样式化组件中的条件呈现

Question

如何在样式组件中使用条件渲染来设置我的按钮类在React中使用样式组件？

在css中，我会这样做：

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

如果我尝试使用'

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

下面是一个简单的TypeScript示例：

import * as React from 'react';
import { FunctionComponent } from 'react';
import styled, { css } from 'styled-components';

interface IProps {
  isProcessing?: boolean;
  isDisabled?: boolean;
  onClick?: () => void;
}

const StyledButton = styled.button<IProps>`
  width: 10rem;
  height: 4rem;
  cursor: pointer;
  color: rgb(255, 255, 255);
  background-color: rgb(0, 0, 0);

  &:hover {
    background-color: rgba(0, 0, 0, 0.75);
  }

  ${({ disabled }) =>
    disabled &&
    css`
      opacity: 0.5;
      cursor: not-allowed;
    `}

  ${({ isProcessing }) =>
    isProcessing &&
    css`
      opacity: 0.5;
      cursor: progress;
    `}
`;

export const Button: FunctionComponent<IProps> = ({
  children,
  onClick,
  isProcessing,
}) => {
  return (
    <StyledButton
      type="button"
      onClick={onClick}
      disabled={isDisabled}
      isProcessing={isProcessing}
    >
      {!isProcessing ? children : <Spinner />}
    </StyledButton>
  );
};

<Button isProcessing={this.state.isProcessing} onClick={this.handleClick}>Save</Button>

Answer 2

我没有注意到任何问题

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

在上述情况下，当使用活动道具呈现StyledYourComponent时，背景将暗显，如果没有提供活动道具，或者是falsy Styled components自动为您生成类名：）

如果要添加多个样式属性，则必须使用css标记，该标记是从样式化组件导入的：

我没有注意到任何问题

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;`
  }
`

或者您也可以使用object来传递样式，但请记住CSS属性应为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

你可以简单地这么做

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

在你的风格中，像这样：

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;
  `}
`;