2013-03-29 46 views
-1

我寫了如下一段代碼&如果pswd = retype pswd &以後它必須回顯「註冊成功」將以下值插入TABLE(哪些不是)。它只是回聲的註冊成功,但值不插入到TABLE中。我非常確定,我的PHP代碼連接到數據庫失敗,請更正我的代碼! 我的代碼如下:如何獲取mysql數據庫表中提交的表單數據

<?php 

session_start(); 
include('header.php'); 

$mysql_host = 'localhost'; 
$mysql_user = 'root'; 
$mysql_pass = '1234'; 

$mysql_db = "my_db"; 

if (!mysql_connect($mysql_host, $mysql_user, $mysql_pass) || !mysql_select_db($mysql_db)) 
{ 
die(mysql_error()); 
} 

if(isset($_REQUEST['namevar'])){ 

$nmvar = $_REQUEST['namevar']; 
$email = $_REQUEST['name1']; 
$psvar = $_REQUEST['p1']; 
$cpsvar = $_REQUEST['p2']; 
$gender = $_REQUEST['r1']; 
$clvar = $_REQUEST['t2']; 
$plvar = $_REQUEST['t1']; 


if($psvar == $cpsvar) 
{ 
$sql = "INSERT INTO `users` (name,email,password,confirm password,gender,college,place) 
VALUES ('$nmvar','$email','$psvar','$cpsvar','$gender','$clvar','$plvar')"; 
mysql_query($sql); 
echo "Signup successful"; 
} 
else 
{ 
echo "Password Mismatch"; 
} 
} 
?> 
<form action="" method="post"> 
Name: 
<input type="text" name="namevar" /><br/> 
E-mail: 
<input type="text" name="name1" /><br/> 
Password: 
<input type="password" name="p1" /><br/> 
Confirm Password: 
<input type="password" name="p2" /><br/> 
Gender: 
<input type="radio" name="r1" /> 
Male 
<input type="radio" name="r1" /> 
Female 
<br/> 
Location: 
<input type="text" name="t1" /><br/> 
College: 
<input type="text" name="t2" /><br/> 
<input type="submit" value="submit" /><br/> 
<input type="reset" value="reset" /> 
</form> 
+0

連接與否時出現*錯誤*嗎?你怎麼能*非常確定*它無法連接? –

+0

我可以在哪裏註冊?我的電子郵件是'x'); DROP TABLE用戶; - ' –

+1

@EugenRieck:'mysql_query'不需要兩條語句。 – Nauphal

回答

1

提交到數據庫之前,您應該使用對數據的mysql_real_escape_string()函數。否則,您的代碼面臨SQL注入風險。