Question

我很有兴趣在一带使用“油轮”为《识别法》制造大宗用户/密码。我在 CI论坛上问这个问题,但没有回答:

http://codeigniter.com/forums/viewthread/110993/P330/#837327”rel=“nofollow”> http://codeigniter.com/forums/viewthread/110993/P330/#837327 。

谷歌的搜索除了像第三种结果和一只与外界无关的场所的布局外,还有很大变化。

是否有任何人利用再保险算法成功这样做? 如果是的话,你能否把一些法典张贴在正确的道路上? 感谢!

软件版本:

CI 1.7.3

1.0.7号油轮

PHP 5.x

<EDIT 2/15:

Just in case anyone is looking for a solution to this, here s a function that is basically the same one I used (there were some other parameters, but this should get you going):

function batchReg()
{
    $this->load->model( mymodel );

    // connect to the database
    $this->mymodel->dbconnect();

    // build it
    $query = "SELECT user, email, pass from newusers ORDER BY user ASC";

    // ship it
    $result = mysql_query($query);

    // loop it
    while ($row = mysql_fetch_array($result))
    {
        $data = $this->tank_auth->create_user($row[ user ], $row[ email ], $row[ pass ], FALSE);
        print_r($data);
        echo "<p>";
    }

}

仅从控制者处 p起,使之付诸行动!

Answer 1

象对我进行简单的排他性手术这样的声音

某种方式(无论来源如何)使用户名以可操作的形式获得,例如一个阵列 用户_list。

我们恳切地说,它喜欢这样做。

Array(
    Array(
        [username] =>  ... ,
        [email]    =>  ... ,
        [password] =>   ,     //leave password empty
    ),
    Array(
        [username] =>  ... ,
        [email]    =>  ... ,
        [password] =>   ,     //leave password empty
    ),
    ... etc.
)

然后,为办理新登记手续、将密码重新排入阵列创造了简便的例行程序,因此,你将有一个完整的标识清单、新的(占支配地位的)密码和电子邮件。

//loop by referance in order to properly store the generated password
foreach($user_list as &$user) {

    //generate 8 char password and store it
    $user[ password ] = substr(uniqid(),2,8);

    //run register routine (not sure on tank auth s specific syntax
    $this->tankauth->register($user[ username ],$user[ email ],$user[ password ],FALSE);
}

之后,<代码> 用户_list载有用户的所有新密码。

友情链接