从Android调用的Firebase Cloud Function可以工作，但总是得到空响应

Question

我正在尝试调用一个Google Firebase Cloud Function，它只是使用事务来检查我的Firestore中是否存在某些内容，然后将其删除。我通过单击按钮从 Android 调用该函数，它确实有效，该条目被删除，但我在 Android 中得到的响应为空，而不是我尝试发回的消息。

云功能：

//Takes in the uid of the user who sent the request and the user who received it,
// and deletes the request
exports.cancelFriendRequest = functions.https.onCall((data, context) => {
//Grab the parameters
const sender = data.sender;
const recipient = data.recipient;
//Ensure parameters are good
if(sender === undefined || sender === "" || recipient === undefined || recipient === "") {
return res.status(400).send("Invalid arguments.");
}
const db = admin.firestore();
//Build document references to check that friend request exists
let receivedFriendRequestDocRef = db.doc("users/"+recipient+"/receivedFriendRequests/"+sender);
let sentFriendRequestDocRef = db.doc("users/"+sender+"/sentFriendRequests/"+recipient);
db.runTransaction(transaction => {
return transaction.getAll(receivedFriendRequestDocRef, sentFriendRequestDocRef).then(docs => {
//Check that the friend request exists
if(!docs[0].exists) {
return Promise.reject(new Error("Friend request does not exist."));
}
//Delete friend requests
transaction.delete(receivedFriendRequestDocRef);
transaction.delete(sentFriendRequestDocRef);
return Promise.resolve("Friend request deleted successfully.");
});
}).then(result => {
//I've also tried return res.status(200).send("Success: " + result);
//But that wasn't working so I thought I'd try this, which I saw in a Google sample git repo
return "Success: " + result;
}).catch(err => {
return err.toString();
});
});

安卓云函数调用：

public static Task<String> cancelFriendRequest(String sender, String recipient) {
FirebaseFunctions mFunctions = FirebaseFunctions.getInstance();
//Create the arguments to the callable function
Map<String, Object> data = new HashMap<>();
data.put("sender", sender);
data.put("recipient", recipient);
return mFunctions
.getHttpsCallable("cancelFriendRequest")
.call(data)
.continueWith(new Continuation<HttpsCallableResult, String>() {
@Override
public String then(@NonNull Task<HttpsCallableResult> task) throws Exception {
//This continuation runs on either success or failure, but if the task
// has failed then getResult() will throw an Exception which will be
// propagated down.
String result = (String) task.getResult().getData();
return result;
}
});
}

Android 中调用云函数的函数：

private static void cancelFriendRequest(String uid) {
String userId = FirebaseAuth.getInstance().getCurrentUser().getUid();
CloudFunctions.cancelFriendRequest(userId, uid).addOnCompleteListener(new OnCompleteListener<String>() {
@Override
public void onComplete(@NonNull Task<String> task) {
String result = task.getResult();
Log.e("Result", "" + result);
Snackbar.make(mRecyclerView, result, Snackbar.LENGTH_LONG).show();
}
});
}

我很乐意提供更多信息。我还发现调用该函数非常慢。如果我将其设置为 onRequest 并通过手动导航到 URL 来调用它，它的工作速度非常快，但是 onCall 和从 Android 调用的速度非常慢。无论如何，这是另一个问题，谢谢！

编辑：此外，Snackbar弹出(空，因为它返回null(，在文档实际从数据库中删除之前的10-15秒。

Answer 1

您错过了 main 函数(在云函数代码上(中的返回值，因此该函数在事务完成执行之前返回。从文档中：

使用这些推荐的方法来管理您的生命周期 功能：

解析执行异步处理的函数(也称为 作为"后台函数"，通过返回 JavaScript 承诺。

尝试返回事务(return db.runTransaction(transaction => {....});(，它应该可以工作，如下所示：

//Takes in the uid of the user who sent the request and the user who received it,
// and deletes the request
exports.cancelFriendRequest = functions.https.onCall((data, context) => {
//Grab the parameters
const sender = data.sender;
const recipient = data.recipient;
//Ensure parameters are good
if(sender === undefined || sender === "" || recipient === undefined || recipient === "") {
return res.status(400).send("Invalid arguments.");
}
const db = admin.firestore();
//Build document references to check that friend request exists
let receivedFriendRequestDocRef = db.doc("users/"+recipient+"/receivedFriendRequests/"+sender);
let sentFriendRequestDocRef = db.doc("users/"+sender+"/sentFriendRequests/"+recipient);
return db.runTransaction(transaction => {
return transaction.getAll(receivedFriendRequestDocRef, sentFriendRequestDocRef).then(docs => {
//Check that the friend request exists
if(!docs[0].exists) {
return Promise.reject(new Error("Friend request does not exist."));
}
//Delete friend requests
transaction.delete(receivedFriendRequestDocRef);
transaction.delete(sentFriendRequestDocRef);
return Promise.resolve("Friend request deleted successfully.");
});
}).then(result => {
//I've also tried return res.status(200).send("Success: " + result);
//But that wasn't working so I thought I'd try this, which I saw in a Google sample git repo
return "Success: " + result;
}).catch(err => {
return err.toString();
});
});