我得到资源#6,#资源7当我打印以下变量:
$salty_password = sha1($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和不知道是什么意思。我正尝试使用我在散列和腌制后注册的明文密码重新登录。我觉得我很亲密,但有些错误。帮助为什么这是行不通的,也将不胜感激。
在此先感谢
+1 Zend引擎中的A *资源*是一个C指针。从[devzone.zend.com](http://devzone.zend.com/article/1024)文章:*“虽然PHP zval可以表示各种各样的内部数据类型,但无法表示一种数据类型完全在一个脚本中是指针,[...]也没有办法用传统的操作符来有效地处理它们,解决这个问题的方法是简单地用一个称为资源的基本上任意的标签来引用指针。 – netcoder 2011-01-28 19:38:43