我收到错误消息,例如“TypeError :(中间值)(中间值).val(...)。在获取选定选项后,prop不是控制台日志中的函数” $范围。请查看我的脚本中的错误:
HTML模板:
<ion-view view-title="Doll List">
<ion-content class="has-header">
<div class="list">
<label class="item item-input item-select">
<div class="input-label">
Product
</div>
<select name='options2' ng-model="item.doll" ng-options="DOLLtype as DOLLtype.name for DOLLtype in dolltype track by DOLLType.id">
</select>
</label>
</div>
</ion-content>
</ion-view>
controller.js
.controller('DollCtrl', ['$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是使用cordova sqlite插件查询数据的services.js的示例名称
答案 0 :(得分:1)
有时,当您忘记分号时可以触发此操作。 你似乎忘记了分号:
$ scope.dolltype.push({id:itemtype.id,name:itemtype.name}); &lt; ---
DOLLTYPE.get(...)。then(...); &lt; ---
和DOLLTYPE.all()。然后(...); &lt; ---
当你缩小时,这有时会破裂