How can I test React component's style with Jest + Enzyme?

Question

Alright I have a component called <TestButton />. Inside the <TestButton /> there are two Semantic UI React component, <Button /> and <Header>.

Basically, when the <Button> is clicked, it toggles display: none; to <Header>.

I want to check (I want to learn) on how to assert <Header>'s display: none; when <Button> is clicked.

TestButton.js

const TestButton = (props) => {
  return (
    <div id='test-button'>
      <Header id='please-hide-me' size='huge'>Please Hide Me</Header>
      <Button
        onClick={
          () => {
            hiding = !hiding;
            let findDOM = document.getElementById(searchForThisID);
            if (findDOM) { findDOM.style.display = hiding ? 'none' : ''; }
            return hiding;
          }
        }
      >
        Sample Toggle
      </Button>
    </div>
  );
};

My unit test is based on How to test style for a React component attribute with Enzyme. It looks like this:

test(`

  `, () => {
    const wrapper = shallow(<TestButton />);
    const button = wrapper.find('Button');
    const header = wrapper.find('Header');
    const headerStyle = header.get(0).style;

    expect(headerStyle).to.have.property('display', '');
    wrapper.find('Button').simulate('click');
    expect(headerStyle).to.have.property('display', 'none');
  }
);

But it has this error:

TypeError: Cannot read property 'have' of undefined

What should I do?

Solution

There are a few mistakes in your provided code:

1. You should not be using DOM element's style property because React does not manage it. Shift the hidden property into the state instead.

2. I believe headerStyle is a shallow copy of the style object. After you simulate click, it does not get updated. You will have to query the element again for the style object.

3. to.have.property is not valid Jest syntax. It should be toHaveProperty.

Please refer to the corrected code here. If you paste the following into create-react-app, it should just work.

app.js

import React, { Component } from 'react';

function Header(props) {
  return <h1 style={props.style}>Header</h1>;
}

function Button(props) {
  return <button onClick={props.onClick}>Click Me</button>;
}

export class TestButton extends React.Component {
  constructor(props) {
    super(props);
    this.state = { hiding: false };
  }

  render() {
    return (
      <div>
        <Header
          id="please-hide-me"
          style={{
            display: this.state.hiding ? 'none' : '',
          }}
        >
          Please Hide Me
        </Header>
        <Button
          onClick={() => {
            this.setState({
              hiding: !this.state.hiding,
            });
          }}
        >
          Sample Toggle
        </Button>
      </div>
    );
  }
}

class App extends Component {
  render() {
    return (
      <div className="App">
        <TestButton />
      </div>
    );
  }
}

export default App;

app.test.js

import React from 'react';
import ReactDOM from 'react-dom';
import App from './App';
import { configure, shallow } from 'enzyme';
import Adapter from 'enzyme-adapter-react-16';

configure({ adapter: new Adapter() });

import { TestButton } from './App';

it('renders without crashing', () => {
  const wrapper = shallow(<TestButton />);
  expect(wrapper.find('Header').get(0).props.style).toHaveProperty(
    'display',
    '',
  );
  wrapper.find('Button').simulate('click');
  expect(wrapper.find('Header').get(0).props.style).toHaveProperty(
    'display',
    'none',
  );
});

Comments

[notalentgeek]

It solves my problem! Thanks. Is there anyway I can have the TestButton just as a const not a class? I learned that most ReactJS devs only use const as a component. Additionally: is using in-line style a good practice is ReactJS?

[Yangshun Tay]

Nope, you need it as a class because you need to store state. Pure vanilla React inline styling is bad, but if you use stuff like CSS modules and jss, they do some post-processing to extract the inline styles out into stylesheets, which will be better.