如何删除Firebase节点中除最新的X个子项外的所有项目？ [英] How to delete all but most recent X children in a Firebase node?

问题描述

给定一个Firebase节点行，其中包含唯一ID子元素（来自 push（）操作），例如这：

  Firebase-- 
 --lines 
 --K3qx02jslkdjNskjwLDK 
 --K3qx23jakjdz9Nlskjja 
 --K3qxRdXhUFmEJdifOdaj 
 --etc ... 
  

我想成为能够删除行 除最近添加的200（或100，或其他）之外的所有子元素。基本上这是一个清理操作。现在我知道我可以通过抓住客户端的行所有子项的快照，计算条目，然后使用 endsAt（totalChildren -numToKeep）来获取相关数据并运行 remove（）。但是我想避免把所有这些数据都抓到客户端。

有没有其他的方法可以解决上面的问题？

用例是棘手的，是您正在计算物品，Firebase确实（有意）没有任何基于计数的操作。因此，您需要检索前N个项目，以知道哪个项目是N + 1。

  ref.child （snapshot）{
 if（snapshot.numChildren（）> MAX_COUNT）{
 var childCount = 0; 
 var updates = { ; 
 snapshot.forEach（function child（child）{
 if（++ childCount< snapshot.numChildren（） -  MAX_COUNT）{
 updates [child.key（）] = null; 
} 
}）; 
 ref.child（'lines'）.update（updates）; 
} 
}）; 
  

这里需要注意几点：

• 这将会下载所有行


• 它会执行一个 update（）

优化此方法的一种方法（除了选择不同的/基于时间的截断策略）之外，还有一个单独的 line bids。

  lineids 
 --K3qx02jslkdjNskjwLDK 
 --K3qx23jakjdz9Nlskjja 
 --K3qxRdXhUFmEJdifOdaj 
  

所以你仍然保持每行的数据在 lines ，但也保留一个只是id的列表。然后删除多余的代码然后变成：
$ b $ $ $ $ $ $ $ $ $ $ $ ref.child（'lineids'）。once（'value'，函数（快照）{
if（snapshot.numChildren（）> MAX_COUNT）{
var childCount = 0;
var updates = {};
snapshot.forEach（function子元素）{
if（++ childCount< snapshot.numChildren（） - MAX_COUNT）{
updates ['lineids /'+ child.key（）] = null;
updates [ /'+ child.key（）] = null;
}
}）;
ref.update（updates）;
}
}）;


这个最后一个片段稍微有点牵扯，但是不必下载所有行数据



有许多变化可以选择，但是我希望这可以成为启动的足够灵感。

Given a Firebase node lines filled with unique-ID children (from push() operations), such as this:

Firebase--
  --lines
    --K3qx02jslkdjNskjwLDK
    --K3qx23jakjdz9Nlskjja
    --K3qxRdXhUFmEJdifOdaj
    --etc...

I want to be able to delete all children of lines except the most recently added 200 (or 100, or whatever). Basically this is a cleanup operation. Now I know I could do this by grabbing a snapshot of all children of lines on the client side, counting the entries, then using an endsAt(totalChildren-numToKeep) to grab the relevant data and run remove(). But I want to avoid grabbing all that data to the client.

Is there an alternative to my idea above?

解决方案

Keep the most recent N items, is one of the trickier use-cases to implement. If you have any option to change it into "keep items from the past N hours", I recommend going that route.

The reason the use-case is tricky, is that you're counting items and Firebase does (intentionally) not have any count-based operations. Because of this, you will need to retrieve the first N items to know which item is N+1.

ref.child('lines').once('value', function(snapshot) {
  if (snapshot.numChildren() > MAX_COUNT) {
    var childCount = 0;
    var updates = {};
    snapshot.forEach(function (child) {
      if (++childCount < snapshot.numChildren() - MAX_COUNT) {
        updates[child.key()] = null;
      }
    });
    ref.child('lines').update(updates);
  }
});

A few things to note here:

• this will download all lines
• it performs a single update() call to remove the extraneous lines

One way to optimize this (aside from picking a different/time-based truncating strategy) is to keep a separate list of the "line ids".

lineids
  --K3qx02jslkdjNskjwLDK
  --K3qx23jakjdz9Nlskjja
  --K3qxRdXhUFmEJdifOdaj

So you'll still keep the data for each line in lines, but also keep a list of just the ids. The code to then delete the extra ones then becomes:

ref.child('lineids').once('value', function(snapshot) {
  if (snapshot.numChildren() > MAX_COUNT) {
    var childCount = 0;
    var updates = {};
    snapshot.forEach(function (child) {
      if (++childCount < snapshot.numChildren() - MAX_COUNT) {
        updates['lineids/'+child.key()] = null;
        updates['lines/'+child.key()] = null;
      }
    });
    ref.update(updates);
  }
});

This last snippet is slightly more involved, but prevents from having to download all lines data by just downloading the line ids.

There are many variations you can choose, but I hope this serves as enough inspiration to get started.

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