什么是“资源#"?
嗨
当我打印以下变量时,我得到了 Resource#6 和 Resource#7:
I am getting Resource#6 and Resource#7 when I print the following variables:
$salty_password = sha1($row['salt'], $_POST['password']);
if(isset($_POST['subSignIn']) && !empty($_POST['email']) && !empty($_POST['password'])) {
$query = "SELECT `salt` FROM `cysticUsers` WHERE `Email` = '" . $_POST['email'] . "'";
$request = mysql_query($query,$connection) or die(mysql_error());
$result = mysql_fetch_array($request);
$query2 = "SELECT * FROM `cysticUsers` WHERE `Email` = '". $_POST['email']."' AND `Password` = '$salty_password'";
$request2 = mysql_query($query2,$connection) or die(mysql_error());
$result = mysql_fetch_array($request2);
print_r($request);
print_r($request2);
if(@mysql_num_rows($request,$request2)) {
$_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;
}
?>
试图解决这个代码块,但不确定这意味着什么.我正在尝试使用经过哈希和加盐处理后注册的明文密码重新登录.我觉得我很接近,但有些地方有点不对劲.也非常感谢您提供有关为什么这不起作用的帮助.
Trying to troubleshoot this code chunk and not sure what that means. I am trying to sign back in with the clear text password I signed up with after its been hashed and salted. I feel like I'm very close but something is slightly wrong. Help on why that is not working would be greatly appreciated as well.
提前致谢
推荐答案
mysql_query()
以 resource
(就 PHP OOP 代码而言,它们不是 objects,但我想不出更好的词了).这些包含只能由某些函数读取的二进制数据,例如 mysql_fetch_*()
函数.
mysql_query()
returns result sets as objects of type resource
(they're not objects in terms of PHP OOP code but I can't think of a better word). These contain binary data that can only be read by certain functions, for example the mysql_fetch_*()
functions.
要调试 MySQL 查询,您应该使用 mysql_error()
和 mysql_errno()
检查错误和/或将 SQL 语句保存在变量中并打印出来.
To debug your MySQL queries you should check for errors using mysql_error()
and mysql_errno()
and/or save your SQL statements in variables and print those.
据我所知,您正在执行两个查询,但覆盖了相同的 $result
变量,而对第一个没有做任何事情.此外,mysql_num_rows()
一次只能计算一个结果集,因此您不能将两个结果集传递给同一个调用.
From what I see, you're performing two queries but overwriting the same $result
variable, without doing anything about the first. Also, mysql_num_rows()
can only count one result set at a time, so you can't pass two result sets into the same call.
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