对话框隐藏（）和显示（）——Jquery？
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Dialog box hide() and show() -- Jquery?
提问by tíń?
Iam using 3 dialogs in my page for 3 diff purpose.
我在我的页面中使用 3 个对话框用于 3 个差异目的。
Iam creating dialog as
我正在创建对话框作为
$(".dialog").dialog({
height: 238,
width: 465,
resizable: false,
title: "Edit"
});
After my action done on dialog iam closing dialog as
在我对对话框完成操作后，我关闭对话框为
$(".ui-dialog").hide();
When i hide this way dialog is not getting opening for 2nd time, So tried showing dialog starting of function like
当我以这种方式隐藏对话框时，第二次没有打开，所以尝试显示对话框启动功能，如
$(".ui-dialog").show();
My problem started here....
我的问题从这里开始......
When i show dialog, Dialogs are getting opened multiple times, STill the first openined dialog is getting Overlapped wth the second dialog,
当我显示对话框时，对话框被多次打开，但第一个打开的对话框与第二个对话框重叠，
Is there any proper way to hide and Show dialog without overlapping or in clean way.
是否有任何正确的方法可以在不重叠或干净的情况下隐藏和显示对话框。
回答by Arun P Johny
回答by Tony Brix
If you just want to hide/show a dialog without closing it you can use
如果您只想隐藏/显示对话框而不关闭它，您可以使用
$(".dialog").parent().hide()
$(".dialog").parent().show()
回答by Harun ERGUL
To show/hidethe dialog we need to use open/closemethods
要显示/隐藏对话框，我们需要使用打开/关闭方法
//To close the dialog use 'close' method.
//It will hide the dialog. Your html and data still exist in the DOM
$("#my_dialog_id").dialog('close');
//Show closed dialog again if it is still exists and not destroyed the
$("#my_dialog_id").dialog('open');
//This method totally destroy your dialog. Element will be returned to pre-init state
$("#my_dialog_id").dialog('destroy');
//To check if the dialog is open or not
var isOpen = $( "#my_dialog_id" ).dialog( "isOpen" );
