基本上点击提交按钮后,我想弹出一个弹出框,表示成功或失败,然后单击确定确认消息.目前我正在弹出一个弹出框“undefined”,然后是失败的消息弹出框.请帮助！
这是脚本
<?PHP
include ('config.PHP');
if (isset($_POST['name'])) {
$name = "name";
$query = "INSERT INTO pop ('id','name') VALUES ('','$name')";
$result = MysqL_query($query,$cn);
if ($result) {
echo "<script type='text/javascript'>alert('submitted successfully!')</script>";
}
else
{
echo "<script type='text/javascript'>alert('Failed!')</script>";
}
}
?>
<html>
<head>
</head>
<body>
<form action="" method="post">
Name:<input type="text" id="name" name="name"/>
<input type="submit" value="submit" name="submit" onclick="alert();"/>
</form>
</body>
你正在回应HTML的body标签之外.
把你的回声放在那里,你应该没事.
把你的回声放在那里,你应该没事.
另外,从你的提交中删除onclick =“alert()”.这是您的第一个未定义的消息的原因.
<?PHP
$posted = false;
if( $_POST ) {
$posted = true;
// Database stuff here...
// $result = MysqL_query( ... )
$result = $_POST['name'] == "danny"; // Dummy result
}
?>
<html>
<head></head>
<body>
<?PHP
if( $posted ) {
if( $result )
echo "<script type='text/javascript'>alert('submitted successfully!')</script>";
else
echo "<script type='text/javascript'>alert('Failed!')</script>";
}
?>
<form action="" method="post">
Name:<input type="text" id="name" name="name"/>
<input type="submit" value="submit" name="submit"/>
</form>
</body>
</html>