我正在尝试实现自己的UrlSerializer类,这是我所做的:

import { UrlSerializer,UrlTree } from '@angular/router';

export class CustomUrlSerializer implements UrlSerializer {

    parse(url: string): UrlTree {
        // Change plus signs to encoded spaces
        url.replace("%20", '-');
        // Use the default serializer that you can import to just do the
        // default parsing now that you have fixed the url.
        return super.parse(url)
    }

    serialize(tree: UrlTree): string {
        // Use the default serializer to create a url and replace any spaces with + signs
        return super.serialize(tree).replace("%20", '-');
    }
}


当我尝试编译时,出现以下错误:

c:/xampp/htdocs/proj/src/app/custom-url-serializer.ts (11,12): 'super' can only be referenced in a derived class.
c:/xampp/htdocs/proj/src/app/custom-url-serializer.ts (16,12): 'super' can only be referenced in a derived class.


怎么了?

最佳答案

我会说问题出在implements关键字。因为它需要一个没有实现的接口,所以您不能调用superUrlSerializer是一个抽象类,因此您可以使用DefaultUrlSerializer

import { DefaultUrlSerializer, UrlTree } from '@angular/router';
class CustomUrlSerializer extends DefaultUrlSerializer {
    parse(url: string) : UrlTree {
        return super.parse(url);
    }
}
new CustomUrlSerializer().parse('http://stackoverflow.com');


它应该工作。

关于angular - Angular 2-实作UrlSerializer,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/43399936/

10-09 22:19