Firebase.once 不是运行页面未首次加载时的回调

Question

我的 Ionic 应用程序有一个视图，它只显示一个 select 下拉菜单，然后提示用户输入一个数字。然后将该号码保存到该特定用户的 object 英里数 object，键是 selected 选项，值是输入的号码。这是控制器的代码并查看。

控制器：

// Get who is logged in
$scope.user = facebook.get();
// Array of airlines
var airRef = ref.child("airlines");
$scope.airlines = $firebaseArray(airRef);
console.log($scope.airlines);
$scope.selectedAir = {};
$scope.miles = {};
// Add program to user
$scope.addProgram = function () {
  if(jQuery.isEmptyObject($scope.user)) {
      var authData = ref.getAuth();
      var theUser = ref.child("users").child(authData.uid);
      var selected = {};

      theUser.child("miles").child($scope.selectedAir.name.$id).once("value", function(snapshot) {
          console.log("apples");

          var exist = snapshot.exists();
          if(!exist) {
              selected[$scope.selectedAir.name.$id] = $scope.miles.num;
              theUser.child("miles").update(selected);
              $state.go("app.saved");
          } else {
              var alertPopup = $ionicPopup.alert({
                title: 'Oops!',
                template: "You already created this airline! Go to the 'Add Ticket' page to add more points."
              });
              alertPopup.then(function(res) {
                console.log("You already created this airline! Go to the 'Add Ticket' page to add more points.");
              });
          }
      })

  } else {
      var theUser = ref.child("users").child($scope.user.id);
      var selected = {};

      theUser.child("miles").child($scope.selectedAir.name.$id).once("value", function(snapshot) {
          var exist = snapshot.exists();
          if(!exist) {
              selected[$scope.selectedAir.name.$id] = $scope.miles.num;
              theUser.child("miles").update(selected);
              $state.go("app.saved");
          } else {
              var alertPopup = $ionicPopup.alert({
                title: 'Oops!',
                template: "You already created this airline! Go to the 'Add Ticket' page to add more points."
              });
              alertPopup.then(function(res) {
                console.log("You already created this airline! Go to the 'Add Ticket' page to add more points.");
              });
          }
      })

  }
}

查看：

<ion-view view-title="Add a Program">
<ion-nav-buttons side="left">
<button class="button back-button buttons button-clear header-item" ng-click="goBack()">
  <i class="icon ion-ios-arrow-back"></i>
</button>
  </ion-nav-buttons>
<ion-content class="padding">
      <div class="list marginZeroA paddingTL width80">
          <p class="mainText"> </p>
          <div class="list">
              <label class="item item-input item-select">
                  <div class="input-label">

                  </div>
                  <select ng-model="selectedAir.name" ng-options="airline.$id for airline in airlines">

                </select>
            </label>
          </div>
          <label class="item item-input loginInputs">
              <input style="color:black"ng-model="miles.num" class="milesPoints" type="number" min="0" max="10000000"
                placeholder="30000" ng-click="revealInput(1)">
          </label>
          <button class="button button-outline button-calm loginButton
            saveTicket" ng-click="addProgram()" disabled> Add </button>
    </div>
 </ion-content>
 </ion-view>

当我加载到该特定视图时，该应用程序运行良好，但如果我像通常那样从登录页面开始，它将无法运行。如果我在这里提供了错误的 child id：

.child($scope.selectedAir.name.$id).once("value",

我会收到一个控制台错误，告诉我它错了，所以我知道它正在尝试运行，但是当传递正确的 child id 时，什么也没有发生。没有回调，没有错误捕获。谢谢

Answer 1

找到我的问题：

当 ref.once('value', ... ) 或 ref.update( ... ) 等 Firebase 函数既不执行任何内部命令也不抛出错误（在错误回调中捕获）时，很可能由于 Firebase.goOffline() 调用，您的应用程序未连接到 Firebase。这是我的情况，我在另一个视图中调用了它，当我切换视图时，从未调用 Firebase.goOnline() 来重新打开连接。

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