Firebase Cloud function-Get document snapshot field which is the URL of a file in firebase storage and delete the file from firebase storage - js

Question

我想從文檔快照中檢索字段值，即 firebase 存儲中文件的 URL 並從 firebase 存儲中刪除文件

我可以使用以下代碼成功刪除過期的 Firestore 文檔：

const functions = require("firebase-functions");
const admin = require("firebase-admin");
const { firestore } = require("firebase-admin");
admin.initializeApp();

exports.removeExpiredDocuments = functions.pubsub.schedule("every 1 hours").onRun(async (context) => {
  const db = admin.firestore();
  const now = firestore.Timestamp.now();
  const ts = firestore.Timestamp.fromMillis(now.toMillis() - 86400000); // 24 hours in milliseconds = 86400000

  const snap = await db.collection("photos").where("timestamp", "<", ts).get();
  let promises = [];
  snap.forEach((snap) => {
    promises.push(snap.ref.delete());
  });
  return Promise.all(promises);
});

但我不知道如何從 forEach 塊中的文檔快照中檢索字段值（文件的 URL）並從 firebase 存儲中刪除文件。

這是firestore數據庫：

photourl 的字段值將被檢索。

Answer 1

我認為代碼看起來像：

//some code ....
    snap.docs.map((doc) => {
    if (doc.exist) {
       var url = doc.data().photourl;
       //do something logic call to firestorage and deleted data base on url get 
       //write logic deleted url firebase after deleted success firestorage           
    }

  });