Welcome to OStack Knowledge Sharing Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
670 views
in Technique[技术] by (71.8m points)

reactjs - conditional rendering in styled components

How can I use conditional rendering in styled-components to set my button class to active using styled-components in React?

In css I would do it similarly to this:

<button className={this.state.active && 'active'}
      onClick={ () => this.setState({active: !this.state.active}) }>Click me</button>

In styled components if I try to use '&&' in the classname it doesn't like it.

import React from 'react'
import styled from 'styled-components'

const Tab = styled.button`
  width: 100%;
  outline: 0;
  border: 0;
  height: 100%;
  justify-content: center;
  align-items: center;
  line-height: 0.2;
`

export default class Hello extends React.Component {
  constructor() {
    super()
    this.state = {
      active: false
    }  
    this.handleButton = this.handleButton.bind(this)
}

  handleButton() {
    this.setState({ active: true })
  }

  render() {
     return(
       <div>
         <Tab onClick={this.handleButton}></Tab>
       </div>
     )
  }}
See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Answer

0 votes
by (71.8m points)

You can simply do this

<Tab active={this.state.active} onClick={this.handleButton}></Tab>

And in your styles something like this:

const Tab = styled.button`
  width: 100%;
  outline: 0;
  border: 0;
  height: 100%;
  justify-content: center;
  align-items: center;
  line-height: 0.2;

  ${({ active }) => active && `
    background: blue;
  `}
`;

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome to OStack Knowledge Sharing Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

...