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>
用此代碼打開彈出窗體。這個沒問題。我想再次發佈表單(../Ticket/New.php)回到這個頁面。我怎樣才能做到這一點?javascript窗體發佈到主窗口
而如何關閉彈出價值? – milesh 2013-02-13 19:20:35
我假設你想在提交表單後關閉它,所以它將是在彈出代碼中添加。 – 2013-02-13 19:56:58
我在彈出的提交按鈕上使用了onclick =「self.close()」。謝謝。 – milesh 2013-02-14 08:58:30