javascript - AngularJS 和 Firebase '$timeout is not defined'

Question

大家好，我刚刚学习了 angular JS 和 Firebase，出于某种原因，当我尝试在以下代码中调用 $timeout 函数时，似乎遇到了引用错误:

'use strict';

/**
 * @ngdoc function
 * @name drivenApp.controller:MainCtrl
 * @description
 * # MainCtrl
 * Controller of the drivenApp
 */
angular.module('drivenApp')
  .controller('MainCtrl', function ($scope) {
    var rootRef = new Firebase('https://vivid-torch-5432.firebaseio.com/');
    var childRef = rootRef.child('message');

    childRef.on('value', function(snapshot){
      $timeout(function() {
        var snapshotVal = snapshot.val();
        console.log(snapshotVal);
        $scope.message = snapshot.val();
      });
    });
  });

我得到这个确切的错误:

Uncaught ReferenceError: $timeout is not defined(anonymous function) @ main.js:16(anonymous function) @ firebase.js:202gc @ firebase.js:52cc @ firebase.js:30dc @ firebase.js:29h.Kb @ firebase.js:221h.Ld @ firebase.js:189Fh.Ld @ firebase.js:179(anonymous function) @ firebase.js:177zh @ firebase.js:171La.onmessage @ firebase.js:170

知道为什么会这样吗？谢谢，尼克

Answer 1

您需要声明 $timeout 才能像这样使用它:

angular.module('drivenApp')
  .controller('MainCtrl', function ($scope, $timeout) {
    var rootRef = new Firebase('https://vivid-torch-5432.firebaseio.com/');
    var childRef = rootRef.child('message');

    childRef.on('value', function(snapshot){
      $timeout(function() {
        var snapshotVal = snapshot.val();
        console.log(snapshotVal);
        $scope.message = snapshot.val();
      });
    });
  });