我正在学习 PHP 来创建反馈表。我按照教程创建了一个带有表单和 php 文件的 html 页面来执行请求并将其保存到数据库中。奇怪的是,我有两个问题。
在我按下提交后,我收到以下错误:未定义变量:第 8 行 /Users/wj/Sites/all_sanbox/VTC-PHP/07/send_feedback.php 中的用户
我用一些数据填写了表格,但是执行的 php 文件好像没有输入数据一样。
我使用的代码如下:
反馈.html
<HTML>
<HEAD>
<TITLE>Vinyl Dealers: feedback </TITLE>
</HEAD>
<BODY>
<H2>Feedback</H2>
<BR />
<FORM ACTION="send_feedback.php" METHOD="POST">
Your name:
<INPUT TYPE=TEXT NAME="user" MAXLENGTH=40 SIZE=40 />
<BR />
Your email:
<INPUT TYPE=TEXT NAME="email" MAXLENGTH=40 SIZE=40 />
<BR /><BR />
Can we keep you updated with news about our site?
<INPUT NAME="spam" TYPE=RADIO VALUE="1" CHECKED />Yes
<INPUT NAME="spam" TYPE=RADIO VALUE="0" />No
<BR /><BR />
Comments:
<BR />
<TEXTAREA COLS=60 ROWS=10 NAME="comments"></TEXTAREA>
<BR />
<INPUT TYPE=SUBMIT VALUE="submit" />
</FORM>
</BODY>
</HTML>
发送反馈.php
<html>
<head>
<title>Vinyl Dealers: we appreciate your feedback</title>
</head>
<body>
<?php
if (!$user || !$email || !$comments){ ?>
<h2>Whoops</h2>
<br>Please fill in all fields<br>
<a href="feedback.html">Click here to go back to the feedback page.</a>
<?php
exit;
}
$user = addslashes($user);
$email = addslashes($email);
$comments = addslashes($comments);
$db = mysql_connect("localhost","root","admin");
mysql_select_db("vinyldealers",$db);
$addfeedback = "INSERT INTO feedback (user, email, spam, comments)
VALUES(" . '$user' . "," . $email . "," . $comments .")";
$result = mysql_query($addfeedback);
?>
<H2>Thank you</H2>
<BR>
We have added your comments to our database.
</body>
</html>