Javascript Firebase云功能超时

使用shell测试时，以下函数运行良好，数据在firestore中创建。 当推入prod时，它返回

函数执行耗时60002毫秒，状态为“超时”

有什么意见吗

exports.synchronizeAzavista = functions.auth.user().onCreate(event => {

    console.log('New User Created');

    const user = event.data;
    const email = user.email;
    const uid = user.uid;

    return admin.database().ref(`/delegates`)
        .orderByChild(`email`)
        .equalTo(email)
        .once("child_added").then(snap => {
            const fbUserRef = snap.key;

            return admin.firestore().collection(`/users`).doc(`${fbUserRef}`).set({
                email: email,
                uid: uid
            }).then(() => console.log("User Created"));
        });
});

编辑

我已经用以下内容更新了我的代码，但我仍然得到

函数返回未定义、预期的承诺或值

，但我无法确定函数返回未定义的位置。为什么我的

getUser（）

函数不返回任何内容

const admin = require('firebase-admin');

admin.initializeApp(functions.config().firebase);

exports.synchronizeAzavista = functions.auth.user().onCreate(event => {

    console.log('New User Created');//This log

    const user = event.data;
    const email = user.email;
    const uid = user.uid;

    console.log('Const are set');//This log

    getUser(email).then(snap => {

        console.log("User Key is " + snap.key);//No log

        const fbUserRef = snap.key;

        return admin.firestore().collection(`/users`).doc(`${fbUserRef}`).set({
            email: email,
            uid: uid
        });
    }).then(() => console.log("User Data transferred in Firestore"));
});

function getUser(email) {

    console.log("Start GetUser for " + email);//This log 

    const snapKey = admin.database().ref(`/delegates`).orderByChild(`email`).equalTo(email).once("child_added").then(snap => {
        console.log(snap.key);//No Log here
return snap;
    });
    return snapKey;
}

---

你没有回复你写给Firestore的承诺

exports.synchronizeAzavista = functions.auth.user().onCreate(event => {
    const user = event.data;
    const email = user.email;
    const uid = user.uid;

    return admin.database().ref(`/delegates`)
        .orderByChild(`email`)
        .equalTo(email)
        .once("child_added").then(snap => {
            const fbUserRef = snap.key;

            return admin.firestore().collection(`/users`).doc(`${fbUserRef}`).set({
                email: email,
                uid: uid
            });
        });
});

---

你没有回复你写给Firestore的承诺

exports.synchronizeAzavista = functions.auth.user().onCreate(event => {
    const user = event.data;
    const email = user.email;
    const uid = user.uid;

    return admin.database().ref(`/delegates`)
        .orderByChild(`email`)
        .equalTo(email)
        .once("child_added").then(snap => {
            const fbUserRef = snap.key;

            return admin.firestore().collection(`/users`).doc(`${fbUserRef}`).set({
                email: email,
                uid: uid
            });
        });
});

---

谢谢@Frank，这是否意味着then（）终止集合（）的承诺？不-问题不在于“then”-您可以包含一个then部分，但您需要返回承诺，因此Firebase知道它必须让您的函数保持活动状态，直到该承诺解决。-然后firebase可以将承诺的结果解释为成功或错误（拒绝承诺）谢谢@Frank，这是否意味着then（）终止了集合的承诺（）？不-问题不在于“then”-您可以包含then部分，但您需要返回承诺，所以Firebase知道它必须让你的功能保持活跃，直到承诺解决为止。-然后firebase可以将承诺的结果解释为成功或错误（拒绝承诺）