我正在PHP中创建一个流程表单,我试图显示反馈,但它不想显示。我有代码是:PHP表单不显示反馈
<html>
<head>
<style type="text/css">
.error{color: #FF0000;}
</style>
</head>
<body>
<h1>Customer Feedback</h1>
<p1>Please tell us what you think</p1><br><br>
<?PHP
$name = trim($_POST[fullname]);
$email = trim($_POST[email]);
$text = trim($_POST[feedback]);
?>
<form method='POST' action='<?php echo htmlspecialchars($_SERVER['PHP_SELF']);?>' >
<p1>Your name:</p1><br>
<input type="text" name="fullname" value="<?php echo $fullname; ?>" required><br><br>
<p1>Your email address:</p1><br>
<input type="text" name="email" value="<?php echo $email; ?>" required><br><br>
<p1>Your feedback:</p1><br>
<textarea rows="5" cols="50" name="feedback"><?php echo nl2br($text);?></textarea>
<textarea><?php echo $text;?></textarea><br><br>
<input type="submit" Value="Send Feedback"><br><br>
</form>
<?php
if(isset($_POST[fullname]) && $_POST[fullname] != "" && !empty($_POST[fullname])) {
echo "Hi " . $name . ".<br>";
}
else{
echo "Please enter a name....";
}
if(isset($_POST[email]) && $_POST[email] != "" && !empty($_POST[email]) {
echo "Your email is " . $email . ".<br>";
}
else{
echo "Please enter a email address.";
}
if(isset($_POST[feedback]) && $_POST[feedback] != "") {
echo "Your feedback is:" . $feedback . "<br>";
}
else{
echo "No feedback.";
}
?>
</body>
</html>
当我运行的页面,它显示的姓名,电子邮件和“你的意见是:”这是进入textarea的,但没有反馈。
编辑 我想在文本框中使用nl2br()函数。
因为'$ feedback'不被任何定义。 –
smitthy你没有表现出对获得答案和标记答案的兴趣,所以我删除了我的答案。 –