不确定问题出在哪里。返回时表单变量的值为"1"，而不是用户放入表单的值。它从哪里得到"1"？我该如何修复它？
<html>
<head>
<title>Feedback Form</title>
<!-- Modified by: Student Name -->
<!-- The page should accept user input in form values, then, after the form
is submitted, hide the form and reveal a confirmation message using
the data entered into the form elements. -->
</head>
<body>
<h1 align="center">We Need You!</h1>
<h2 align="center">Please provide us with your valuable feedback!</h2>
<hr>
<?php
if (!(isset($myName) || isset($myAge) || isset($myFav) || isset($myComments)))
{
$myName = "anonymous";
$myAge = "unspecified";
$myFav = "unspecified";
$myQuestion = "unspecified";
}
$mySubmit = isset($_POST['btnSubmit']);
?>
<form name="frmFeedback" id="frmFeedback" action="sendFeedback.php" method="post"
<?php if ($mySubmit == "Send Feedback!") { echo ' style="display: none"'; } ?>>
Name: <input type="text" name="txtName">
<br>
<br>
Age: <select name="mnuAge">
<option value="youth">Youth</option>
<option value="teen">Teen</option>
<option value="adult">Adult</option>
<option value="senior">Senior</option>
</select>
<br>
<br>
What was your favorite page?
<br>
<input type="radio" name="radFav" value="ASP tutorial">ASP Tutorial
<br>
<input type="radio" name="radFav" value="JavaScript tutorial">JavaScript Tutorial
<br>
<input type="radio" name="radFav" value="PHP tutorial"> PHP Tutorial
<br>
<br>
Which pages did you visit?
<br>
<input type="checkbox" name="chkView[]" value="ASP tutorial">ASP Tutorial
<br>
<input type="checkbox" name="chkView[]" value="JavaScript tutorial">JavaScript Tutorial
<br>
<input type="checkbox" name="chkView[]" value="PHP tutorial"> PHP Tutorial
<br>
<br>
Do you have any additional scripting questions?
<br>
<textarea name="txaQuestions" wrap="soft" cols="50" rows="10">
</textarea>
<br>
<br>
<input type="submit" name="btnSubmit" value="Send Feedback!">
</form>
<?php
//Once the form elements have been filled in, extract data from form and store in
//variables
$myName = $_POST['txtName'];
$myAge = $_POST['mnuAge'];
$myFav = $_POST['rdFav'];
$myQuestion = $_POST['txaQuestions'];
if ($mySubmit == "Send Feedback!")
{
//hide form
//$myFormDisp = "none";
//display message
print("<h3 align='center'>Thank you!!</h3>");
print("Hello, ".$myName."!");
print("Thank you very much for your feedback on our tutorial site.");
print("The ".$myAge." age group is one of our most critical market segments,")
print("so we really appreciate the time you took to fill out our form. ");
print("Active web visitors like yourself are what make these pages possible. ");
print("We are very glad you enjoyed the ".$myFav." page.");
if (isset($_POST['chkView']))
{
print(", and hope that you found the other pages you viewed (");
foreach($_POST['chkView'] as $myView)
{
print("".$myView.", ");
}
print("etc.) to be just as helpful.");
}
else
{
print(". The next time you visit we hope you have a chance to view");
print("our other tutorials also.</p>");
}
print("<p>We will respond to your question: '"".$myQuestion."'" ");
print("just as soon as we can</p>");
print("<h3 align='center' Thanks for stopping by!</h3>");
}
else
{
//set form to display
//$myFormDisp = "block";
}
?>
</body>
</html>
isset返回一个布尔值（在字符串中表示为1或0），如果设置了变量，则返回1（true），如果没有设置，则返回0（false）。
因此，当你这样做时：
//Once the form elements have been filled in, extract data from form and store in
//variables
$myName = isset($_POST['txtName']);
$myAge = isset($_POST['mnuAge']);
$myFav = isset($_POST['rdFav']);
$myQuestion = isset($_POST['txaQuestions']);
如果已设置所有变量，则将它们设置为1；如果未设置，则将其设置为0。
您可以按如下方式修复代码：
//Once the form elements have been filled in, extract data from form and store in
//variables
if(isset($_POST['txtName']) {
$myName = $_POST['txtName'];
};
// etc
这些变量不应该是吗
$myName = isset($_POST['txtName']);
$myAge = isset($_POST['mnuAge']);
$myFav = isset($_POST['rdFav']);
$myQuestion = isset($_POST['txaQuestions'])
这些：
$myName = $_POST['txtName'];
$myAge = $_POST['mnuAge'];
$myFav = ['rdFav'];
$myQuestion = $_POST['txaQuestions'];
否则，您只是存储它们是否已设置，而不是它们的值。
isset()测试该值是否存在。要获得实际值，只需要执行$var = $_GET['var'];之类的语句。
$myName = isset($_POST['txtName']);
这返回1，因为isset()返回布尔值；1（如果设置了值），0（字符串形式的TRUE或FALSE）（如果没有设置值），如手册所述：
如果var存在并且值不是NULL，则返回TRUE，否则返回FALSE。
尝试使用：
if (isset($_POST['txtName'])) {
$myName = $_POST['txtName'];
}
您做错了两件事（while和isset），使用这种方法可能可以修复和简化这两件事：
// Prepare defaults for unset fields:
$defaults = array(
"myName" => "anonymous",
"myAge" => "unspecified",
"myFav" => "unspecified",
"myQuestion" => "unspecified"
);
// make local variables
extract(array_merge($defaults, array_intersect_key(array_filter($_POST), $defaults)));
# $myName, $myAge, ...
这避免了您偶然发现的isset测试，并且只从POST中提取您真正感兴趣的四个变量（并且也应用了默认值）。
而不是
while (!(isset($myName) || isset($myAge) || isset($myFav) || isset($myComments)))
{
$myName = "anonymous";
$myAge = "unspecified";
$myFav = "unspecified";
$myQuestion = "unspecified";
}
这是完全错误的，因为没有循环，为什么你现在检查isset，然后再做一次？
你可以使用：
$myName = isset($_POST['txtName']) ? $_POST['txtName'] : 'anonymous';
$myAge = isset($_POST['mnuAge']) ? $_POST['mnuAge'] : 'unspecified';
$myFav = isset($_POST['radFav']) ? $_POST['radFav'] : 'unspecified';
$myQuestion = isset($_POST['txaQuestions']) ? $_POST['txaQuestions'] : 'unspecified';
正如其他用户所说，isset返回一个布尔值（true/false，但也返回1/0），这是您得到的1，而不是实际的$_POST变量。