我想在单击时更改 div 的显示 -React Styled Component

Question

I am creating a div using styled component. I want to change the visibility of the div on button clicked,

const Category = () => {
  const [showCategory, setShowCategory] = useState(false)


  useEffect(() => {
    setShowCategory(false)
  }, [])
  return (
<button onClick={() => {  setShowCategory(true)}}>
   New Category
</button>
        <AdminInputStyle>
          <form>
            <form-group>
              <label>Add Category</label>
              <input type='text' />
            </form-group>
            <button>Submit</button>
          </form>
        </AdminInputStyle>

  )

}

Here's the styled component

const AdminInputStyle = styled.div`
  display: ${(d) => (d.showCategory ? 'show' : 'hidden')};
`

Answer 1

I have an example, but in the case we will use a Button. Clicking it will alter the visibility.

1. You must pass a property to the styled component if you want it to be visible based on that prop. In your example, you don't pass a prop to the styled component in this scenario, which is why the component cannot detect if it should be visible or not.

2. You will need to / can use the css function from the styled-components library. This can help you return styles based on the properties your styled-component will have. In this example, our property that we pass to the button will be called visible .

 import React from 'react'; import PropTypes from 'prop-types'; import styled, { css } from 'styled-components/macro'; const StyledButton = styled.button` border-radius: 3px; color: white; background-color: green; cursor: pointer; width: 100px; height: 50px; ${({ visible }) => { return css` visibility: ${visible? 'visible': 'hidden'}; `; }} `; export default function Button({ children, visible, onClick }) { return ( <StyledButton visible={visible} onClick={onClick}> {children} </StyledButton> ); } Button.propTypes = { children: PropTypes.node, visible: PropTypes.bool, onClick: PropTypes.func, };

You can see that passing the visible prop will enable the button to alter its' styles based on whether that property is true or false. We utilize a function within the component that returns the css function and this will control the visibility css property.

Here is how we utilize the button and pass props to it from another component; in this example just the App.js file:

 import React, { useState } from 'react'; import './App.css'; import Button from './components/Button'; function App() { const [visible, setVisible] = useState(true); function handleClick() { setVisible(;visible); } return ( <div className="App"> <Button visible={visible} onClick={handleClick}> Click </Button> </div> ); } export default App;

Answer 2

You can try something like this too, show when you need to show the add category when you press add category

return (
    <>
      <button
        onClick={() => {
          setShowCategory(true);
        }}
      >
        New Category
      </button>
      {showCategory && (
        <AdminInputStyle>
          <form>
            <form-group>
              <label>Add Category</label>
              <input type="text" />
            </form-group>
            <button>Submit</button>
          </form>
        </AdminInputStyle>
      )}
    </>
  );