什么是“资源＃”？ [英] What are "Resource#'s"?

问题描述

当我打印下列变量时，获得Resource＃6和Resource＃7：

  $ salty_password = sha1（$ row ['salt']，$ _POST ['password']）; （$ _ POST ['subSignIn']）&&！empty（$ _ POST ['email']）&&！empty（$ _ POST ['password']）
 
） ）{
 
 
 $ query =SELECT`salt` from`cysticUsers` WHERE`Email` =''。 $ _POST ['email']。 ; 
 $ request = mysql_query（$ query，$ connection）或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）或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']; 
 
 //更新用户的最后活动
 $ query =更新``cysticUsers` SET`LastActivity` ='。日期（Y-m-d）。 。日期（g：i：s）。 'WHERE`id` ='。 mysql_real_escape_string（$ _ SESSION ['CLIFE'] ['ID']）。 '限制1; 
 mysql_query（$ query，$ connection）; 
 
 
 if（！empty（$ _ POST ['return']））{
 header（Location：。$ _POST ['return']）; 
 $ b $} else {
 header（Location：CysticLife-Dashboard.php？id =。$ _SESSION ['CLIFE'] ['ID']）; 
} 
} 
 
} else {
 
 $ _SESSION ['CLIFE'] ['AUTH'] = false; 
 $ _SESSION ['CLIFE'] ['ID'] = false; 
 
} 
 
 
？> 
  

试着解决这个代码块并且不确定这意味着什么。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，我们将不胜感激。

预先致谢

解决方案

mysql_query（）返回结果集作为 resource （根据PHP OOP代码，它们不是对象但我想不出一个更好的词）。它们包含只能被某些函数读取的二进制数据，例如 mysql_fetch _ *（）函数。

为了调试你的MySQL查询，你应该使用 mysql_error（）和 mysql_errno（）来检查错误和/或保存你的SQL 从我看到的，你正在执行两个查询，但覆盖相同的 $ result 变量，而不对第一个做任何事情。此外， mysql_num_rows（）一次只能计算一个结果集，因此您无法将两个结果集传递给相同的调用。

HI

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;

    }


?>

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.

Thanks in advance

解决方案

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.

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.

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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