Question

我是React的新手和一般的测试，所以请原谅这个问题的天真。我有一个React表单组件，输入onChance运行一个函数handleChange。试图用Jest测试它但不能使它工作。

这是“登录”组件：

class Login extends React.Component {

  constructor() {
    super();
    this.state = {username: '', password: ''}
    this.disableSubmit = this.disableSubmit.bind(this);
    this.handleChange = this.handleChange.bind(this);
  }

  handleChange(e) {
    this.setState({
      [e.target.name]: e.target.value
    });
  }

  render() {

    return(
      <div className="login">
        <form>
          <h3 className="login__title">LOGIN</h3>
          <div className="input-group">
            <input onChange={this.handleChange} value={this.state.username} className="form-control login__input username" type="text" placeholder="user name" name={'username'} autoFocus/>
          </div>
          <div className="input-group">
            <input onChange={this.handleChange} value={this.state.password} className="form-control login__input password" type="password" placeholder="password" name={'password'}/>
          </div>
          <div>
            <button className="btn btn-primary btn-block login__button" type="submit">Login</button>
          </div>
        </form>
      </div>

    )
  }
}

export default Login;

这是我的测试：

import React from 'react'
import { shallow, mount } from 'enzyme'
import { shallowToJson } from 'enzyme-to-json'


import {Login} from '../../../src/base/components/index'


describe('Given the Login component is rendered', () => {

  describe('Snapshots', () => {
    let component

    beforeEach(() => {
      component = shallow(<Login />)
    })

    it('should be as expected', () => {
      expect(shallowToJson(component)).toMatchSnapshot()
    })
  })

})


test('Submitting the form should call handleSubmit', () => {

  const startState = {username: ''};
  const handleChange = jest.fn();
  const login = mount(<Login />);
  const userInput = login.find('.username');

  userInput.simulate('change');

  expect(handleChange).toBeCalled();

})

快照测试通过正常，但在最后一次尝试中，我的功能测试失败了：

TypeError: Cannot read property 'target' of undefined

猜猜我需要传递一些东西给函数？有点困惑！

提前感谢您的帮助。

更新：

更改了测试，如下所示，但测试失败：expect(jest.fn()).toBeCalled() Expected mock function to have been called.

测试更新：

test('Input should call handleChange on change event', () => {

  const login = mount(<Login />);
  const handleChange = jest.spyOn(login.instance(), 'handleChange');
  const userInput = login.find('.username');
  const event = {target: {name: "username", value: "usertest"}};

  userInput.simulate('change', event);

  expect(handleChange).toBeCalled();

})

Answer 1

是的，您需要将事件对象传递给simulate函数。

  const event = {target: {name: "special", value: "party"}};

  element.simulate('change', event);

编辑：哦，你还需要做类似的事情：

jest.spyOn(login.instance(), 'handleChange')

但这与您的错误无关

Answer 2

handleChange目前尚未被嘲笑。几种方法：

将更改事件处理程序作为支持传递给Login组件。

<div className="input-group">
  <input 
    onChange={this.props.handleChange} 
    value={this.state.username}
    className="form-control login__input username" 
    type="text"
    placeholder="user name"
    name={'username'}
    autoFocus
    />
</div>

<强> login.spec.js

...
const handleChange = jest.fn();
const login = mount(<Login handleChange={handleChange}/>);
...

用mock函数替换handleChange。

...
const handleChange = jest.fn();
const login = mount(<Login />);
login['handleChange'] = handleChange // replace instance
...
expect(handleChange).toBeCalled();

使用jest spyOn创建一个包装原始函数的模拟函数。

...
const handleChange = jest.spyOn(object, 'handleChange') // will call the original method
expect(handleChange).toBeCalled();

使用模拟函数替换Login组件上的handleChange。 ... const handleChange = jest.spyOn（object，＆＃39; handleChange＆＃39;）。mock //将调用原始方法期望（handleChange）.toBeCalled（）;

Answer 3

在此处找到解决方案：Enzyme simulate an onChange event

test('Input should call handleChange on change event', () => {

  const event = {target: {name: 'username', value: 'usertest'}};
  const login = mount(<Login />);
  const handleChange = jest.spyOn(login.instance(), 'handleChange');
  login.update(); // <--- Needs this to force re-render
  const userInput = login.find('.username');

  userInput.simulate('change', event);

  expect(handleChange).toBeCalled();

})

为了工作需要这个login.update();！

感谢大家的帮助！

在React Js中用Jest测试函数

3 个答案: