我有一个PHP程序,允许用户弹出消息,以供用户确认。如下所示是我使用的链接。Javascript/PHP弹出页面问题
从inbox.php:
echo "<a href='inbox.php' onclick=\"popup('acknowledge.php?id=$id')\"><font size=1px color=maroon>acknowledge</font></a></td>";
从acknowledge.php:
if ($_POST['no']) {
header("location: inbox.php");
}
?>
<body bgcolor=skyblue>
<center><form name=form1 method=post>
<b><u> Acknowledge Message </u></b><br><br>
Are you sure yout want to acknowledge this message?<br><br>
<input type=button name=yes value="Yes">
<input type=submit name="no" value="No">
</form>
</center>
</body>
的问题是,每次我点击 “否”,去从前页返回。它将页面大小设置为与弹出页面相同。它也变小了。有什么问题?非常感谢答案。
这里是弹出()的代码:
<script language="JavaScript" type="text/JavaScript">
<!-- function popup(url) {
var width = 500;
var height = 135;
var left = (screen.width - width)/2;
var top = (screen.height - height)/2;
var params = 'width='+width+', height='+height;
params += ', top='+top+', left='+left;
params += ', directories=no';
params += ', location=no';
params += ', menubar=no';
params += ', resizable=no';
params += ', scrollbars=no';
params += ', status=no';
params += ', toolbar=no';
newwin=window.open(url,'windowname5', params);
if (window.focus) {newwin.focus()}
return false;
}
// -->
</script>
函数名称为'popup'的代码是什么? – Alsciende 2010-03-26 09:47:43
嗨Alsciende,我把我的代码弹出。希望你能帮我弄明白。 :) – Suezy 2010-03-26 10:10:37