2017-08-16 36 views
0

我经历了一个艰难的过程,弄清楚如何在配置文件中绑定正确的地址,并将mysql连接到远程服务器。昨天它使用不同的代码,现在它不连接。我得到了死亡“连接失败:”但它没有向我显示connect_error,因为我甚至找不到这个问题?有人看到我的代码有问题吗?注意:我知道这是不安全的,我通常会准备好声明,但现在只是尝试与db连接问题一起工作。Mysql配置正确的远程连接,但PHP代码失败连接?

<?php 
$server = "174.---.--.187"; 
$username = "dylanto"; 
$pass = "------"; 
$db = "survey"; 
//$port = 3306; 
//create connection 
$conn = new mysqli($server, $username, $pass, $db); 
//check connection 
if (!$conn->connect_error) { 
die("Connection failed: " . $conn->connect_error);} 

$user = $_POST['user']; 
$pass = $_POST ['pass']; 

$sql = "insert into login (user, pass) values ('$user','$pass')"; 

if ($conn->query($sql)==TRUE) { 
echo "Account created";} 

else {echo "something went wrong";} 
$con->close; 
?> 

HTML代码:

<html><head><title>Log-in</title> 
<link rel="stylesheet" type="text/css" href="sytle.css"></head> 
<body> 
<center><u><strong><h2>Login</h2></u></strong></center> 
<br /> 
<center> 
<form action="signup_process.php" method="POST"> 
Pick Username: <br> 
<input type ="text" name = "user"><br> 
Pick Password:<br> 
<input type ="password" name ="pass"><br> 
<input type="submit" name="submit" value="Sign-up"> 
</form></center> 
</body> 
</html> 
+2

'如果(!$ conn-> connect_error){'这个=不存在连接错误删除'!' – cmorrissey

+1

此外,您的代码对于需要学习PDO或mysqli的SQL注入攻击是开放的。 – cmorrissey

+1

您已经开放了SQL注入。由于您使用的是mysqli,请利用[prepared statements](http://php.net/manual/en/mysqli.quickstart.prepared-statements.php)和[bind_param](http://php.net/手动/ EN/mysqli的-stmt.bind-param.php)。 **这将照顾任何可能发生讨厌的报价问题。** – aynber

回答

0

if (!$conn->connect_error) { 
die("Connection failed: " . $conn->connect_error); 
} 

到:

if ($conn->connect_error) { 
die("Connection failed: " . $conn->connect_error); 
} 

由于如果($ conn-> connect_error)=如果任何错误,并且如果( !$ conn-> connect_error)如果没有错误。

准备好的语句:

来源:

$sql = "insert into login (user, pass) values ('$user','$pass')"; 

if ($conn->query($sql)==TRUE) { 
echo "Account created";} 

else {echo "something went wrong";} 

要:

//sanityze POST 
$user = trim(mysqli_real_escape_string($conn, htmlspecialchars($_POST['user'], ENT_QUOTES, 'UTF-8'))); 
$pass = trim(mysqli_real_escape_string($conn, htmlspecialchars($_POST['pass'], ENT_QUOTES, 'UTF-8'))); 

//use prepared 

$stmt = $conn->prepare("insert into login (user, pass) values (?,?)"); 

//bind parameter 
$stmt->bind_param("ss", $user, $pass) 

$stmt->execute(); 

//then check 
+0

是的,评论中的这个人已经告诉我这个,但是谢谢,现在我理所当然地错过了。 – do734

+2

如果你打算包含一个准备好的语句,那么'mysqli_real_escape_string()','htmlspecialchars()'就没有用处。 –

+0

是的,我喜欢使用准备好的语句 – do734