Question

在Karma和Angular2中遇到令人沮丧的问题，我试图将其简化为最简单的演示：考虑这个TypeScript类，同时使用实例成员方法和静态成员方法来执行相同的操作：

export class Echo {

    static classLog = console.log;

    instanceLog = console.log;

    static classEcho(msg: string) {
        this.classLog(msg);
    }

    instanceEcho (msg: string) {
        this.instanceLog(msg);
    }
}

...然后是一个简单的纯TypeScript测试：

import {Echo} from './echo';

var e = new Echo();
e.instanceEcho('instanceEcho!');
Echo.classEcho('classEcho!');

..结果：

instanceEcho!
classEcho!

好。

然而，尝试从Jasmine验证相同的事情：

import {it} from '@angular/core/testing';
import {Echo} from './echo';

describe('Logging', ()=> {
    it('instanceEcho will invoke \'console.log()\'', () => {
        spyOn(console, 'log');
        expect(console.log).not.toHaveBeenCalled();
        var e = new Echo();
        e.instanceEcho('test instanceEcho');
        expect(console.log).toHaveBeenCalledWith('test instanceEcho');
    });

    it('classEcho will invoke \'console.log()\'', () => {
        spyOn(console, 'log');
        expect(console.log).not.toHaveBeenCalled();
        Echo.classEcho('test classEcho');
        expect(console.log).toHaveBeenCalledWith('test classEcho');
    });
});

..结果：

START:
✔ instanceEcho will invoke 'console.log()'
LOG: 'test classEcho'
✖ classEcho will invoke 'console.log()'

SUMMARY:
✔ 1 tests completed
✖ 1 test failed

FAILED TESTS:
    ✖ classEcho will invoke 'console.log()'
      Chrome 50.0.2661 (Mac OS X 10.11.5)
    Expected spy log to have been called with [ 'test classEcho' ] 
    but it was never called.

由于静态方法确实可以正常工作，为什么Jasmine没有捕获它呢？

Answer 1

我现在意识到这不是Jasmine监视TypeScript静态成员的问题，而只是一个普通的JavaScript问题，即失去了对this的引用。如果切换到，Jasmine SpyOn将正确捕获console.log（）的调用，例如：

export class Echo2 {

    static classLogger = console;
    instanceLogger = console;

    static classEcho(msg: string) {
        this.classLogger.log(msg);
    }

    instanceEcho (msg: string) {
        this.instanceLogger.log(msg);
    }
}

不能Jasmine SpyOn TypeScript静态成员

1 个答案: