我正在为以下组件编写单元测试:
<template>
<sub-component
@foo="bar"
/>
</template>
<script>
import SubComponent from './SubComponent';
export default {
name: 'MyComponent',
components: { SubComponent },
methods: {
bar(payload) {
this.$emit('baz', ...payload);
}
}
}
</script>
测试将是:
import { shallowMount } from '@vue/test-utils';
import _ from 'lodash';
import MyComponent from '../../components/MyComponent';
describe('MyComponent.vue', () => {
let wrapper;
beforeEach(() => {
wrapper = shallowMount(MyComponent);
});
it('should emit baz on subcomponent foo', () => {
const subComp = wrapper.find('sub-component-stub');
expect(subComp.exists()).toBe(true); // passes
subComp.vm.$emit('foo');
return wrapper.vm.$nextTick().then(() => {
expect(wrapper.emitted().baz).toBeTruthy(); // does not pass;
// upon logging:
console.log(_.isEqual(wrapper, subComp)); // => true
})
})
})
该示例过于简化,但是这里的原理是我想要一个可重用的<sub-component>
(一个模式)和围绕它的各种功能包装(与模式类型执行的一项特定任务有关),以映射其他功能。我不需要父组件中的功能,因为它会违反DRY-我必须将其放在包含特定类型的模态的每个组件中。
如果<sub-component>
不是<template>
的直接子代,则可以正常工作。以某种方式,看来wrapper
和subComp
托管在同一元素上。
应如何正确测试?
答案 0 :(得分:1)
另一种可能性是在dom中找到您的元素并检查根组件发出的值。
import { shallowMount } from '@vue/test-utils'
import MyComponent from './MyComponent.vue'
import SubComponent from './SubComponent.vue'
describe('MyComponent', () => {
it('should emit baz on subcomponent foo', () => {
const wrapper = shallowMount(MyComponent)
const subComponent = wrapper.find(SubComponent)
expect(subComponent.exists()).toBe(true)
expect(wrapper.emitted('baz')).toBeUndefined()
subComponent.vm.$emit('foo', ['hello'])
expect(wrapper.emitted('baz')[0]).toEqual(['hello'])
// or expect(wrapper).toEmit('baz', 'hello') cf. below for toEmit
})
})
如果要为Jest自定义匹配器:
toEmit(received, eventName, data) {
if (data) {
expect(received.emitted()[eventName][0]).toEqual([data])
} else {
expect(received.emitted()[eventName][0]).toEqual([])
}
return { pass: true }
}