如何在Firebase中以升序或降序方式检索分页的子项？ [英] How to retrieve paginated children in ascending or descending order in Firebase?
问题描述
看起来，以相反顺序显示Firebase记录的唯一方法是将其全部提取出来，然后反向对其进行迭代。
对于大型数据集，特别是对于移动客户端来说，这可能会很难实现。
更好的方法？什么是从Firebase查询分页数据的通用和首选解决方案，按升序或降序排列？ strong>更新回答
@Tyler回答 stackoverflow.com/ a / 38024909/681290 是一个更新和更方便的答案这个问题
TL; DR
如果你想要做的是没有解决方案是：
$ $ p $ code> ref .orderBy（field，'DESC'）。offset（n）.limit（x）
在Firebase github中，有不受支持的工具进行分页，但仅限于升序。。
$ b 否则，这里是迄今为止我找到的最接近的解决方案。我故意将这个问题解释为通用的，而不仅仅是OP所要求的时间字段。
使用优先级我们的目标是使用 具有负数据的冗余子字段 例如，我们可以使用负时间戳 问题： 使用limitToLast（）和 和多余的字段： 这对于时间戳字段应该是非常有效的，因为通常这是一个唯一的字段。 由此产生的项目阵列只需在使用前倒转即可。 （只要记住 以下项目的示例包含 首页查询得到最好的得分项目： 结果： 注意，子集的第一项有一个 使用该查询， code> b，c，d 项目，而不是根据API文档预期的 解决方法 这可以通过不期望每个子集保持相同的记录量来缓解，并放弃那些已经被加载。 我没有看到这个用例的任何解决方法，我猜Firebase不适用于这个： ） 编辑：最后，我使用 Algolia 为所有搜索相关的目的。这是一个非常好的API，我希望谷歌最终获得Algolia整合Firebase&阿尔戈利亚，因为它们的互补性接近100％！免责声明：不拥有股份！ ： - ） Suppose I'm using firebase for a commenting system and I want to retrieve the comments for a given topic, but there are so many comments in one topic that I don't want to retrieve them all at once. I also want the newest comments to be displayed on top. It seems that the only way to display firebase records in reverse order is to retrieve them all and then iterate over them in reverse. This could get very unwieldy on large datasets, especially for mobile clients. Is there any better way? What is the generic and preferred solution to query paginated data from Firebase, in ascending or descending order? Updated answer @Tyler answer stackoverflow.com/a/38024909/681290 is a more up to date and more convenient answer for this question TL;DR There is no solution if what you are looking to do is : Also, in Firebase github there are unsupported tools that do pagination, although only in ascending order only. Otherwise, here are the closest possible solutions, that I have found myself, or on the web so far. I deliberately interpreted the question as generic, not only about time fields as asked by the OP. Use priorities The goal is to use Issues : Redundant child fields with negative data For example, we can use a negative timestamp Issues : Using limitToLast() and This let us avoid negative and redundant field : This should be pretty effective with timestamp fields, because usually this is a unique field. The resulting array of items simply need to be reversed before use. (just remember Issues : Example with the following items holding a First page query to get the best scoring items : Results : Note the first item of the subset has a With that query, we get Workaround This can be mitigated by not expecting each subset to hold the same amount of record, and discarding those which have already been loaded. But, if the first million users of my app have a I don't see any workaround for this use case, I guess Firebase is not intended for this :-) Edit: In the end, I am using Algolia for all search-related purpose. It's gt a really nice API, and I hope google ends up acquiring Algolia to integrate both Firebase & Algolia, as their complementarity is close to 100%! Disclaimer: no shares owned!! :-) 这篇关于如何在Firebase中以升序或降序方式检索分页的子项？的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！ setWithPriority（）和 setPriority（） 问题：
优先级字段的优势相反？ （实际上，优先级存储为一个名为 .priority 的基础字段，您可以在导出json数据时看到）
timeRev 索引字段除了时间以便能够最先获得最新的项目。
ref.orderByChild（'timeRev'）
.limitToFirst（100）;
score 字段可以一次更新，不知道是否有可能与两个字段，一个正面和一个负面）
限制（），但是现在我们可以使用 limitToFist（）， limitToLast ）和范围查询（见下面）
$ b endAt（）范围查询
ref.orderBy（'time'）
.limitToLast（100）
Array.prototype.reverse（）是可变的，所以它会改变你的数组）
startAt 或 endAt 边界。如果大量项目共享相同的值，那么数据集不能以固定长度偏移量进行分割。
得分值：
{
物品：{
a：{score：0}，
b：{score：0}，
c：{score：1}，
d：{score：1}，
e：{score： 2}，
f：{score：5}
}
}
$ $ $ $ $ $ $ $ ref.child（'items'）。orderByChild（'score' ）
.limitToLast（3）
pre $ $ b $ {
d：{score：1}，
e：{score：2}，
f：{score：5}
$ b $ p
$ b 1 得分，所以我们通过选择得分 1 或更少的所有项目来尝试获得前一页面：
ref.child（'items'）。orderByChild（'score'）
.endAt（1）
.limitToLast（3）
a，b，c 给定 endAt（1）是包含性的，所以它会尝试获得1的所有分数，并且无法对之前已经返回的排序进行排序。
但是，如果我的应用程序的第一百万用户有一个 0 得分，这个子集不能分页，因为 endAt 偏移量是无用的，因为它是基于值记录数量。
ref.orderBy(field, 'DESC').offset(n).limit(x)
setWithPriority() and setPriority() on children to be able to get ordered data later with orderByPriority().
priority field instead ? (in fact, priority is stored as an underlying field called .priority, that you can see when exporting json data)
timeRev indexed field in addition to time to be able to get the newest items first. ref.orderByChild('timeRev')
.limitToFirst(100);
score field can be updated at once, not sure if it's possible with two fields, one positive and one negative)limit() in the Firebase API, but is kinda obsolete now that we can use limitToFist(), limitToLast(), and range queries (see below)
endAt() range queriesref.orderBy('time')
.limitToLast(100)
Array.prototype.reverse() is mutable, so it will change your array)
startAt or endAt boundary. If a lot of items share the same value, the dataset cannot be split in fixed length offsets.score value : {
items: {
a: {score: 0},
b: {score: 0},
c: {score: 1},
d: {score: 1},
e: {score: 2},
f: {score: 5}
}
}
ref.child('items').orderByChild('score')
.limitToLast(3)
{
d: {score: 1},
e: {score: 2},
f: {score: 5}
}
1 score, so we try to get the previous page by selecting all items with score of 1 or less :ref.child('items').orderByChild('score')
.endAt(1)
.limitToLast(3)
b,c,d items, instead of a,b,c items, which is expected as per the API docs, given the endAt(1) is inclusive, so it will try to get all scores of 1, and has no way to sort which were already returned before.0 score, this subset cannot be paginated, because the endAt offset is useless, as it is based on the value instead of the number of records.
