我有以下jQuery代码显示模式弹出窗口:-
$(document).ready(function () {
$(function () {
$.ajaxSetup({ cache: false });
$(document).on('click', 'button[data-modal]', function (e) {
$('#myModalContent').css({ "margin": "5px", "max-height": screen.height * .82, "max-width": screen.height * .82, "overflow-y": "auto" }).load($(this).attr("data-url"), function () {
$('#myModal').modal({
height: 1000,
width: 2200,
resizable: true,
keyboard: true,
backdrop: 'static',
draggable: true
}, 'show');
});
return false;
});
});
});
和以下HTML:-
<div id='myModal' class='modal fade in'>
<div class="modal-dialog">
<div class="modal-content">
<div id='myModalContent'></div>
</div>
</div>
</div>
当前,当我在模式弹出窗口之外单击时,模式弹出窗口将不会被隐藏/关闭..所以我可以将此行为强加给我的jQuery模式弹出窗口吗?
答案 0 :(得分:2)
您需要删除backdrop: 'static',