Firebase

Question

这是我的数据

{
  "deck" : {
    "-JkpwAnieKjQVsdtPD4m" : {
      "deck" : "Deck 1",
      "user" : "simplelogin:1"
    },
    "-Jkq4unexm-qwhO_U2YO" : {
      "deck" : "Deck 2",
      "user" : "simplelogin:1"
    },
    "-Jkq5-II1q5yM6w3ytmG" : {
      "deck" : "Deck 3",
      "user" : "simplelogin:6"
    },
    "-Jks5mbMHmPB9MwnnOCj" : {
      "deck" : "Deck 4",
      "user" : "simplelogin:1"
    }
  }
}

如果我要补充：

      cards: {
        "-GeneratedKey":{
          "title":"foo",
          "text":"bar",
        }
      }

用甲板 "Deck 2" 说物品，我如何 select 将那个物体推到它上面。最终结果将是：

{
  "deck" : {
    "-JkpwAnieKjQVsdtPD4m" : {
      "deck" : "Deck 1",
      "user" : "simplelogin:1"
    },
    "-Jkq4unexm-qwhO_U2YO" : {
      cards: {
        "-GeneratedKey":{
          "title":"foo",
          "text":"bar",
        }
      }
      "deck" : "Deck 2",
      "user" : "simplelogin:1"
    },
    "-Jkq5-II1q5yM6w3ytmG" : {
      "deck" : "Deck 3",
      "user" : "simplelogin:6"
    },
    "-Jks5mbMHmPB9MwnnOCj" : {
      "deck" : "Deck 4",
      "user" : "simplelogin:1"
    }
  }
}

这是我尝试过的：

  deckRef.orderByChild('deckName').equalTo('Deck 2').push({
    card: {
      title: 'foo',
      text: 'bar'
    }
  });

但这只是返回了一个错误。我该怎么做才能做到这一点？

Answer 1

deckRef.orderByChild('deckName').equalTo('Deck 2') returns 查询，不是参考。一个查询可以匹配多个节点。即使在您的情况下它只匹配一个，您也需要先将该节点捕获到一个 ref 中才能 push 到它。

var query = deckRef.orderByChild('deckName').equalTo('Deck 2');
query.once('child_added', function(snapshot) {
    snapshot.ref().child('cards').push({
        title: 'foo',
        text: 'bar'
    });
});

Firebase - 推送到推送的数据

Firebase - Push to pushed data

javascript