从MySQL翻译的PHP PDO问题 [英] Php PDO issues translating from mysql

问题描述

好吧，即使经过2天尝试转换所有内容，我也仍然很难掌握PDO.

Okay, I'm really struggling to get to grasps with PDO, even after 2 days of trying to convert everything.

我现在处于为$user_data['???']创建数组的阶段，这就是我所拥有的.

I'm now at the stage of creating an array for the $user_data['???'] And here's what I've got.

if (logged_in() === true) {
    $session_user_id = $_SESSION['user_id'];
    $user_data = user_data($session_user_id, 'id', 'username', 'password', 'email', 'active', 'coins');
    $user_id = $user_data['id'];
    if (user_active($user_data['username'] === false) {
        session_destroy();
        header('Location: index.php');
        exit();
    }
}

所以这就是我获取$user_data['???']数据的方法.

So that's my way of getting the data for $user_data['???'] The functions to go with it are..

function user_data($user_id){
    $data = array(); 
    $user_id = (int)$user_id;

    $func_num_args = func_num_args();
    $func_get_args = func_get_args();

    if ($func_num_args > 1) {
        unset($func_get_args[0]);

        $fields = '`' . implode('`, `', $func_get_args) . '`';
        $data = //mysql_fetch_assoc(//mysql_query("SELECT $fields FROM `users` WHERE `user_id` = $user_id"));

        return $data;
    }
}

function user_active($username) { 
    $username = sanitize($username);
    $query = //mysql_query("SELECT COUNT(`user_id`) FROM `users` WHERE `username` = '$username' AND `active` = 1");
    return (//mysql_result($query, 0) == 1) ? true : false;
}

我正在拔头发，试图弄清楚如何将其转换为PDO，有人可以给我任何帮助吗?

I'm pulling my hair out trying to figure out how to convert this into PDO, can anyone give me any help?

此外，在将其转换为PDO之后.说简单的Welcome <?php $user_data['username'] ?>, Hope you enjoy your stay!消息欢迎用户，还是现在需要使用一种完全不同的方法?

Also, after I do convert it to PDO. Would it be as simple for to say welcome the user with a simple message of Welcome <?php $user_data['username'] ?>, Hope you enjoy your stay! or would I need to use a completely different method now?

提前谢谢！

推荐答案

因此，通过示例代码的外观，我得到的印象是您正从不赞成使用的mysql扩展迁移到PDO.

So by the looks of the example code I get the impression that you're migrating from the deprecated mysql extension over to PDO.

mysql扩展确实使您不寒而栗，这是因为如果您未将mysql链接资源指定为mysql_query，它将使用由mysql_connect创建的最后创建的链接资源.一旦开始使用PDO，您将必须在user_data和user_active函数中使用PDO连接.最简单的方法是在每个功能中创建一个PDO连接，这是非常重复的，根本不是一个好的解决方案，但它可以工作.

Something the mysql extension does that may be tripping you up is that if you don't specify the mysql link resource to mysql_query it uses the lastly created link resource that was created by mysql_connect. Once you move to using PDO you're going to have to have the PDO connection available in the user_data and user_active functions. The simplest approach would be to create a PDO connection in each function, it's terribly repetitive and not a good solution at all, but it works.

代码中的 sql注入漏洞也存在一些问题.除此之外，您可以编写如下功能:

There are also some issues with sql injection vulnerabilities in the code. Aside from that you could write the functions like this:

<?php

function user_data($user_id){
  $data = array(); 
  $user_id = (int)$user_id;

  $func_num_args = func_num_args();
  $func_get_args = func_get_args();

  if ($func_num_args > 1) {
    unset($func_get_args[0]);

    // connect to the DB
    $dsn = 'mysql:dbname=<your_db_name>;host=127.0.0.1';
    $user = '<your_db_user>';
    $password = '<your_db_user_password>';
    $dbh = new PDO($dsn, $user, $password);

    // request the data
    $fields = '`' . implode('`, `', $func_get_args) . '`';
    $sql = sprintf('select %s from users where user_id = ? limit 1', $fields);
    $stmt = $dbh->prepare($sql);
    $stmt->execute(array($userid));
    $data = $stmt->fetch(PDO::FETCH_ASSOC);

    return $data;
  }
}

function user_active($username) { 
  $username = sanitize($username);

  // connect to the DB
  $dsn = 'mysql:dbname=<your_db_name>;host=127.0.0.1';
  $user = '<your_db_user>';
  $password = '<your_db_user_password>';
  $dbh = new PDO($dsn, $user, $password);

  $sql = 'select count(user_id) from users where username = ? and active';
  $stmt = $dbh->prepare($sql);
  $stmt->execute(array($username));

  return $stmt->fetchColumn() == 1;      

}  

希望有帮助.

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