reactjs中状态改变后如何重新渲染？

Question

import React from 'react'

export default function App(){
    const[flag,setFlag] =React.useState({1:false,2:false})

    const change=(id)=>{
        flag[id] = !flag[id]
        setFlag(flag)
        console.log(flag)
    }

    return(
        <div>
            {flag[1] ? <p>this is true</p> : <p>this is false</p>}
            <button onClick={()=>change(1)}>Change</button>
        </div>
    )
}

I want to change my output based on flag state. While flag JSON is changing, but it's not re-rendering.

Answer 1

your code is working fine.

const change=(id)=>{
   console.log(flag[id]);
        flag[id] = !flag[id]
        setFlag(flag)
        console.log(flag)
    }

Could you check the render method.

Check here https://codepen.io/palash924332931/pen/VwYpwOq?editors=1111

Answer 2

You should be using setFlag() to change the state instead of changing flag directly. By using the setter for the flag state, you can call render() if you use setFlag() . In order to use setFlag() with your current setup, you can use computed property names like so:

const change = id => {
  setFlag(currentFlag => ({[id]: !currentFlag}));
}

Answer 3

The state updater returned by useState will not rerender the component's children if you set a new value that equals the current value. https://reactjs.org/docs/hooks-reference.html#bailing-out-of-a-state-update

    function App() {
  const [flag, setFlag] = React.useState({ 1: false, 2: false });

  const change = id => {
    var newFlag = Object.assign({}, flag);
    newFlag[id] = !flag[id];
    setFlag(newFlag);
  };
  return (
    <div>
      {flag[1] ? (
        <p>this is true</p>
      ) : (
        <p>this is false</p>
      )}
      <button onClick={() => change(1)}>Change</button>
    </div>
  );
}

codepen example