JavaScript Firebase：查询快照始终为空

问题描述

无论我做什么，我似乎无法找到一种方法来访问子节点onSite，当我记录snapshot.val（）时，它显示在那里，但我无法弄清楚如何访问它。

No matter what I do I can't seem to figure out a way to access the child "onSite", which shows as being there when I log snapshot.val(), but I cannot figure out how to access it.

代码：

firebase.database().ref().child("users").orderByChild('facebook_id').equalTo(fbID).once("value").then(function(snapshot) {
    console.log(snapshot.val());
    console.log(snapshot.child("onSite").val());
});

以下是回复：

Here is the response:

它不应该为空，它应该是假的。我不能做孩子（4mUUjF ...）。child（onSite）。val（）因为我不知道在查询之前ID是什么。使用每个循环不起作用，它只循环通过第一级，即ID。

It shouldn't be null, it should be false. I can't do child("4mUUjF...").child("onSite").val() because I don't know what the ID is before the query. Using an each loop doesn't work, it only loops through the first level, which is the ID.

推荐答案

执行时针对Firebase数据库的查询，可能会有多个结果。因此快照包含这些结果的列表。即使只有一个结果，快照也会包含一个结果列表。

When you execute a query against the Firebase Database, there will potentially be multiple results. So the snapshot contains a list of those results. Even if there is only a single result, the snapshot will contain a list of one result.

您的代码需要使用来处理列表Snapshot.forEach（）：

firebase.database().ref().child("users").orderByChild('facebook_id').equalTo(fbID)
.once("value").then(function(result) {
    result.forEach(function(snapshot) {
        console.log(snapshot.val());
        console.log(snapshot.child("onSite").val());
    });
});

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