【发布时间】:2015-11-06 06:38:14
【问题描述】:
如何在控制器中获取 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 ui-select 显示分配给 $scope.states 的数组状态中的值,但是当尝试显示字段时,控制器中的作业类型显示错误作业类型未定义。
【问题讨论】:
-
我尝试在提交按钮时将值保存到字段中,但字段.jobtype 似乎未定义
-
假设您的 html 代码包含在具有
ng-controller="mycontroller"的元素中,则所选值将是$scope.field.jobtype -
如果我使用 $scope.field.jobtype 是否有效
标签: angularjs html jquery-select2 ui-select