implementing CSS class names in React

What is the recommended approach for using className in react, especially with multiple class names? I have been reviewing the documentation, but I am not finding a clear answer. I have come across examples like:

const divStyle = {
  color: 'blue',
  backgroundImage: 'url(' + imgUrl + ')',
};

function HelloWorldComponent() {
  return <div style={divStyle}>Hello World!</div>;
}

However, is there a way for me to achieve something like this?

import React from 'react';
import ReactDOM from 'react-dom';
import './css/landing.css';
import './css/w3.css';


class Home extends React.Component {
  const homeClasses = 'bgimg-1 w3-display-container w3-opacity-min';

  render() {
    return (
      <div className={homeClasses}>
        <h1>SUP</h1>
      </div>
    );
  }
}


ReactDOM.render(
  <Home />,
  document.getElementById('root')
);

Or is there a way to simply list them out in the class name section?

Answer №1

When deciding how to style your component, the functionality it needs to have plays a crucial role.

If your component is mostly static, using a built-in style like in the first example is recommended:

const mystyle = {
width: '100%',
...
}

<div style={mystyle}>...</div>

However, for more dynamic reusability, consider using a class method to generate styles based on passed props, as shown in this render function:

render() {
// User's preference or Default Style:
const defaultStyle =  {
        width: this.props.style.width || '100%',
        height: this.props.style.height || '100%',
      }
//if this.props.className doesn't exist use mycssclass
const defaultClassName = this.props.className || 'mycssclass' 

return (
    <div className={defaultClassName} style={defaultStyle}>...</div> )

You can also implement conditional styling using the ternary operator based on props, such as using an isActive state property to determine the class to apply.

render() {
const activeClassName = this.props.className + ' mycomponent-active'

return (
   <div className={this.props.isActive ? activeClassName : this.props.className} style={ this.props.style } 
   </div>);
}

An alternative approach is to create a function that returns a specific style object for your component, similar to the initial example.

The key decision lies in whether you want your component's style to be customizable by designers/users or consistent across different implementations. If customization is desired, expose the CSS class name through props or set a default one:

<div className={this.props.className || 'someclassName'}>...</div>

Otherwise, refer back to the previous examples.

Answer №2

Achieving this is totally possible! Check out the code snippet provided below:

class Example extends React.Component {
  cssClasses = 'demo demo2';
  render() {
    return ( 
       <div className = { this.cssClasses }>
           Hello World 
       </div>
    );
  }
}

ReactDOM.render( <Example/> , document.getElementById('app'));
.demo {
  color: blue
}

.demo2 {
  font-size: 20px;
}
<script src="https://cdnjs.cloudflare.com/ajax/libs/react/15.1.0/react.min.js"></script>
<script src="https://cdnjs.cloudflare.com/ajax/libs/react/15.1.0/react-dom.min.js"></script>

<div id='app'></div>

The mistake you made was in defining homeClasses. It should not be declared like this:

const homeClasses = '...';

In your component, homeClasses should not be defined with const. Simply do this:

homeClasses = '...';

Also, remember to use the this reference since homeClasses is a property of your component.

<div className={this.homeClasses}>
   <h1>SUP</h1>
</div>

Answer №3

One solution is readily available for this particular issue. By accessing and extracting the information from a file (such as data.json), you can utilize this data as properties of your choice.

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