这是我对此的第二次编辑。我已经苦思冥想了几天,感觉我已经很接近了。我尝试了许多不同版本的第三个代码块,但就是无法得到正确结果。你有什么想法,我做错了什么(更改的是第三个代码块)。
if(!$error) {
$alpha = "ABCDEFGHIJKLMNOPQRSTUVWXYZabcedfghijklmnopqrstuvwxyz1234567890";
$rand = str_shuffle($alpha);
$salt = substr($rand,0,40);
$hashed_password = sha1($salt . $_POST['Password']);
$query = "INSERT INTO `Users` (
`FirstName`,
`LastName`,
`Email`,
`Password`,
`salt`,
`RelationshipToCF`,
`State`,
`Gender`,
`Birthday`,
`Status`
)VALUES(
'" . mysql_real_escape_string($_POST['firstName']) . "',
'" . mysql_real_escape_string($_POST['lastName']) . "',
'" . mysql_real_escape_string($_POST['email']) . "',
'" . $hashed_password . "',
'" . $salt . "',
'" . mysql_real_escape_string($_POST['RelationToCF']) . "',
'" . mysql_real_escape_string($_POST['State']) . "',
'" . mysql_real_escape_string($_POST['sex']) . "',
'" . mysql_real_escape_string($_POST['DateOfBirth_Year'] . "-" . $_POST['DateOfBirth_Month'] . "-" . $_POST['DateOfBirth_Day']) . "',
'pending'
)";
mysql_query($query, $connection);
这是我正在使用的更新现有密码的方法:
$query = "SELECT * FROM `Users`";
$request = mysql_query($query,$connection);
while($result = mysql_fetch_array($request)) {
$alpha = "ABCDEFGHIJKLMNOPQRSTUVWXYZabcedfghijklmnopqrstuvwxyz1234567890";
$rand = str_shuffle($alpha);
$salt = substr($rand,0,40);
$hashed_password = sha1($salt . $result['Password']);
$user = $result['id'];
$query2 = "UPDATE `Users` SET `salt` = '$salt' WHERE `id` = '$user'";
$request2 = mysql_query($query2,$connection) or die(mysql_error());
$query3 = "UPDATE `Users` SET `encrypted_passwords` = '$hashed_password' WHERE `id` = '$user'";
$request3 = mysql_query($query3,$connection) or die(mysql_error());
}
现在,我希望允许用户使用他们注册时的密码进行登录,但目前只能使用哈希密码进行登录。显然,这还没有应用到真正的数据库中。
以下是我需要修改的登录页面查询:
if(isset($_POST['subSignIn']) &&
!empty($_POST['email']) &&
!empty($_POST['password'])) {
$email = mysql_real_escape_string($_POST['email']);
$password = mysql_real_escape_string($_POST['password']);
$query = "SELECT
`id`,`email`,`password` FROM `Users`
WHERE `Email` = '" . $email . "' AND
`Password` = '" . $password . "' &&
`Status` = 'active' LIMIT 1";
$request = mysql_query($query,$connection) or die(mysql_error());
if(@mysql_num_rows($request)) {
$result = mysql_fetch_array($request);
$_SESSION['LIFE']['AUTH'] = true;
$_SESSION['LIFE']['ID'] = $result['id'];
$query = "UPDATE `Users` SET` LastActivity` = '" . date("Y-m-d") ." " . date("g:i:s") . "' WHERE `id` ='" .mysql_real_escape_string($_SESSION['LIFE']['ID']) . "' LIMIT 1";
mysql_query($query,$connection);
if(!empty($_POST['return'])) {
header("Location: " . $_POST['return']);
}else{
header("Location: Dashboard.php?id=" . $_SESSION['LIFE']['ID']);
}
}else{
$_SESSION['LIFE']['AUTH'] = false;
$_SESSION['LIFE']['ID'] = false;
}
我一直在搜索网络上的方法,但是想要利用这里所有伟大的思维来寻求建议/方法/教程/指引。
<== 在原先的5个答案之后,这是我的尝试 ==>
i
f(isset($_POST['subSignIn']) && !empty($_POST['email']) && !empty($_POST['password'])) {
$query = "SELECT id FROM cysticUsers WHERE Email = '$email' AND Password = SHA1(CONCAT(salt,'$password')) AND Status = 'active' LIMIT 1";
$request = mysql_query($query,$connection) or die(mysql_error());
if(@mysql_num_rows($request)) {
$row = mysql_fetch_assoc($request);
if (sha1($row['salt'] . $_POST['password']) === $row['password']) {
$_SESSION['CLIFE']['AUTH'] = true;
$_SESSION['CLIFE']['ID'] = $result['id'];
// UPDATE LAST ACTIVITY FOR USER
$query = "UPDATE `cysticUsers` SET `LastActivity` = '" . date("Y-m-d") . " " . date("g:i:s") . "' WHERE `id` = '" . mysql_real_escape_string($_SESSION['CLIFE']['ID']) . "' LIMIT 1";
mysql_query($query,$connection);
if(!empty($_POST['return'])) {
header("Location: " . $_POST['return']);
}else{
header("Location: CysticLife-Dashboard.php?id=" . $_SESSION['CLIFE']['ID']);
}
}
}else{
$_SESSION['CLIFE']['AUTH'] = false;
$_SESSION['CLIFE']['ID'] = false;
}
}
?>