61

我正在尝试从 Firebase 数据库中添加/删除条目。我想将它们列出在要添加/修改/删除的表中(前端),但我需要一种唯一标识每个条目的方法,以便修改/删除。Firebase 在使用 push() 时默认添加一个唯一标识符,但我没有看到任何有关如何在 API 文档中选择此唯一标识符的内容。这甚至可以做到吗?我应该使用 set() 来创建唯一 ID 吗?

我使用他们的教程将这个快速示例放在一起:

<div id='messagesDiv'></div>
<input type='text' class="td-field" id='nameInput' placeholder='Name'>
<input type='text' class="td-field" id='messageInput' placeholder='Message'>
<input type='text' class="td-field" id='categoryInput' placeholder='Category'>
<input type='text' class="td-field" id='enabledInput' placeholder='Enabled'>
<input type='text' class="td-field" id='approvedInput' placeholder='Approved'>
<input type='Button' class="td-field" id='Submit' Value="Revove" onclick="msgRef.remove()">

<script>
var myDataRef = new Firebase('https://unique.firebase.com/');

  $('.td-field').keypress(function (e) {
    if (e.keyCode == 13) {
      var name     = $('#nameInput').val();
      var text     = $('#messageInput').val();
      var category = $('#categoryInput').val();
      var enabled  = $('#enabledInput').val();
      var approved = $('#approvedInput').val();
      myDataRef.push({name: name, text: text, category: category, enabled: enabled, approved: approved });
      $('#messageInput').val('');
    }
  });
  myDataRef.on('child_added', function(snapshot) {
    var message = snapshot.val();
    displayChatMessage(message.name, message.text, message.category, message.enabled, message.approved);
  });
  function displayChatMessage(name, text, category, enabled, approved, ) {
    $('<div/>').text(text).prepend($('<em/>').text(name+' : '+category +' : '+enabled +' : '+approved+ ' : ' )).appendTo($('#messagesDiv'));
    $('#messagesDiv')[0].scrollTop = $('#messagesDiv')[0].scrollHeight;
  };
</script>

现在假设我有三行数据:

fred : 1 : 1 : 1 : test message 1
fred : 1 : 1 : 1 : test message 2
fred : 1 : 1 : 1 : test message 3

如何唯一标识第 2 行?

在 Firebase 数据库中,它们看起来像这样:

-DatabaseName
    -IuxeSuSiNy6xiahCXa0
        approved: "1"
        category: "1"
        enabled: "1"
        name: "Fred"
        text: "test message 1"
    -IuxeTjwWOhV0lyEP5hf
        approved: "1"
        category: "1"
        enabled: "1"
        name: "Fred"
        text: "test message 2"
    -IuxeUWgBMTH4Xk9QADM
        approved: "1"
        category: "1"
        enabled: "1"
        name: "Fred"
        text: "test message 3"
4

8 回答 8

66

对于任何发现此问题并使用的人Firebase 3+,在推送后获取自动生成的对象唯一 ID 的方式是使用Promise 快照上的key属性(而不是方法):

firebase
  .ref('item')
  .push({...})
  .then((snap) => {
     const key = snap.key 
  })

在Firebase 文档中阅读有关它的更多信息。

附带说明一下,那些考虑生成自己的唯一 ID 的人应该三思而后行。它可能具有安全性和性能影响。如果您不确定,请使用 Firebase 的 ID。它包含一个时间戳,并具有一些开箱即用的简洁安全功能。

更多关于它的信息

push() 生成的唯一键按当前时间排序,因此生成的项目列表将按时间顺序排序。密钥也被设计成不可猜测的(它们包含 72 个随机位的熵)。

于 2016-10-26T18:27:30.060 回答
44

要获取任何快照的“名称”(在本例中为 push() 创建的 ID),只需像这样调用 name():

var name = snapshot.name();

如果你想得到 push() 自动生成的名字,你可以在返回的引用上调用 name(),像这样:

var newRef = myDataRef.push(...);
var newID = newRef.name();

注意: snapshot.name()已被弃用。查看其他答案。

于 2013-05-19T17:33:06.210 回答
33

snapshot.name()已被弃用。改为使用key。任何 DataSnapshot 上的key属性(代表 Firebase 根的属性除外)都将返回生成它的位置的键名。在您的示例中:

myDataRef.on('child_added', function(snapshot) {
    var message = snapshot.val();
    var id = snapshot.key;
    displayChatMessage(message.name, message.text, message.category, message.enabled, message.approved);
});
于 2015-03-23T19:43:07.667 回答
12

为了得到uniqueIDpush()必须使用这个变体:

// Generate a reference to a new location and add some data using push()
 var newPostRef = postsRef.push();
// Get the unique key generated by push()
var postId = newPostRef.key;

Ref当你push()使用这个 ref 时,你会生成一个新的,你.key可以获得uniqueID.

于 2017-10-13T02:03:11.547 回答
5

正如@Rima 指出的那样,key()这是将 ID firebase 分配给您的push().

但是,如果您希望去掉中间人,Firebase 发布了一个带有 ID 生成代码的要点。它只是当前时间的一个函数,这就是它们如何保证唯一性,即使没有与服务器通信。

有了它,您可以使用generateId(obj)set(obj)复制push()

这是 ID 函数

/**
 * Fancy ID generator that creates 20-character string identifiers with the following properties:
 *
 * 1. They're based on timestamp so that they sort *after* any existing ids.
 * 2. They contain 72-bits of random data after the timestamp so that IDs won't collide with other clients' IDs.
 * 3. They sort *lexicographically* (so the timestamp is converted to characters that will sort properly).
 * 4. They're monotonically increasing.  Even if you generate more than one in the same timestamp, the
 *    latter ones will sort after the former ones.  We do this by using the previous random bits
 *    but "incrementing" them by 1 (only in the case of a timestamp collision).
 */
generatePushID = (function() {
  // Modeled after base64 web-safe chars, but ordered by ASCII.
  var PUSH_CHARS = '-0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ_abcdefghijklmnopqrstuvwxyz';

  // Timestamp of last push, used to prevent local collisions if you push twice in one ms.
  var lastPushTime = 0;

  // We generate 72-bits of randomness which get turned into 12 characters and appended to the
  // timestamp to prevent collisions with other clients.  We store the last characters we
  // generated because in the event of a collision, we'll use those same characters except
  // "incremented" by one.
  var lastRandChars = [];

  return function() {
    var now = new Date().getTime();
    var duplicateTime = (now === lastPushTime);
    lastPushTime = now;

    var timeStampChars = new Array(8);
    for (var i = 7; i >= 0; i--) {
      timeStampChars[i] = PUSH_CHARS.charAt(now % 64);
      // NOTE: Can't use << here because javascript will convert to int and lose the upper bits.
      now = Math.floor(now / 64);
    }
    if (now !== 0) throw new Error('We should have converted the entire timestamp.');

    var id = timeStampChars.join('');

    if (!duplicateTime) {
      for (i = 0; i < 12; i++) {
        lastRandChars[i] = Math.floor(Math.random() * 64);
      }
    } else {
      // If the timestamp hasn't changed since last push, use the same random number, except incremented by 1.
      for (i = 11; i >= 0 && lastRandChars[i] === 63; i--) {
        lastRandChars[i] = 0;
      }
      lastRandChars[i]++;
    }
    for (i = 0; i < 12; i++) {
      id += PUSH_CHARS.charAt(lastRandChars[i]);
    }
    if(id.length != 20) throw new Error('Length should be 20.');

    return id;
  };
})();
于 2016-03-19T18:30:06.457 回答
4

您可以使用with.then()之后返回的承诺更新添加 ObjectID 的记录:.push()snapshot.key

const ref = Firebase.database().ref(`/posts`);
ref.push({ title, categories, content, timestamp})
   .then((snapshot) => {
     ref.child(snapshot.key).update({"id": snapshot.key})
   });
于 2018-10-01T16:17:55.460 回答
0

如果您想push()在写入数据库时​​或写入数据库之后获取由 firebase 方法生成的唯一键,而无需再次调用,您可以这样做:

var reference = firebaseDatabase.ref('your/reference').push()

var uniqueKey = reference.key

reference.set("helllooooo")
.then(() => {

console.log(uniqueKey)



// this uniqueKey will be the same key that was just add/saved to your database



// can check your local console and your database, you will see the same key in both firebase and your local console


})
.catch(err =>

console.log(err)


});

push()方法有一个key属性,它提供刚刚生成的密钥,您可以在写入数据库之前、之后或期间使用它。

于 2019-06-09T04:32:35.653 回答
-2

我是怎么做的:

FirebaseDatabase mFirebaseDatabase = FirebaseDatabase.getInstance();
DatabaseReference ref = mFirebaseDatabase.getReference().child("users").child(uid); 

String key = ref.push().getKey(); // this will fetch unique key in advance
ref.child(key).setValue(classObject);

现在您可以保留密钥以供进一步使用..

于 2017-01-19T09:49:35.803 回答