我无法获得除第一个之外的角度选项卡在加载时处于活动状态。我正在传递一个url参数,例如?location = Tucson来做出这个决定。我希望第二个选项卡在加载时显示为活动状态。奇怪的是,第一个选项卡被自动设置为活动,也许这是默认情况。 我在用 *棱角带 * @version v2.1.4 - 2014-11-26
如何通过控制器将第二个标签设置为活动状态? 这是我的设置
<section class="mainbar" data-ng-controller="contactus as vm">
<div>
<tabset>
<tab class="tab-pane" id="Tampa" heading="Tampa">
<tab-heading>Tampa</tab-heading>
<div class="actionBtn">
<a id="tpabtn" class="btn btn-default ldbtn request" data-ng-click="vm.getcontactformF()">
<div><span class="glyphicon glyphicon-envelope"></span></div>
<div>
<div class="title">Have Questions?</div>
<div>Send us a message</div>
</div>
</a>
</div>
</tab>
<tab class="tab-pane" id="Tucson" active="active">
<tab-heading>Tucson</tab-heading>
<div class="actionBtn">
<a href="tel:8885210206" class="btn btn-default ldbtn">
<div><span class="glyphicon glyphicon-earphone"></span></div>
<div>
<div class="title">RV Service</div>
<div class="phone">888.521.0206</div>
</div>
</a>
</div>
</tab>
</tabset>
</div>
<script type="text/ng-template" id="_contactfservice.html">
<div data-ng-include data-src="'/app/form/contactusform.html'"></div>
</script>
</section>
JS
(function () {
'use strict';
var controllerId = "contactus";
angular
.module('app')
.controller(controllerId, ['$location', '$window', '$rootScope', '$scope', '$sce', '$q', 'common', 'config', 'bootstrap.dialog', 'datacontext', contactus]);
function contactus($location, $window, $rootScope, $scope, $sce, $q, common, config, bsDialog, datacontext) {
/* jshint validthis:true */
var vm = this;
vm.highlightLocation = highlightLocation;
activate();
function activate() {
common.activateController([getLocationHours()], controllerId).then(function () {
common.postLoad();
log('Activated contact us View test test');
$window.ga('send', 'event', $location.path(), 'form-viewed');
highlightLocation();
});
}
function highlightLocation() {
var location = common.getQueryStringParameter('location', document.location.href);
console.log(location);
if (location == "Tucson") {
$scope.tabs = [{active: false}, {active: true}];
}
}
}
})();
答案 0 :(得分:0)
bootstrap ui的一些奇怪的问题,这是有效的。
$scope.data = {};
$timeout(function () {
$scope.vm.data.static2 = true;
}, 0)