当用户在输入姓名和年龄后提交时，代码不起作用。页面应提交到同一页面，HTML 表单也应与下面的结果一起显示。请帮帮我！！
<html>
<head>
<title>My first PHP page</title>
</head>
<body>
<form action="<?php echo $_SERVER['PHP_SELF']; ?>" method="post"> <!-- $_SERVER['PHP_SELF'] array -->
Name: <input type="text" name="name"/>
Age: <input type="text" name="age"/>
<input type="submit"/>
</form>
</body>
</html><?php
/* because both the HTML form data and the PHP processing code are in the same script, you will need a conditional check to
see if the form has been submitted */
if ( isset($_POST['submit']) ) { //was the form submitted?
print "Raveen";
//echo "Welcome ". $_POST["name"] . "<br>";
//echo "You are $_POST["age"] years old<br>";
//echo "The path to this file is: $_SERVER[PHP_SELF]";
}
?>
你需要给名字才能得到帖子名称
<input name="yourformname" ...
之后，您可以检查帖子名称
like: if ( isset($_POST['yourformname']) ) {
一定是这样的
<input name="submit" type="submit" />
还有另一种方法可以了解如何获得帖子。 外汇 在您的表单中创建一个隐藏的输入，例如：
<html>
<head>
<title>My first PHP page</title>
</head>
<body>
<form action="<?php echo $_SERVER['PHP_SELF']; ?>" method="post"> <!-- $_SERVER['PHP_SELF'] array -->
Name: <input type="text" name="name"/>
Age: <input type="text" name="age"/>
<input type="hidden" name="gotit"/>
<input type="submit"/>
</form>
</body>
</html>
之后，您可以检查
<?php
/* because both the HTML form data and the PHP processing code are in the same script, you will need a conditional check to
see if the form has been submitted */
if ( isset($_POST['gotit']) ) { //was the form submitted?
print "Raveen";
//echo "Welcome ". $_POST["name"] . "<br>";
//echo "You are $_POST["age"] years old<br>";
//echo "The path to this file is: $_SERVER[PHP_SELF]";
}
?>