javascript - 如何仅从 firebase 中提取节点/值而不是键？

Question

如何使用 javascript 仅从 firebase 中提取节点而不提取 key ？换句话说，我只想要来自下面的 firebase 的键值对的值，这意味着我不想要下面的唯一键，只想要下面的东西。

目前，我的代码是..

function PullFirebase() {
    new Firebase('https://myfirebase.firebaseIO.com/quakes').on('value', function (snapshot) {
        var S = snapshot.val();

        function printData(data) {
            var f = eval(data);
            console.log(data + "(" + f.length + ") = " + JSON.stringify(f).replace("[", "[\n\t").replace(/}\,/g, "},\n\t").replace("]", "\n]"));
        }
        printData(S);
    });
}
PullFirebase();

这会在控制台中产生以下内容

[object Object](undefined) = {"-JStYZoJ7PWK1gM4n1M6":{"FID":"quake.2013p618454","agency":"WEL(GNS_Primary)","depth":"24.5703","latitude":"-41.5396","longitude":"174.1242","magnitude":"1.7345","magnitudetype":"M","origin_geom":"POINT (174.12425 -41.539614)","origintime":"2013-08-17T19:52:50.074","phases":"17","publicid":"2013p618454","status":"automatic","type":"","updatetime":"2013-08-17T19:54:11.27"},
    "-JStYZsd6j4Cm6GZtrrD":{"FID":"quake.2013p618440","agency":"WEL(GNS_Primary)","depth":"26.3281","latitude":"-38.8725","longitude":"175.9561","magnitude":"2.6901","magnitudetype":"M","origin_geom":"POINT (175.95611 -38.872468)","origintime":"2013-08-17T19:45:25.076","phases":"13","publicid":"2013p618440","status":"automatic","type":"","updatetime":"2013-08-17T19:48:15.374"},...

但我只想拥有字典，例如

[{"FID":"quake.2013p618454","agency":"WEL(GNS_Primary)","depth":"24.5703","latitude":"-41.5396","longitude":"174.1242","magnitude":"1.7345","magnitudetype":"M","origin_geom":"POINT (174.12425 -41.539614)","origintime":"2013-08-17T19:52:50.074","phases":"17","publicid":"2013p618454","status":"automatic","type":"","updatetime":"2013-08-17T19:54:11.27"},{"FID":"quake.2013p597338","agency":"WEL(GNS_Primary)","depth":"5.0586","latitude":"-37.8523","longitude":"176.8801","magnitude":"2.2362","magnitudetype":"M","origin_geom":"POINT (176.88006 -37.852307)","origintime":"2013-08-10T00:21:54.989","phases":"17","publicid":"2013p597338","status":"automatic","type":"","updatetime":"2013-08-10T03:42:41.324"}...]

Answer 1

如果我没理解错的话，您想要获取 quakes 下的所有子对象。

这里通常有两种方法:

1. 获取父节点的值并循环子节点
2. 监控子节点添加/更新/删除到父节点

您的方法与第 1 条相符，所以我会先回答第 1 条。我还将给出方法 #2 的示例，当您的数据集发生变化时，它会更有效。

迭代 Firebase ref 的子元素

在您的 on('value', 处理程序中，您可以使用 forEach 跳过唯一 ID:

new Firebase('https://myfirebase.firebaseIO.com/quakes').on('value', function (snapshot) {
    var quakes = [];
    snapshot.forEach(function(childSnapshot) {
        quakes.push(childSnapshot.val());
    });
    var filter = new crossfilter(quakes);

});

forEach 函数是同步的，所以我们可以简单地等待循环完成，然后创建交叉过滤器。

监控 Firebase ref 的 child

在那种情况下，最好的构造是:

var quakes = new Firebase('https://myfirebase.firebaseIO.com/quakes');
var quakeCount = 0;
quakes.on('child_added', function (snapshot) {
    var quake = snapshot.val();
    quakeCount++;
    console.log("quakeCount="+quakeCount+", FID="+quake.FID);
});
quakes.on('child_removed', function (old_snapshot) {
    var quake = old_snapshot.val();
    quakeCount--;
    console.log("quakeCount="+quakeCount+", removed FID="+quake.FID);
});

使用此代码结构，您可以主动监听添加和删除的地震。您仍然需要保留所有地震的数组，然后在 child_added、child_changed 和 child_removed 中对其进行修改。

他们如何比较

当您第一次运行代码时，监视子项将产生与 on('value', 构造相同的数据。但是当稍后添加/删除子项时 on(' value', 将再次接收所有地震，而 on('child_added', 和 on('child_removed', 将仅针对相关地震调用.