如何仅从firebase而不是键中提取节点/值？

Question

你如何只使用firebase从firebase而不是键中拉出节点？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,而只需要下面的内容.

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();

这会在控制台中生成以下内容

[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"},...

但我想只有字典,比如

[{"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

如果我理解正确,你想要获得所有子对象quakes.

你通常有两种方法:

1. 获取父节点的值并循环遍历子节点
2. 监视子项被添加/更新/删除到父节点

你的方法与#1匹配,所以我先回答那个问题.我还将举例说明方法#2,当数据集发生变化时效率更高.

迭代Firebase参考的孩子

在on('value',处理程序中,您可以使用forEach以下命令跳过唯一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);

});

该forEach函数是同步的,因此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参考的儿童

在这种情况下,最好的结构是:

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);
});

使用此代码构造,您可以主动侦听添加和删除的地震.你仍然必须保留所有的地震,然后您可以修改的数组child_added,child_changed和child_removed.

他们如何比较

首次运行代码时,监视子项将导致与on('value',构造相同的数据.但是当孩子被添加/移除后,on('value',将再次收到所有地震,on('child_added',而且on('child_removed',只会被召唤到有问题的地震.