Question

我想知道这个php编码的错误，只是想知道这个编码中的错误在哪里我跳过了html部分页面输入设计，因为我只想知道php部分，，我试图用这个用户输入用户名和密码登录网站，这个特定的网站应该是密码保护密码

<?php>
if isset($_POST=['submit']));
{
$inputuser = $_POST['user'];
$inputpass = $_POST['pass'];

$user = "root";
$password = "";
$database = "Tutorial";

$connect = mysql_connect("localhost",$user,$password);
@mysql_select_db($database) or ("database not found");

$query = " SELECT * FROM 'users' WHERE 'user' =                             i   $inputuser";//for query specific data
$querypass = "SELECT * FROM 'users' WHERE 'user' =            $        i   $inputpass'";

$result = mysql_query($query);
$resultpass = mysql_query($querypass);

$row = mysql_fetch_array($result);
$rowpass = mysql_fetch_array($resultpass);

$serveruser = $row["user"];
$serverpass = $row["password"];

if($serveruser&&$serverpass){
if (!$result) {
die("username and password is invalid");
}

echo "<br> <center>database output</b></center><br><br>";
mysql_close();
echo $inputpass;
echo $serverpass;
if ($inputpass == $serverpass) {
header('location: Home.php');
} else {
header('location: fail.php');
}
}

Answer 1

尝试这种方式：

$query="SELECT * FROM users WHERE user='" . mysql_real_escape_string( $inputuser ) . "' AND psw='" . mysql_real_escape_string( $inputpsw ) . "'";
$qr=mysql_query($query) or die (mysql_error());
if(mysql_num_rows($qr)>0) //admitting that usernames and psw are unique
{
  //success
  header('location:home.php');
}
else //no rows=no username responding to $inputusername
{
  header('location:fail.php');
}

php mysql用户名密码验证

1 个答案: