TypeError: Cannot read property '$filter' of undefined

Question

当用户单击 link 时，我在 UI 上显示 json 对象，下面是它显示的异常

TypeError: Cannot read property '$filter' of undefined

演示：http://plnkr.co/edit/5NOBQ3rzE3EFwTnuLbXG?p=preview

示例 js 代码：

app.controller('BaseCtrl', ['$scope','$http','$filter', function($scope,$http,$filter) {

$scope.myList = [{"sId":100,"spread":"21x","owner":"Michael","labels":"deffered incomplete"},
{"sId":101,"spread":"34","owner":"Steve","labels":"complete"},
{"sId":102,"spread":"90s","owner":"John","labels":"tested"},
{"sId":103,"spread":"332","owner":"Dex","labels":"complete deffered"}  
];

//$scope.myListObj;
$scope.showList = function(myListObj){
  console.log("myListObj " + JSON.stringify(myListObj));
  //  $scope.defferedList = $scope.myListObj.filter(function( obj ) {
    $scope.defferedList =  JSON.stringify($scope.myListObj).filter(function( obj ) {
      console.log(obj.labels.includes('deffered'));
      return obj.labels.includes('deffered');
}); 
console.log("defferedList :: " +  $scope.defferedList);
}
}]);

我在 controller.Any 输入中包含了 '$http','$filter'？

Answer 1

你的问题不在于 $filter 我看到的错误是

Cannot read property 'filter' of undefined

这是指 javascript 方法 filter 而不是 AngularJS 的“$filter”。 Filter 是一种可以在数组上调用的 javascript 方法。似乎也没有必要在这里做 JSON.stringify 。试试这个

$scope.defferedList =  $scope.myListObj.filter(function( obj ) {
    console.log(obj.labels.includes('deffered'));
    return obj.labels.includes('deffered');
});

Answer 2

我做了一个快速的改变，因为我认为这就是你真正想要的。

基本上，您没有使用 $filter，而是试图过滤其原型中没有该方法的类型。如果您不进行字符串化，它将按预期工作。

var app = angular.module('myApp', []);
app.controller('BaseCtrl', ['$scope','$http', function($scope,$http) {
  
  $scope.myList = [{"sId":100,"spread":"21x","owner":"Michael","labels":"deffered incomplete"},
{"sId":101,"spread":"34","owner":"Steve","labels":"complete"},
{"sId":102,"spread":"90s","owner":"John","labels":"tested"},
{"sId":103,"spread":"332","owner":"Dex","labels":"complete deffered"}  
];

//$scope.myListObj;
$scope.showList = function(myListObj){
  console.log("myListObj " + JSON.stringify(myListObj));
  //  $scope.defferedList = $scope.myListObj.filter(function( obj ) {
    $scope.defferedList =  myListObj.filter(function( obj ) {
   console.log(obj.labels && obj.labels.includes('deffered'));
      return obj.labels && obj.labels.includes('deffered');
}); 
console.log("defferedList :: ", $scope.defferedList);
}
}]);

<!DOCTYPE html>
<html>

<head>
  <link rel="stylesheet" href="https://maxcdn.bootstrapcdn.com/bootstrap/3.3.6/css/bootstrap.min.css">
  <script src="//ajax.googleapis.com/ajax/libs/angularjs/1.4.8/angular.min.js"></script>
 
  <script src="script.js"></script>
  <style>
    .my-col-50{float:left;}
  </style>
</head>

<body ng-app="myApp" ng-controller="BaseCtrl">
  <div class="container">
   
    <a href="javascript:void(0)" ng-click="showList(myList)"> 
    Click here</a>
 </div>
</body>
</html>

Answer 3

这可以通过如下编写 showList 方法来纠正

  $scope.showList = function(myListObj) {
  console.log("myListObj " + JSON.stringify(myListObj));
  $scope.defferedList = myListObj.filter(function(obj) {
      console.log(obj.labels.includes('deffered'));
      return obj.labels.includes('deffered');
    });
    console.log("defferedList :: " + $scope.defferedList);
  }

该错误基本上是因为您试图在 $scope.myListObj 的字符串化版本而不是实际数组上应用适用于数组类型的过滤操作。

此外，由于您正在使用数组对象的过滤方法，因此您不需要“$filter”作为模块的依赖项。

TypeError: Cannot read property '$filter' of undefined

TypeError: Cannot read property '$filter' of undefined

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angularjs

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