number,
        public firstName?: string,
        public lastName?: string,
        public email?: string,
        public address?: Address,
    ) { }
}

现在您可以轻松地实例化用户对象 让 user = new User()

Working demo link