Wat*_*ont 4 javascript oop class typescript angular
方法返回一个类类型:Widget。
我想用以下代码创建这种类型的对象:
const wType = def.getWidgetType(); // returns Widget as type
const obj = new wType('foo'); // use the const like a normal type with parameters
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getWidgetType()
public getWidgetType(): any {
return TextWidget;
}
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错误
error TS2351: Cannot use 'new' with an expression whose type lacks a call or construct signature.
是否有一个“好的”版本(没有 eval)来创建具有给定类类型的对象?
假设getWidgetType返回的是构造函数,您可以调用new wType('foo')提供的签名getWidgetType显式声明它返回构造函数签名。
例如,以下代码是有效的:
class Definition<T> {
// Takes in a constructor
constructor(public ctor: new (p: string) => T) {
}
// returns a constructor (aka a function that can be used with the new operator)
// return type annotation could be inferred here, was added for demonstrative purposes
getWidgetType() : new (p: string) => T{
return this.ctor;
}
}
const def = new Definition(class {
constructor(p: string) {}
});
const wType = def.getWidgetType();
const obj = new wType('foo')
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