$firebaseArray将对象从索引移动到索引-$firebaseArraymoveobjectfromindextoindex

To order your items, leave it to Firebase. For example say that you want the transactions to be ordered on creation date, you could:

要订购商品,请将其留给Firebase。例如,假设您希望在创建日期订购交易,您可以:

var ref = fb.child("members")
            .child(fbAuth.uid)
            .child("accounts")
            .child($scope.AccountId)
            .child("transactions")
            .orderByChild("creationDate");
$scope.transactiOns= $firebaseArray(ref);

If that does not cover what you want, please update your question to be clearer on what you're trying to sort on. Right now, I have no clue what you're trying to sort on.

如果这不能满足您的需求,请更新您的问题,以便更清楚地了解您要排序的内容。现在,我不知道你想要点什么。

I just noticed your update and that indeed clarifies what you're trying to accomplish.

我刚刚注意到你的更新,这确实澄清了你想要完成的任务。

In your first data snippet, the data is ordered by customindex. So you can retrieve them in that order by:

在您的第一个数据代码段中,数据按customindex排序。因此,您可以按以下顺序检索它们:

var ref = fb.child("members")
            .child(fbAuth.uid)
            .child("accounts")
            .child($scope.AccountId)
            .child("transactions")
            .orderByChild("customindex");
$scope.transactiOns= $firebaseArray(ref);

In your last data snippet, the transactions are no longer ordered by any property. So when the user drags the Starbucks transaction from its last place and drops it in second place, you'll need to update the customindex property of all transactions after it. If you update those values, AngularFire will automatically ensure that the items are in the correct order in the $firebaseArray again.

在您的上一个数据代码段中,任何属性都不再对事务进行排序。因此,当用户从最后一个位置拖动星巴克交易并将其放在第二位时,您需要更新之后所有交易的customindex属性。如果更新这些值,AngularFire将自动确保$ firebaseArray中的项目顺序正确。

Note that this requires updating 4 transactions, which may not be very efficient. Alternatively you can:

请注意,这需要更新4个事务,这可能不是非常有效。或者你可以:

1. keep the transaction details (what you initially had) and the transaction order in separate nodes. That way you'll only have to update the transaction_order node.

保留事务详细信息(您最初拥有的)和单独节点中的事务顺序。这样你只需要更新transaction_order节点。

2. keep some space between your initial indexes. So instead of [1,2,3,4,5], start with [10,20,30,40,50]. Then when the user moves Starbucks up between 20 and 30, you can simply update the Starbucks transaction's customindex to 25: [10,20,25,30,40,50]. This approach is a bit hacky, but is simpler to implement and will normally work pretty well.

在初始索引之间保留一些空间。因此,而不是[1,2,3,4,5],以[10,20,30,40,50]开头。然后,当用户将星巴克在20到30之间移动时,您只需将星巴克交易的customindex更新为25:[10,20,25,30,40,50]。这种方法有点笨拙,但实现起来更简单,并且通常可以很好地工作。