Firebase Firestore-获取最近的趋势帖子

Question

如何使用Firestore查询“最近的趋势帖子”？

如果在过去的4个小时内上传，我将其视为“最新”。 如果它具有2000多个喜欢。

，我将其视为“趋势”。

我尝试了以下操作：

const MIN_NUMBER_OF_LIKES_FOR_BEING_TRENDING = 2000;

async function getRecentTrendingPosts(
  limit = 10,
  minimumPostsDate = diffDate(new Date(), 4, "hours"),
) {
  const query = firestore
    .collectionGroup("userPosts")
    .where("date", ">=", minimumPostsDate)
    .where("totalLikes", ">=", MIN_NUMBER_OF_LIKES_FOR_BEING_TRENDING)
    .orderBy(date, "desc");

  const querySnapshot = await query.limit(limit).get();

  const posts = await Promise.all(
    querySnapshot.docs.map((postDoc) => parsePost(postDoc))
  );

  return posts;
}

但是它不起作用，而且我很确定这是因为将“＆gt; =”应用于两个不同的字段。

在复合查询中，范围（＆lt;，＆lt; =，＆gt;，＆gt; =），而不是等于（！=，not-In）比较必须在同一字段上滤波。

实施此类查询的其他想法吗？

---

另外，我正在考虑添加

.where("__name__", ">=", uuidv4())

查询，只是为了增加一定程度的随机性。

Answer 1

不是最好的解决方案，而是对我有用：

import _ from "lodash";

const MIN_NUMBER_OF_LIKES_FOR_BEING_TRENDING = 2000;
const MAX_TRENDING_POSTS_TO_RETRIEVE = 10;

async function getTrendingPosts(
  currentUserId
  limit = MAX_TRENDING_POSTS_TO_RETRIEVE,
) {
  const minLikes = _.random(
    MIN_NUMBER_OF_LIKES_FOR_BEING_TRENDING,
    MIN_NUMBER_OF_LIKES_FOR_BEING_TRENDING * _.random(1, 10)
  );

  const maxLikes = _.random(
    minLikes + 1, 
    (minLikes + 1) * _.random(1, 100)
  );

  const query = firestore
    .collectionGroup("userPosts")
    .where("totalLikes", ">=", minLikes)
    .where("totalLikes", "<=", maxLikes)
    .orderBy("totalLikes", "desc")
    .orderBy("date", "desc");

  const querySnapshot = await query.limit(limit).get();

  const posts = await Promise.all(
    querySnapshot.docs.map((postDoc) => parsePost(postDoc, currentUserId))
  );

  return posts;
}