获得孩子的价值

Question

我想获取image的值。

我已经具有用户的UID，但是当我的代码运行时，出现此错误：Error: No such object: XXXXX.appspot.com/https://XXXXX.firebaseio.com/images/-M9OeGPXB-KqEVqxZYCb/image

我不明白为什么，因为该值不为空。

我的数据库的屏幕截图：

我的代码：

// The Cloud Functions for Firebase SDK to create Cloud Functions and setup triggers.
const functions = require('firebase-functions');

// The Firebase Admin SDK to access Cloud Firestore.
var admin = require("firebase-admin");

var serviceAccount = require("./serviceAccountKey.json");

admin.initializeApp({
  credential: admin.credential.cert(serviceAccount),
  databaseURL: "https://XXXXXX.firebaseio.com",
  storageBucket: "gs://XXXXX.appspot.com",
});

var defaultStorage = admin.storage();

// Cut off time. Child nodes older than this will be deleted.
const CUT_OFF_TIME = 24 * 60 * 60 * 1000; // 24 Hours in milliseconds.

/**
 * This database triggered function will check for child nodes that are older than the
 * cut-off time. Each child needs to have a `timestamp` attribute.
 */
exports.deleteOldItems = functions.database.ref('/images/{pushId}').onWrite(async (change) => {
  const ref = change.after.ref.parent; // reference to the parent
  const now = Date.now();
  const cutoff = now - CUT_OFF_TIME;
  const oldItemsQuery = ref.orderByChild('timestamp').endAt(cutoff);
  const snapshot = await oldItemsQuery.once('value');


  snapshot.forEach((child) => {

        const uid = child.key;
        const re = admin.database().ref('images/'+uid);
        const img = re.child('image');

        const bucket = defaultStorage.bucket();
        const file = bucket.file(img);

    // Delete the file
        return file.delete();
    })


  // create a map with all children that need to be removed
  const updates = {};
  snapshot.forEach(child => {
    updates[child.key] = null;
  });
  // execute all updates in one go and return the result to end the function
  return ref.update(updates);
});

预先感谢您的帮助。

Answer 1

img变量只是对数据库中属性的引用。如果您在上调用toString()，则会返回该引用的完整路径，即https://yourprojct.firebaseio.com/path/to/property。

因此，您将需要实际使用数据库中的值，并使用其中的 来查找存储中的文件：

snapshot.forEach((child) => {
    const uid = child.key;
    const img = child.child('image');
    console.log(img.val());

现在，这将记录图像的路径，因为它存储在数据库中。