conditional rendering in styled components

Question

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>
     )
  }}

Answer 1

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;
  `}
`;

Answer 2

I didn't notice any && in your example, but for conditional rendering in styled-components you do the following:

// Props are component props that are passed using <StyledYourComponent prop1="A" prop2="B"> etc
const StyledYourComponent = styled(YourComponent)`
  background: ${props => props.active ? 'darkred' : 'limegreen'}
`

In the case above, background will be darkred when StyledYourComponent is rendered with active prop and limegreen if there is no active prop provided or it is falsy Styled-components generates classnames for you automatically :)

If you want to add multiple style properties you have to use css tag, which is imported from styled-components:

I didn't notice any && in your example, but for conditional rendering in styled-components you do the following:

import styled, { css } from 'styled-components'
// Props are component props that are passed using <StyledYourComponent prop1="A" prop2="B"> etc
const StyledYourComponent = styled(YourComponent)`
  ${props => props.active && css`
     background: darkred; 
     border: 1px solid limegreen;`
  }
`

OR you may also use object to pass styled, but keep in mind that CSS properties should be camelCased:

import styled from 'styled-components'
// Props are component props that are passed using <StyledYourComponent prop1="A" prop2="B"> etc
const StyledYourComponent = styled(YourComponent)`
  ${props => props.active && ({
     background: 'darkred',
     border: '1px solid limegreen',
     borderRadius: '25px'
  })
`

Answer 3

If your state is defined in your class component like this:

class Card extends Component {
  state = {
    toggled: false
  };
  render(){
    return(
      <CardStyles toggled={this.state.toggled}>
        <small>I'm black text</small>
        <p>I will be rendered green</p>
      </CardStyles>
    )
  }
}

Define your styled-component using a ternary operator based on that state

const CardStyles = styled.div`
  p {
    color: ${props => (props.toggled ? "red" : "green")};
  }
`

it should render just the <p> tag here as green.

This is a very sass way of styling