我正在尝试使用crypt()
将我当前的数据库从纯文本密码更新为哈希密码..我正在尝试这样做而无需用户更改密码(这是一种不稳定的方法)我的代码是这样的:
$Query = $Database->prepare("SELECT ID,Username,Password FROM userlist");
$Query->execute();
$Query->bind_result($ID,$Username,$Password);
while ($Query->fetch()){
$Hashed = $FrameWork->Hash_Password($Password);
$Secondary_Query = $Database->prepare("UPDATE userlist SET Password=?, Salt=? WHERE ID=?");
$Secondary_Query->bind_param('ssi', $Hashed['Password'],$Hashed['Salt'],$ID);
$Secondary_Query->execute();
$Secondary_Query->close();
}
$Query->close();
我收到了错误:
致命错误:在非对象中调用成员函数bind_param() 第24行的C:\ inetpub \ www \ AdminChangeTextPass.php
现在。我知道我的列名称与我的数据库名称是100%匹配。我也知道我的变量设置正确。
调试
调试:
$Query = $Database->prepare("SELECT ID,Username,Password FROM userlist");
$Query->execute();
$Query->bind_result($ID,$Username,$Password);
while ($Query->fetch()){
echo $Password."<br>";
}
$Query->close();
// Returns:
//test
//test
Then:
$Query = $Database->prepare("SELECT ID,Username,Password FROM userlist");
$Query->execute();
$Query->bind_result($ID,$Username,$Password);
while ($Query->fetch()){
print_r($FrameWork->Hash_Password($Password));
}
$Query->close();
/*
Returns:
Array ( [Salt] => ÛûÂÒs8Q-h¸Ý>c"ÿò [Password] => Ûûj1QnM/Ui/16 )
Array ( [Salt] => ÛûÂÒs8Q-h¸Ý>c"ÿò [Password] => Ûûj1QnM/Ui/16 )
*/
数据库架构
CREATE TABLE IF NOT EXISTS `userlist` (
`ID` int(255) NOT NULL AUTO_INCREMENT,
`Username` varchar(255) NOT NULL,
`Password` varchar(255) NOT NULL,
`Salt` text NOT NULL,
PRIMARY KEY (`ID`)
) ENGINE=InnoDB DEFAULT CHARSET=latin1 AUTO_INCREMENT=9 ;
--
-- Dumping data for table `userlist`
--
INSERT INTO `userlist` (`ID`, `Username`, `Password`, `Salt`) VALUES
(1, 'test', 'test', ''),
INSERT INTO `userlist` (`ID`, `Username`, `Password`, `Salt`) VALUES
(2, 'test', 'test', '');
让我的代码看起来像这样:
$Secondary_Query = $Database->prepare("UPDATE userlist SET Password=? WHERE ID=?");
$Query = $Database->prepare("SELECT ID,Username,Password FROM userlist LIMIT 1");
var_dump($Secondary_Query);
#$Query->execute();
#$Query->bind_result($ID,$Username,$Password);
# while ($Query->fetch()){
# $Hashed = $FrameWork->Hash_Password($Password);
# $Secondary_Query = $Database->prepare("UPDATE userlist SET Password=? WHERE ID=?");
# $Secondary_Query->bind_param('ssi', $Hashed['Password'],$Hashed['Salt'],$ID);
# $Secondary_Query->execute();
# $Secondary_Query->close();
# }
#$Query->close();
var_dump($Secondary_Query);
返回:
object(mysqli_stmt)#3 (10) { ["affected_rows"]=> int(-1) ["insert_id"]=> int(0) ["num_rows"]=> int(0) ["param_count"]=> int(2)
[ “场计数”] =&GT; int(0)[“errno”] =&gt; int(0)[“error”] =&gt; string(0)“” [ “error_list”] =&GT; array(0){} [“sqlstate”] =&gt; string(5)“00000” [ “ID”] =&GT; int(1)}
var_dump($Query);
返回:
object(mysqli_stmt)#4(10){[“affected_rows”] =&gt; INT(-1) [ “INSERT_ID”] =&GT; int(0)[“num_rows”] =&gt; int(0)[“param_count”] =&gt; INT(0) [ “场计数”] =&GT; int(3)[“errno”] =&gt; int(0)[“error”] =&gt; string(0)“” [ “error_list”] =&GT; array(0){} [“sqlstate”] =&gt; string(5)“00000” [ “ID”] =&GT; int(2)}
因为我还没有提交答案..我的工作代码如下:
$Query = $Database->prepare("SELECT ID,Username,Password FROM userlist");
$Query->execute();
$Query->bind_result($ID,$Username,$Password);
$Query->store_result();
while ($Query->fetch()){
$Hashed = $FrameWork->Hash_Password($Password);
$Secondary_Query = $Database->prepare("UPDATE userlist SET Password=?, Salt=? WHERE ID=?");
$Secondary_Query->bind_param('ssi', $Hashed['Password'],$Hashed['Salt'],$ID);
$Secondary_Query->execute();
$Secondary_Query->close();
}
$Query->close();