如何模拟react-router上下文

时间:2016-06-30 11:12:41

标签: reactjs react-router enzyme

我有相当简单的反应组件(链接包装器,如果路由处于活动状态,则添加'活动'类):

import React, { PropTypes } from 'react';
import { Link } from 'react-router';

const NavLink = (props, context) => {
  const isActive = context.router.isActive(props.to, true);
  const activeClass = isActive ? 'active' : '';

  return (
    <li className={activeClass}>
      <Link {...props}>{props.children}</Link>
    </li>
  );
}

NavLink.contextTypes = {
  router: PropTypes.object,
};

NavLink.propTypes = {
  children: PropTypes.node,
  to: PropTypes.string,
};

export default NavLink;

我该怎么测试呢?我唯一的尝试是:

import NavLink from '../index';

import expect from 'expect';
import { mount } from 'enzyme';
import React from 'react';

describe('<NavLink />', () => {
  it('should add active class', () => {
    const renderedComponent = mount(<NavLink to="/home" />, { router: { pathname: '/home' } });
    expect(renderedComponent.hasClass('active')).toEqual(true);
  });
});

它不起作用并返回TypeError: Cannot read property 'isActive' of undefined。它肯定需要一些路由器模拟,但我不知道如何编写它。

5 个答案:

答案 0 :(得分:19)

感谢@Elon Szopos的答案,但我设法写了一些更简单的内容(https://github.com/airbnb/enzyme/pull/62之后):

import NavLink from '../index';

import expect from 'expect';
import { shallow } from 'enzyme';
import React from 'react';

describe('<NavLink />', () => {
  it('should add active class', () => {
    const context = { router: { isActive: (a, b) => true } };
    const renderedComponent = shallow(<NavLink to="/home" />, { context });
    expect(renderedComponent.hasClass('active')).toEqual(true);
  });
});

我必须将mount更改为shallow,以便不评估Link,这会给我一个与react-router TypeError: router.createHref is not a function相关的错误。

我宁愿拥有&#34;真实&#34; react-router不仅仅是一个对象,但我不知道如何创建它。

答案 1 :(得分:13)

对于react router v4,您可以使用<MemoryRouter>。 AVA和酶的例子:

import React from 'react';
import PropTypes from 'prop-types';
import test from 'ava';
import { mount } from 'enzyme';
import sinon from 'sinon';
import { MemoryRouter as Router } from 'react-router-dom';

const mountWithRouter = node => mount(<Router>{node}</Router>);

test('submits form directly', t => {
  const onSubmit = sinon.spy();
  const wrapper = mountWithRouter(<LogInForm onSubmit={onSubmit} />);
  const form = wrapper.find('form');
  form.simulate('submit');

  t.true(onSubmit.calledOnce);
});

答案 2 :(得分:5)

测试依赖于上下文的组件可能有点棘手。我所做的是写一个我在测试中使用的包装器。

您可以在下面找到包装器:

import React, { PropTypes } from 'react'

export default class WithContext extends React.Component {
  static propTypes = {
    children: PropTypes.any,
    context: PropTypes.object
  }

  validateChildren () {
    if (this.props.children === undefined) {
      throw new Error('No child components were passed into WithContext')
    }
    if (this.props.children.length > 1) {
      throw new Error('You can only pass one child component into WithContext')
    }
  }

  render () {
    class WithContext extends React.Component {
      getChildContext () {
        return this.props.context
      }

      render () {
        return this.props.children
      }
    }

    const context = this.props.context

    WithContext.childContextTypes = {}

    for (let propertyName in context) {
      WithContext.childContextTypes[propertyName] = PropTypes.any
    }

    this.validateChildren()

    return (
      <WithContext context={this.props.context}>
        {this.props.children}
      </WithContext>
    )
  }
}

您可以在此处查看示例用法:

  <WithContext context={{ location: {pathname: '/Michael/Jackson/lives' }}}>
    <MoonwalkComponent />
  </WithContext>

  <WithContext context={{ router: { isActive: true }}}>
    <YourTestComponent />
  </WithContext>

它应该像你期望的那样工作。

答案 3 :(得分:3)

您可以将https://github.com/pshrmn/react-router-test-context用于此目的

“创建一个复制React Router的context.router结构的伪上下文对象。这对于使用Enzyme的浅单元测试非常有用。”

安装后,您将能够执行类似

的操作
describe('my test', () => {
  it('renders', () => {
    const context = createRouterContext()
    const wrapper = shallow(<MyComponent />, { context })
  })
})

答案 4 :(得分:-1)

您需要了解mount&amp; shallowofficial documentation解释了 Shallow Render

  

在为React编写单元测试时,浅渲染可能会有所帮助。   浅渲染使您可以渲染一个组件&#34;一个深度&#34;和   断言其渲染方法返回的事实,而不用担心   关于未实例化的子组件的行为   渲染。这不需要DOM。

然后,没有复杂性,也没有上下文:

   const renderedComponent = shallow(<NavLink to="/home" />);

而不是

   const renderedComponent = mount(<NavLink to="/home" />, { router: { pathname: '/home' } });

它会起作用!