为什么从 fetch 函数返回的对象是它之外的函数？

Question

I don't get why data type is an object inside my function but outside is a function... Any ideas why ? Thanks.

const fetchAllUsersData = async () => {
    setLoading(true);
    const userFromFirebase = await firebase.database().ref("users").orderByChild("nickname").equalTo(nickname).on("child_added", ((snapshot) => {
        const data = snapshot.val();
        console.log("data", typeof data); // show object
        return data;
    }));
    console.log("userFromFirebase",typeof userFromFirebase); // show function
}

fetchAllUsersData();

Answer 1

Read the documentation: https://firebase.google.com/docs/reference/node/firebase.database.Reference#on

The .on method will return anything, not your DataSnapshot. A better approach to this would be to use callbacks instead. Here is an example:

const fetchAllUsersData = async (callback) => {
    setLoading(true);
    const userFromFirebase = await firebase.database().ref("users").orderByChild("nickname").equalTo(nickname).on("child_added", ((snapshot) => {
        const data = snapshot.val();
        console.log("data", typeof data); // show object
        callback(data);
    }));
}

fetchAllUsersData((data) => {
        console.log("data", typeof data); // still shows object
});