在 ReactJS 中单击按钮状态更改后，React 不会重新呈现（preact）

Question

我有以下代码可以正常工作，只是它在按下按钮后不会重新呈现页面。

根据用户是否在房间内，我会显示两个不同的按钮。如果用户是房间的一部分，他们可以单击 LEAVE，这将执行 API 调用。然后我希望组件重新加载并让按钮显示 JOIN（因为它们不再是那个房间的一部分）。

import JoinedRooms from '../../components/matrix_joined_rooms';
import JoinRoom from '../../components/matrix_function_join_room';
import LeaveRoom from '../../components/matrix_function_leave_room';
import { useEffect, useState } from 'preact/hooks';

const JoinLeaveButton = ({ name, roomId }) => {
  const joinedRooms = JoinedRooms();
  const [x, setX] = useState(5);

  useEffect(() => console.log("re-render because x changed:", x), [x])

  const handleXClick = (number) => {
    setX(number)
  }

  if (joinedRooms.includes(name)) {
    return <button name={roomId} onClick={() => {
      LeaveRoom(roomId);
      handleXClick(10);
    }
    }>LEAVE</button>
  } else {
    return <button name={roomId} onClick={() => {
      JoinRoom(roomId);
      handleXClick(20);
    }
    }>JOIN</button>
  }
}

export default JoinLeaveButton;

我的 JoinRoom 和 LeaveRoom 组件是一个简单的 API 调用，如下所示：

const JoinRoom = (roomId) => {
  fetch(`https://example.com/_matrix/client/r0/rooms/${roomId}/join`, {
    method: "POST",
    headers: {
      Authorization: `Bearer ${localStorage.getItem("mx_access_token")}`
    }
  });
}

export default JoinRoom;

按钮本身的功能有效，问题是我必须手动重新加载页面才能正确显示。

我已经放置了一个虚拟状态，只要您按下按钮就会执行该状态，并且它还会正确地记录到控制台。我的印象是，更改状态应该在 React（或在本例中为 preact）中重新渲染组件。

谢谢！

Answer 1

本质上：您需要在某处存储已加入房间的状态，并在用户每次加入或离开房间时更新该状态。

我在这里说得太过分了，但是像这样的自定义挂钩很有意义：

// api calls
const fetchRooms = async userid => { ... }
const joinRoom = async (userId,roomId) => { ... }
const leaveRoom = async (userId,roomId) => { ... }

// custom hook
const useRooms = (userId) => {
   const [fetching,setFetching] = useState(true);
   const [error,setError] = useState(false);

   // joinedRooms state is an empty array when this hook is first used
   // it will be updated later using the useEffect hook
   // or via the join or leave functions below
   const [joinedRooms,setJoinedRooms] = useState([]);

   // when the component that uses this hook is mounted
   // or the user id changes, update the state
   useEffect(() => {
       let canceled;

       setFetching(true);
       (async() => {
          try {
             const rooms = await fetchRooms(userId);
             canceled || setJoinedRooms(rooms);
          } catch(err) {
             canceled || setError(error);
          } finally {
             canceled || setFetching(false);
          }
       })();
       return () => canceled = true;
   },[userId]);

   const leave = async roomId => {
     try {
        await leaveRoom(userId,roomId)
        // alternatively you could fetch all the user rooms again here
        setJoinedRooms(joined => joined.filter(r => r !== roomId));
     } catch(err) {
       // couldn't leave the room - what do you want to do with the state?
     }
   }

   const join = async roomId => {
     try {
        await joinRoom(userId,roomId);
        // alternatively you could fetch all the user rooms again here
        setJoinedRooms(joined => [...joined,roomId]);
     } catch(err) {
       // couldn't join the room - what do you want to do with the state?
     }
   }

   return {
      fetching,
      error,
      joinedRooms,
      leave,
      join
   }


}

在组件中，您可以像这样使用它：

const Rooms = (userId,listOfAllYourRoomIds) => {
   const { joinedRooms, fetching, error, join, leave } = useRooms(userId);

   // the `join` and `leave` functions are what you'll call 
   // when a user wants to join or leave a room, the joinedRooms prop will get
   // updated according, and everything will "just work"

   return listOfAllYourRoomIds.map(roomId => <SomeRoomComponent roomId={roomId}/>)
   
}

在 ReactJS 中单击按钮状态更改后，React 不会重新呈现（preact）

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