[英]Typescript constructor shorthand when parameters are passed as an object
I know we can make constructor short hand when we pass the parameters in a traditional way like我知道当我们以传统方式传递参数时,我们可以简化构造函数,例如
class Foo {
private name: string;
private age: number;
constructor(name: string, age: number) {
this.name = name;
this.age= age;
}
}
So the equivalent shorthand constructor notation for this class will be所以这个 class 的等效简写构造函数符号将是
class Foo {
constructor(private name: string, private age: number) {}
}
Similarly, how can I do the same shorthand when the constructor parameters are passed in as objects like below.同样,当构造函数参数作为如下对象传入时,我该如何做同样的速记。
class Foo {
private name: string;
private age: number;
private group: string;
constructor({
name,
age,
group,
}: {
name: string;
age: number;
group: string;
}) {
this.name= name;
this.age= age;
this.group= group;
}
}
You can do like this:你可以这样做:
class Foo {
constructor(public obj : { name: string, age: number, group: string}) {
}
}
let foo = new Foo({name: 'name', age: 42, group: 'answers'});
alert(foo.obj.name);
I am not aware of a shorter way.我不知道更短的方法。 If you ignore the constructors and just try to assign that object to three variables name
, age
, and group
, then this is really a question about how to declare types while destructuring an object:如果您忽略构造函数并尝试将 object 分配给三个变量name
、 age
和group
,那么这确实是一个关于如何在解构 object 时声明类型的问题:
const { name, age, group } = {
name: 'name',
age: 42,
group: 'answers',
};
TypeScript does not have a special notation for regular destructuring (lots of previous answers and blog posts end up using the same style as you did), so it does not have such a notation for destructuring inside a function definition. TypeScript 没有用于常规解构的特殊符号(许多先前的答案和博客文章最终使用与您相同的样式),因此在 function 定义中没有用于解构的符号。
You can keep the code a little cleaner by making the type definition an interface .通过将类型定义设为接口,您可以使代码更简洁。
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