如何在TypeScript中为对象动态分配属性？

最新推荐文章于 2025-03-17 21:36:13 发布

翻译最新推荐文章于 2025-03-17 21:36:13 发布 · 2.3k 阅读

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原文链接：https://oldbug.net/q/RkGB/How-do-I-dynamically-assign-properties-to-an-object-in-TypeScript

文章标签：

#typescript

本文探讨了在TypeScript中动态为对象分配属性的方法。由于TypeScript的静态类型检查，直接赋值会引发错误。解决方案包括使用any类型、对象字面量的动态属性名和可选属性标记。通过这些技巧，开发者可以在保持类型安全的同时实现动态属性赋值。

本文翻译自：How do I dynamically assign properties to an object in TypeScript?

If I wanted to programatically assign a property to an object in Javascript, I would do it like this: 如果我想以编程方式将属性分配给Javascript中的对象，则可以这样做：

var obj = {};
obj.prop = "value";

But in TypeScript, this generates an error: 但是在TypeScript中，这会产生一个错误：

The property 'prop' does not exist on value of type '{}' 类型“ {}”的值不存在属性“ prop”

How am I supposed to assign any new property to an object in TypeScript? 我应该如何在TypeScript中为对象分配任何新属性？

#1楼

参考：https://stackoom.com/question/RkGB/如何在TypeScript中为对象动态分配属性

#2楼

Although the compiler complains it should still output it as you require. 尽管编译器抱怨它仍应按您的要求输出。 However, this will work. 但是，这将起作用。

var s = {};
s['prop'] = true;

#3楼

您可以添加此声明以使警告静音。

declare var obj: any;

#4楼

Or all in one go: 或一劳永逸：

  var obj:any = {}
  obj.prop = 5;

#5楼

I tend to put any on the other side ie var foo:IFoo = <any>{}; 我倾向于把any放在另一边，即var foo:IFoo = <any>{}; So something like this is still typesafe: 所以像这样的东西仍然是类型安全的：

interface IFoo{
    bar:string;
    baz:string;
    boo:string;     
}

// How I tend to intialize 
var foo:IFoo = <any>{};

foo.bar = "asdf";
foo.baz = "boo";
foo.boo = "boo";

// the following is an error, 
// so you haven't lost type safety
foo.bar = 123;

Alternatively you can mark these properties as optional: 另外，您可以将这些属性标记为可选：

interface IFoo{
    bar?:string;
    baz?:string;
    boo?:string;    
}

// Now your simple initialization works
var foo:IFoo = {};

Try it online 在线尝试

#6楼

Store any new property on any kind of object by typecasting it to 'any': 将任何新属性存储在任何类型的对象上，方法是将其类型转换为“ any”：

var extend = <any>myObject;
extend.NewProperty = anotherObject;

Later on you can retrieve it by casting your extended object back to 'any': 稍后，您可以通过将扩展对象转换回'any'来检索它：

var extendedObject = <any>myObject;
var anotherObject = <AnotherObjectType>extendedObject.NewProperty;