更改 ng-model 时显示按钮 AngularJS

Question

我只想在 modelCheck 发生变化时显示按钮 post。我的代码： HTML :

  <table>
  <tr>
    <th>name</th>
     <th><button class="btn" ng-if="modelCheck === {}" ng-click="post()">post</button></th>
      <th><button class="btn disabled" ng-if="modelCheck !== {}" ng-click="post()">post</button></th>
  </tr>
  <tr ng-repeat="x in messages">
    <td>{{x.name}}</td>
    <td><input type="checkbox" ng-model="modelCheck[x.name]"/></td>
  </tr>
</table>

JS:

  $scope.modelCheck = {};

  $scope.messages = 
    [
        {
            "name": "eva",
        },
        {
            "name": "ben",
    },
 ];

$scope.post = function(){
    var data = [];
       for (var k in $scope.modelCheck) {
           if ($scope.modelCheck.hasOwnProperty(k) && $scope.modelCheck[k]) {
            data.push({'name': k});
           }
       }
       console.log(data);
       // do with data as you wish from here
    };

笨蛋：http://next.plnkr.co/edit/xQ3AGFW03qGWKyxI 我的代码不起作用。提前感谢您的回答和帮助！

Answer 1

使用此条件检查空对象

 <th><button class="btn" ng-if="(modelCheck | json) != ({} | json)" ng-click="post()">post</button></th>

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