我创建了一个提交表单的弹出窗口,提交后必须关闭,但我首先想要在不同的页面中弹出窗口中的信息,而不显示其他页面。我该怎么办?
我的popupwindow包含此代码
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN"
"http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<script type="text/javascript">
function closeSelf(){
self.close();
return true;
}
</script>
<title>Add Activity</title>
</head>
<body>
<form action="./addact.php" method="post" onsubmit ="return closeSelf()">
<table width="500" border="1"><br/>
<tr>
<td>Activity Name</td>
<td>Assigned Person</td>
<td>Deadline</td>
</tr>
<tr>
<td> <input name="activities" type="text" size="40%"/></td>
<td><input name="name" type="text" size="40%"/></td>
<td><input type="date" name="deadline" size="20%"/></td>
</tr>
</table>
<input type="submit" name = "saved" id="saved"/>
</form>
</body>
</html>
我的其他页面包含此
<?php session_start(); ?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN"
"http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Untitled Document</title>
</head>
<?php
include('config.php');
$actname= $_POST['activities'];
$assigned = $_POST['name'];
$deadline = $_POST ['deadline'];
$sql = "INSERT INTO ".$_SESSION['pname']."_activities
(actname, assigned, deadline)
VALUES
('$actname', '$assigned', '$deadline')
";
$query = mysql_query($sql);
echo $_SESSION['pname'];
?>
<body>
</body>
</html>
答案 0 :(得分:0)
您应该在返回响应时关闭窗口,而不是在提交页面时关闭窗口。正如@Marc B指出的那样,你有严重的安全问题!