嘗試在php中使用輸入值填充數據庫連接參數。給這個消息

<html> 
<body> 
index.php 
<form action="config.php" method="Post"> 
<table> 
    <tr><td>DB Host: <input type="text" name="dbhost"></></tr> 
    <tr><td>DB Name: <input type="text" name="dbname"></td></tr> 
    <tr><td>DB pass: <input type="text" name="dbpass"></td></tr> 
    <tr><td><button type="submit" name="submit"> Submit</button></td></tr> 
</table> 
</form> 

<?php 
config.php 
$dbhost = $dbname = $dbpass = ""; 
if ($_SERVER["REQUEST_METHOD"] == "POST") { 
    $dbname = test_input($_POST["dbname"]); 
    $dbpass = test_input($_POST["dbpass"]); 

    $con = mysql_connect('$dbhost', '$dbname', '$dbpass'); 
    if (!$con) { 
     echo "could not connect" . mysql_error(); 
    } else { 
     echo "Processing..."; 
    } 
} 
    function test_input($data)![enter image description here][1] 
    { 
     $data = trim($data); 
     $data = stripslashes($data); 
     $data = htmlspecialchars($data); 
     return $data; 
    } 
    ?>

這是錯誤：嘗試在php中使用輸入值填充數據庫連接參數。給這個消息

could not connectphp_network_getaddresses: getaddrinfo failed: No such host is known.

來源

2014-10-08 Chrys Ugwu

變化

$con = mysql_connect('$dbhost', '$dbname', '$dbpass');

到

$con = mysql_connect($dbhost, $dbname, $dbpass);

來源

2014-10-08 05:31:35

嘗試在php中使用輸入值填充數據庫連接參數。給這個消息

回答

相關問題