React Hooks + Firebase Firestore onSnapshot - 正确使用带有react-hooks的 Firestore 侦听器

IT技术 javascript reactjs react-native google-cloud-firestore react-hooks
2021-05-01 09:49:40

问题

想象一下,您有一个带有数据库侦听器的屏幕,该侦听器是在 useEffect 中创建的,并且此侦听器的目的是增加屏幕中的计数器,如下所示:

(只使用 useEffect 钩子而不依赖)

function MyScreen(props) {
     const [magazinesCounter, setMagazinesCounter] = useState(0);

     const handleMagazinesChanges = (snapshot) => {
         const changes = snapshot.docChanges();

         let _magazinesCounter = magazinesCounter;
       
         changes.forEach((change) => {
            if (change.type === "added") {
               _magazinesCounter  += 1;
            }
            if (change.type === "removed") {
               _magazinesCounter  -= 1;
            }
         });
         
         setMagazinesCounter(_magazinesCounter);  
     };

     useEffect(() => {
         const query = props.db.collection("content")
                        .where("type", "==", "magazine");
         
         // Create the DB listener
         const unsuscribe = query.onSnapshot(handleMagazinesChanges, (err) => {});
         return () => {
            unsuscribe();
         }
     }, []);
}

正如您在此处看到的,这将不起作用,因为在状态更新时不会重新创建 useEffect 侦听器中使用的 handleMagazinesChanges...

因此,我尝试修复将 magazinesCounter 作为依赖项传递给 useEffect 的问题,如下所示:

(使用 useEffect 钩子和修改后的状态作为依赖)

 useEffect(() => {
     // ... the same stuff
 }, [magazinesCounter]);

但是这样,我们将进入无限循环,因为将重新创建侦听器,并且

 if (change.type === "added") {
     _magazinesCounter  += 1;
 }
 ...
 setMagazinesCounter(_magazinesCounter);

将被再次执行,并再次......

Pd:另外,将handleMagazinesChanges 函数包装在useCallback 中,并以magazinesCounter 作为依赖项,然后将该函数作为依赖项传递给useEffect,将具有相同的效果...

那么,我该如何解决这种情况呢?我知道,如果我们用 useRef 对相同数据使用辅助引用,我们可以成功执行此操作,避免无限循环。但是,有没有其他更好的方法来做到这一点?我的意思是,没有状态 + 对最新数据的引用:

(无依赖的 useEffect + useRef)

// This works good, but is there any other better solution to this problem?
function MyScreen(props) {
     const [magazinesCounter, setMagazinesCounter] = useState(0);

     const magazinesCounterRef = useRef(magazinesCounter);

     const handleMagazinesChanges = (snapshot) => {
         const changes = snapshot.docChanges();

         changes.forEach((change) => {
            if (change.type === "added") {
               magazinesCounterRef.current  += 1;
            }
            if (change.type === "removed") {
               magazinesCounterRef.current  -= 1;
            }
         });

         setMagazinesCounter(magazinesCounterRef.current);  
     };

     useEffect(() => {
         const query = props.db.collection("content")
                        .where("type", "==", "magazine");

         // Create the DB listener
         const unsuscribe = query.onSnapshot(handleMagazinesChanges, (err) => {});
         return () => {
            unsuscribe();
         }
     }, []);
}

有任何想法吗?谢谢你。

Pd:也许这是要走的路,但我想有更好的方法而不创建辅助变量。

1个回答

为了实现这一目标,我有两个建议:

  1. 将回调移动到内部useEffect以避免每次渲染时重新创建函数
  2. 在设置状态时使用回调来获取当前值(请参阅此处React 文档)以避免在状态更改时需要重新创建回调

将这些与您当前的代码一起使用:

function MyScreen(props) {
     const [magazinesCounter, setMagazinesCounter] = useState(0);

     useEffect(() => {
         // Moved inside "useEffect" to avoid re-creating on render
         const handleMagazinesChanges = (snapshot) => {
             const changes = snapshot.docChanges();

             // Accumulate differences
             let difference  = 0;
             changes.forEach((change) => {
                if (change.type === "added") {
                   difference  += 1;
                }
                if (change.type === "removed") {
                   difference  -= 1;
                }
             });
             
             // Use the setState callback 
             setMagazinesCounter((currentMagazinesCounter) => currentMagazinesCounter + difference);  
         };

         const query = props.db.collection("content")
                        .where("type", "==", "magazine");
         
         // Create the DB listener
         const unsuscribe = query.onSnapshot(handleMagazinesChanges, 
  err => console.log(err));
         return () => {
            unsuscribe();
         }
     }, []);
}