从 Firestore 和 Firebase 函数获取超时

Question

我有一个简单的功能，我正在模拟器上进行本地测试。

    exports.sendNotification = functions.https.onRequest(() => {
  firebase
    .firestore()
    .collection("Users")
    .get()
    .then((snapshot) => {
      snapshot.docs.forEach((doc) => {
        console.log(doc.id, doc.data());
      });
    });
});

当我执行该函数时，它工作正常，但在我收到此消息后不久：

     functions: Your function timed out after ~60s. To configure this timeout, see
      https://firebase.google.com/docs/functions/manage-functions#set_timeout_and_memory_allocation.
>  C:\Users\myusername\AppData\Roaming\npm\node_modules\firebase-tools\lib\emulator\functionsEmulatorRuntime.js:618
>              throw new Error("Function timed out.");
>              ^
>
>  Error: Function timed out.
>      at Timeout._onTimeout (C:\Users\myusername\AppData\Roaming\npm\node_modules\firebase-tools\lib\emulator\functionsEmulatorRuntime.js:618:19)
>      at listOnTimeout (node:internal/timers:557:17)
>      at processTimers (node:internal/timers:500:7)

我的期望是函数会正常结束，因为它在函数的末尾，这条消息是说我需要做一些事情来表明函数已经完成吗？我是否需要更改某些内容，尤其是在投入生产时？

Answer 1

您必须通过发回响应来terminate a HTTP function，否则它将一直运行直到超时。尝试像这样重构您的代码：

exports.sendNotification = functions.https.onRequest((req, res) => {
  // add req, res in functions parameters             ^^^
  return firebase
    .firestore()
    .collection("Users")
    .get()
    .then((snapshot) => {
      snapshot.docs.forEach((doc) => {
        console.log(doc.id, doc.data());
      });
      // Return response
      return res.json({ data: snapshot.docs.map(d => d.data()) })
    });
});