3
當鼠標在glyphicon標誌上盤旋時,顯示一些信息。代碼現在正在工作,但僅適用於add content
。我想使用相同的JQuery代碼進行編輯,刪除..ect。我不想一次又一次地重複Jquery代碼。當鼠標懸停在glyphicon glyphicon-info-sign上時顯示數據內容
如何做到這一點?
HTML:
<input name="add" value="1" type="checkbox" />
Add Contents
<a href="#" title="Add" data-content="1- Problem Report. 2- Maintenance Report. 3- Expeses Report. 4- Add Contract. 5-Items">
<span class="glyphicon glyphicon-info-sign" id="add_inf"></span>
</a>
<input name="edit" value="1" type="checkbox" />
Edit Contents
<a href="#" title="Edit" data-content="1- Problem Report. 2- Maintenance Report. 3- Expeses Report. 4- Contract.">
<span class="glyphicon glyphicon-info-sign" id="edit_inf"></span>
</a>
<input name="delete" value="1" type="checkbox" />
Delete Content
<a href="#" title="Delete" data-content="1- Problem Report. 2- Maintenance Report. 3- Expeses Report. 4- Contract. 5-Items">
<span class="glyphicon glyphicon-info-sign" id="delete_inf"></span>
</a>
JQuery的:
var $btn2 = $('#add_inf');
$btn2.data('state', 'hover');
var enterShow = function() {
if ($btn2.data('state') === 'hover') {
$btn2.popover('show');
}
};
var exitHide = function() {
if ($btn2.data('state') === 'hover') {
$btn2.popover('hide');
}
};
var clickToggle = function() {
if ($btn2.data('state') === 'hover') {
$btn2.data('state', 'pinned');
} else {
$btn2.data('state', 'hover')
$btn.popover('hover');
}
};
$btn2.popover({ trigger: 'manual' })
.on('mouseenter', enterShow)
.on('mouseleave', exitHide)
.on('click', clickToggle);