0
<button class="Button" onclick="NewTicket()">New Ticket</button>
<script language=javascript>
function OpenPopup(url, winname, features)
{
if(winname==''){
window.open(url, winname, features, false);
return;
}
if (!findWindow(url, winname, features))
{
var handle = window.open(url, winname, features, false);
if (handle != null) handle.focus();
}
}
function findWindow(url, winname, features)
{
var handle = window.open('', winname, features, false);
if (handle != null)
{
if ((handle.location != 'about:blank') && (handle.location != ''))
{
handle.focus();
return true;
}
}
return false;
}
function NewTicket(){
OpenPopup('../Ticket/New.php', 'NewTicket', 'channelmode=0, directories=0, fullscreen=0, width=430, height=360, location=0, menubar=0, resizable=0, scrollbars=1, status=0, titlebar=1, toolbar=0', false);
}
</script>
用此代码打开弹出窗体。这个没问题。我想再次发布表单(../Ticket/New.php)回到这个页面。我怎样才能做到这一点?javascript窗体发布到主窗口
而如何关闭弹出价值? – milesh 2013-02-13 19:20:35
我假设你想在提交表单后关闭它,所以它将是在弹出代码中添加。 – 2013-02-13 19:56:58
我在弹出的提交按钮上使用了onclick =“self.close()”。谢谢。 – milesh 2013-02-14 08:58:30