从$ scope中获取选择的选项后,控制台日志上出现了错误消息,如“ TypeError:(intermediate value)(intermediate value).val(...)。prop不是一个函数”。请检查我的脚本出了什么问题:

HTML模板:

<ion-view view-title="Doll List">
    <ion-content class="has-header">
      <div class="list">
        <label class="item item-input item-select">
          <div class="input-label">
           Product
          </div>
          <select name='options2' ng-model="item.doll" ng-options="DOLLtype as DOLLtype.name for DOLLtype in dolltype track by DOLLType.id">
          </select>
        </label>
        </div>
    </ion-content>
  </ion-view>


controller.js

.controller('DollCtrl', ['$scope', '$stateParams', '$location', '$http', '$window', '$q', '$localStorage', '$rootScope', '$filter', 'DOLLTYPE','$timeout', function($scope, $stateParams, $location, $http, $window, $q, $localStorage, $rootScope,  $filter, DOLLTYPE, $timeout) {

 $scope.item = [];
 $scope.item = {
 id: $stateParams.id
 };

 $scope.dolltype = [];
 DOLLTYPE.get($scope.item.id).then(function(itemtype) {
 $scope.dolltype.push({id:itemtype.id, name:itemtype.name})
 $scope.item.doll =  $scope.dolltype[0];
 console.log($scope.item.doll);
 })
 DOLLTYPE.all().then(function(dolltype){
 $scope.dolltype=dolltype;
 console.log($scope.dolltype);
 })


     }}

 ])


我该如何解决?

注意:DOLLTYPE是用于使用cordova s​​qlite插件查询数据的services.js的示例名称

最佳答案

有时,当您忘记分号时,可以触发此操作。
您似乎忘了放分号:

$ scope.dolltype.push({id:itemtype.id,name:itemtype.name});
DOLLTYPE.get(...)。then(...);
和DOLLTYPE.all()。then(...);
缩小时有时会破裂

关于javascript - TypeError :(中间值)(中间值).val(…).prop不是函数( ionic ),我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/31096897/

10-11 13:58