我正在学习PHP以创建反馈表单.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:未定义的变量:第8行的/Users/wj/Sites/all_sanbox/VTC-PHP/07/send_feedback.php中的用户
我在表单中填写了一些数据,但是php文件的执行就好像没有输入数据一样.
我使用的代码如下:
<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>
<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>
我猜你运行的PHP没有register_globals选项,这是安全的.这就是为什么你需要使用$_POST['user']和类似的方式访问提交的值...
如何添加:
$user = $_POST['user']
$email = $_POST['email']
// etc...