Question

这是我的数据库结构

{
  "main" : {
    "fake data1" : {
      "abc" : "1001",
      "high" : "100",
      "low" : "150",
      "mid" : "200",
      "name" : "fake data1"
    },
    "fake data2" : {
      "abc" : "1023",
      "high" : "1020",
      "low" : "1020",
      "mid" : "1501",
      "name" : "fake data2"
    },
    "fake data3" : {
      "abc" : "1001",
      "high" : "100",
      "low" : "150",
      "mid" : "200",
      "name" : "fake data3"
    },
    "fake data4" : {
      "abc" : "1023",
      "high" : "1020",
      "low" : "1020",
      "mid" : "1501",
      "name" : "fake data4"
    },
    "fake data5" : {
      "abc" : "1001",
      "high" : "100",
      "low" : "150",
      "mid" : "200",
      "name" : "fake data5"
    },
    "fake data6" : {
      "abc" : "1023",
      "high" : "1020",
      "low" : "1020",
      "mid" : "1501",
      "name" : "fake data6"
    }
  }
}

我想从上面检索所有名字，但我无法做到。

我正在尝试使用以下查询：

   var datafire = firebase.database().ref();
   var namedb = datafire.child('main')
   var query = namedb.orderByChild('name')
   query.on("value", function(datasnap) {
                        var data = (datasnap.val());
                        console.log(data.name);

Answer 1

对Firebase数据库执行查询时，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的列表。

您的代码需要使用Snapshot.forEach()循环这些项来处理快照包含多个项目这一事实：

var datafire =firebase.database().ref();
var namedb = datafire.child('main')
var query = namedb.orderByChild('name')
query.on("value",function(datasnap){
  datasnap.forEach(function(child) {
    var data=child.val();
    console.log(child.key, data.name);
  });
});

Firebase数据检索：我无法从Firebase数据库中创建的json中检索数据

1 个答案: