Firebase函数类型脚本带有特定uid的initializeAdmin() [英] Firebase Functions Typescript initializeAdmin() with specific uid

问题描述

我已经找到了使用唯一UID初始化firebase函数的这种方法的参考，目的是用数据库规则来限制它，但是我不断收到错误消息 Error: Can't determine Firebase Database URL

I have found reference to this method of initializing firebase functions with a unique UID for the purpose of constraining it with database rules, but I keep getting the error Error: Can't determine Firebase Database URL

我意识到我可以从环境变量中检索DatabaseURL，但是在此之前，我想知道自己是否做错了，因为我所看到的示例不必针对Firebase Functions进行操作.

I realize that I can retrieve the DatabaseURL from environment variables, but before I do that, I want to know if I'm doing something wrong because the examples I have seen have not had to do that for Firebase Functions.

import * as functions from "firebase-functions";
import * as admin from "firebase-admin";

admin.initializeApp({
  ...functions.config().firebase,
  databaseAuthVariableOverride: { uid: "functions" }
});

推荐答案

functions.config().firebase已弃用.相反，您可以改用process.env.FIREBASE_CONFIG的内容.像这样(未经测试):

functions.config().firebase is deprecated. Instead, you can build upon the contents of process.env.FIREBASE_CONFIG instead. Something like this (untested):

const config = JSON.parse(process.env.FIREBASE_CONFIG)
config.databaseAuthVariableOverride = { uid: "uid" }
admin.initializeApp(config)

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