我有一个带有一对单选按钮的角度模态,并且根据quicklink
的值,我想要检查某个按钮,但是在用鼠标点击之前不会检查这两个按钮。
<label>
<input name="quicklink" rows="3" type="radio" value="1"
ng-model="entity.quicklink" id="radio-1" />
yes</label>
<label>
<input name="quicklink" rows="3" type="radio" value="0"
ng-model="entity.quicklink" id="radio-2" />
no</label>
$scope.entity = data;
if ($scope.entity['quicklink'] == '1') {
$("#radio-1").prop("checked", true)
}
if ($scope.entity['quicklink'] == '0') {
$("#radio-2").prop("checked", true)
}
答案 0 :(得分:0)
为了最佳实践,我们应该避免直接从Angular使用jQuery调用DOM。
理想情况下,如果数据具有有效的 quicklink 值{quicklink: '0'}
,则无需执行任何操作。
http://plnkr.co/edit/98rog2BMCLXGfNtoWy1o
angular.module("app", [])
.controller("myController", function($scope){
$scope.entity = {quicklink: '0'};
});
答案 1 :(得分:0)
适合我(没有jQuery):
var app = angular.module('app', []);
app.controller('controller', ['$scope', function($scope) {
$scope.entity = { quicklink: 1 };
}]);
&#13;
<script src="https://ajax.googleapis.com/ajax/libs/angularjs/1.2.23/angular.min.js"></script>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<div ng-app="app" ng-controller="controller">
<div class="form-group">
<label>quicklink label</label>
<label class="radio-inline">
<input name="quicklink" rows="3" type="radio" value="1" class="form-control" ng-model="entity.quicklink" id="radio-1"/>
yes</label>
<label class="radio-inline">
<input name="quicklink" rows="3" type="radio" value="0" class="form-control" ng-model="entity.quicklink" id="radio-2"/>
no</label>
</div>
<div>Value of quicklink is: {{entity.quicklink}}</div>
</div>
&#13;