刷新以加载新数据并滚动以分页数据Flutter Firestore [英] Refresh to load new data and scroll to paginate data Flutter Firestore

问题描述

我有一个名为"Feeds"的Firestore集合，其中包含给定用户的feed(例如Instagram)的帖子列表.视图加载后，我会按时间调用未来和订单

I have a Firestore collection called 'Feeds' that contains a list of posts for a given user's feed (like Instagram). When the view loads, I call a future and order by time

var userPostsDocumentSnapshot = await _feeds
          .document(userUid)
          .collection(items)
          .orderBy('postDateTime', descending: true)
          .getDocuments();

我最初使用的是流，但是当新数据添加到用户的feed中时，我不希望它自动显示，我希望下拉菜单加载新数据.但是，我还想通过在向下滚动(从最新到最旧)时对数据进行分页来提高效率，以使其更有效率.滚动时如何既能下拉加载新数据又能分页数据?

I was initially using a stream, but when new data is added to the user's feed, I don't want it to show automatically, I want a pull-down-to-load new data. However, I also want to make it efficient by paginating the data as you scroll down (from newest to oldest) to make it a little more efficient. How can I achieve both pull-down-to-load new data and also paginate data when scrolling?

推荐答案

对于下拉菜单，您只想再次执行查询.

For the pull-down-to-load, you'll just want to execute your query again.

对于分页，我使用了一个游标，如此处所述.例如，根据您使用UI的方式，您可以将查询限制为特定数量的文档.

For the pagination, I used a cursor, as documented here. For example, depending on how you're doing the UI, you can limit your query by a particular number of documents.

因此，对于您的查询，这样的操作可能会起作用:

So for your query, something like this may work:

var userPostsDocumentSnapshot = _feeds
          .document(userUid)
          .collection(items)
          .orderBy('postDateTime', descending: true)
          .limit(10);

然后，当您实际抓取文档时，请存储对最后一个文档的引用:

Then when you actually grab the documents, store a reference to the last one:

var docsnaps = await userPostsDocumentSnapshot.getDocuments();
var last = docsnaps.docs[documentSnapshots.docs.length-1];

然后在分页时，对查询使用一些细微变化:

Then when you paginate, use a slight variation on your query:

      ...  _feeds
          .document(userUid)
          .collection(items)
          .orderBy('postDateTime', descending: true)
          .startAfter(last);

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