如何在angularjs对象属性中获取值

Question

html ##如何获取控制器

中field.jobtype的值

•     var app=angular.module('myapp',['ui.bootstrap','ui.select']);
      app.controller('mycontroller',function($scope){
          $scope.onStateSelected = function (selectedItem) {
              $scope.selectedState=selectedItem.name;
          }
          $scope.oncarearchange=function(item){
              $scope.selectedLocation=item.name;
          }
  
     var state=[{"name":"entry"},{"name":"intermediate"},{"name":"senier"},{"name":"teamlead"}];
    var carearlevel=[{"name":"entry"},{"name":"intermediate"},{"name":"senier"}{"name":"teamlead"}];
  
          $scope.states=state;
          $scope.carear=carearlevel;
  
      });

  <ui-select ng-model="field.jobtype"  theme="select2"  ng-disabled="disabled" class="form-control input-md c-square" on-select="onStateSelected($item)">
                                                         
    <ui-select-match placeholder="Select a jobtype...">
      
  {{$select.selected.name}}
    
    </ui-select-match>
    
  <ui-select-choices repeat="state.name as state in states | filter: $select.search">
    
  {{state.name}}
    
  </ui-select-choices>
    
  </ui-select>

• 这里将数组状态的ui-select显示值分配给$ scope.states，但是当尝试显示字段时，控制器中的jobtype显示错误，工作类型未定义

Answer 1

当我使用$ scope.field时它工作。 ------在我的angularjs控制器中