Question

在这里摆弄http://jsfiddle.net/prantikv/1nvdzv24/9/

我有一些不均匀的数据，如此

[{
   "fname": "Tonja", //common
    "lname": "Mize",
    "tel": "(963)784-1098",
    "address": "3999 Quis Ln",
    "city": "Sebring",
    "state": "MI",
    "zip": 76593
},
{
    "fname": "Stella", //common
    "Othername": "Lester",
    "mobile": "(936)898-2886"
}];

请注意， fname 属性在两个对象之间是通用的

所以当我这样做时

  <li ng-repeat="(key,val) in populationList | filter:name"> 
 {{ val.**fname**}} 
</li>

我确实得到了fname，但数据不均匀，所以我无法弄清楚如何遍历每个对象。对象的长度也是不同的。

我想要做的是通过选择列表过滤数据

<select ng-model="name">
    <option value="Tonja" selected="Tonja">Tonja</option>
    <option value="Stella">Stella</option>
</select>

但我无法找到显示对象不匹配属性的方法有没有办法动态获得子数据上的所有关键：值对？

Answer 1

WORKING DEMO

你的Html，

<div ng-app='app'>
 <div ng-controller="DemoCtrl">
    <select ng-options="item.fname for item in populationList | fieldList:'fname'" ng-model="myItem" ng-change="changeSelection(myItem)">
    </select>

   <li ng-repeat="key in availableKeys"> 
         {{selectedObject[key]}} 
    </li>
 </div>
</div>

JS

angular.module('filters',[]).
  filter('fieldList', function() {
      return function(populationList, parameter) {
        var filteredArray = [];
        angular.forEach(populationList, function(value, index) {
             if(value.hasOwnProperty(parameter)) {
                 filteredArray.push(value);
             }
        });
        return filteredArray;
      };
    });


angular.module('app',['filters'])
  .controller('DemoCtrl', function($scope) {
  $scope.changeSelection = function(item) {
    $scope.selectedObject = item;
    $scope.availableKeys = Object.keys($scope.selectedObject);
    };
  $scope.populationList = [{
    "fname": "Tonja", //common
    "lname": "Mize",
    "tel": "(963)784-1098",
    "address": "3999 Quis Ln",
    "city": "Sebring",
    "state": "MI",
    "zip": 76593
     },
     {
    "fname": "Stella", //common
    "Othername": "Lester",
    "mobile": "(936)898-2886"
   }];
});

使用ng-repeat解析不均匀的数据/ json对象

1 个答案: