使用ASP.NET管理模板Siminta中包含的引导选项卡时遇到问题。每当我点击下一次点击回发事件时,它将返回活动标签。我尝试了许多解决方案,但徒劳无功。 asp.net bootstrap Keep current active tab after post-back event
<ul class="nav nav-pills">
<li class="active">
<a href="#EmpProfile" data-toggle="tab"><i class="fa fa-home"></i> Profile</a>
</li>
<li>
<a href="#EmpEducation" data-toggle="tab"><i class="fa fa-leanpub"></i> Education</a>
</li>
</ul>
............
<div class="tab-content">
<div class="tab-pane fade in active" id="EmpProfile">
.........
</div
</div>
引导程序中的代码:
+function ($) {
'use strict';
// TAB CLASS DEFINITION
// ====================
var Tab = function (element) {
this.element = $(element)
}
Tab.prototype.show = function () {
var $this = this.element
var $ul = $this.closest('ul:not(.dropdown-menu)')
var selector = $this.data('target')
if (!selector) {
selector = $this.attr('href')
selector = selector && selector.replace(/.*(?=#[^\s]*$)/, '') //strip for ie7
}
if ($this.parent('li').hasClass('active')) return
var previous = $ul.find('.active:last a')[0]
var e = $.Event('show.bs.tab', {
relatedTarget: previous
})
$this.trigger(e)
if (e.isDefaultPrevented()) return
var $target = $(selector)
this.activate($this.parent('li'), $ul)
this.activate($target, $target.parent(), function () {
$this.trigger({
type: 'shown.bs.tab',
relatedTarget: previous
})
})
}
Tab.prototype.activate = function (element, container, callback) {
var $active = container.find('> .active')
var transition = callback
&& $.support.transition
&& $active.hasClass('fade')
function next() {
$active
.removeClass('active')
.find('> .dropdown-menu > .active')
.removeClass('active')
element.addClass('active')
if (transition) {
element[0].offsetWidth // reflow for transition
element.addClass('in')
} else {
element.removeClass('fade')
}
if (element.parent('.dropdown-menu')) {
element.closest('li.dropdown').addClass('active')
}
callback && callback()
}
transition ?
$active
.one($.support.transition.end, next)
.emulateTransitionEnd(150) :
next()
$active.removeClass('in')
}
// TAB PLUGIN DEFINITION
// =====================
var old = $.fn.tab
$.fn.tab = function ( option ) {
return this.each(function () {
var $this = $(this)
var data = $this.data('bs.tab')
if (!data) $this.data('bs.tab', (data = new Tab(this)))
if (typeof option == 'string') data[option]()
})
}
$.fn.tab.Constructor = Tab
// TAB NO CONFLICT
// ===============
$.fn.tab.noConflict = function () {
$.fn.tab = old
return this
}
// TAB DATA-API
// ============
// $(document).on('click.bs.tab.data-api', '[data-toggle="tab"], [data-toggle="pill"]', function (e) {
// e.preventDefault()
// $(this).tab('show')
// })
$('[data-toggle="tab"],[data-toggle="pill"]').click(function () {
$(this).parent().children('.active[data-toggle="tab"],.active[data-toggle="pill"]').removeClass('active');
$(this).addClass('active');
});
}(jQuery);
那么,我该如何解决呢?
答案 0 :(得分:0)
我遵循了这条指令,但它仍然不起作用:Bootstrap Tabs Maintain Selected (Active) Tab on PostBack in ASP.Net,使用相同的bootstrap版本不同之处只是我将DataGrid与Paging一起使用,每当我转到下一个选项卡时,单击DataGrid的页面然后它回到Active Tab(第一个): 添加:
<asp:HiddenField ID="TabName" runat="server" />
然后使用此脚本:
<script type="text/javascript">
$(function () {
var tabName = $("[id*=TabName]").val() != "" ? $("[id*=TabName]").val() : "EmpProfile";
$('#Tabs a[href="#' + tabName + '"]').tab('show');
$("#Tabs a").click(function () {
$("[id*=TabName]").val($(this).attr("href").replace("#", ""));
});
});
</script>
然后,把后面的代码:
protected void Page_Load(object sender, EventArgs e)
{
if (this.IsPostBack)
{
TabName.Value = Request.Form[TabName.UniqueID];
}
}
我也不会忘记添加这个:
protected void grdEmployee_PageIndexChanged(object source, DataGridPageChangedEventArgs e)
{
grdEmployee.CurrentPageIndex = e.NewPageIndex;
TabName.Value = Request.Form[TabName.UniqueID];
BindGrid();
}
}
答案 1 :(得分:0)
<!--Basic Tabs -->
<ul class="nav nav-pills">
<li class="active">
<a href="#EmpProfile" data-toggle="tab"><i class="fa fa-home"></i> Profile</a>
</li>
.............
</ul>
<div class="tab-content">
<div class="tab-pane fade in active" id="EmpProfile">
<asp:UpdatePanel ID="updatePanelEmployee" runat="server">
<ContentTemplate>
<asp:Datagrid ID="grdEmployee" runat="server" ..................
</asp:Datagrid>
</ContentTemplate>
</asp:UpdatePanel>
有效!