无法使用REGEX获取PHP脚本名称

Question

我面临很大的挑战。我有函数where()，该函数可跟踪用户在网站上的访问位置，并获取脚本文件名，并提供描述并插入数据库。

该脚本一直有效，直到最近，并且在从MySQL 5.6.40到5.6.47的最近一次小更新之后，仍未意识到此功能的失败。

我不确定这是否与它有任何关系，但是几天后我发现它不再起作用了。

我们的职能：

function where($scriptname = "index", $userid, $update=1){
    if (!is_valid_id($userid))
        die;

    if (preg_match("/details.php/i", $scriptname))
        $where = "Browsing File Details (ID $_GET[id])";
    elseif (preg_match("/files.php/i", $scriptname))
        $where = "Browsing Files";
    elseif (preg_match("/account-info.php/i", $scriptname))
        $where = "Browsing Account Info (ID $_GET[id])";
    elseif (preg_match("/upload.php/i", $scriptname))
        $where = "Uploading File";
    elseif (preg_match("/account.php/i", $scriptname))
        $where = "Browsing User Control Panel";
    elseif (preg_match("/search.php/i", $scriptname))
        $where = "Searching For Files";
    elseif (preg_match("/forums.php/i", $scriptname))
        $where = "Viewing Forums";
    elseif (preg_match("/index.php/i", $scriptname))
        $where = "Browsing Homepage";
    elseif (preg_match("/mailbox.php/i", $scriptname))
        $where = "Viewing Messages";
    elseif (preg_match("/comments.php/i", $scriptname))
        $where = "Viewing Comments";
    elseif (preg_match("/recover.php/i", $scriptname))
        $where = "Recovering Account";
    elseif (preg_match("/bookmarks.php/i", $scriptname))
        $where = "Viewing Bookmarks";
    elseif (preg_match("/getfile.php/i", $scriptname))
        $where = "Downloaded File (ID $_GET[id])";
    elseif (preg_match("/faq.php/i", $scriptname))
        $where = "Reading FAQ Page";
    elseif (preg_match("/friends.php/i", $scriptname))
        $where = "Viewing Friends";
    elseif (preg_match("/admin.php/i", $scriptname))
        $where = "Managing Admin Panel";
    else
        $where = "Unknown Location";

    if ($update) {
        // Worked until a few days ago. No site changes were made prior for quite some time. 
        //$query = sprintf("UPDATE users SET page=".sqlesc($where)." WHERE id ='%s'", mysql_real_escape_string($userid)); 
        // Now using line below, which does insert into row if I use my own variable. 
        $query = "UPDATE users SET last_access='" . get_date_time() . "', page=" . sqlesc($where) . " WHERE id=" . $userid;
        $result = SQL_Query_exec($query);
    }
        return $where;
}
现在，我已经尝试使用以下代码将其范围缩小到导致它的原因。当前，以上函数仅插入

未知位置

对于数据库，无论查看什么页面。

我明白了这一点：

$stringtest = "This inserts into database!";
$query = "UPDATE users SET last_access='" . get_date_time() . "', page=" . sqlesc($stringtest) . " WHERE id=" . $userid;
这很好，但是没有通过任何$where条件。

我尝试过使用各种正则表达式来匹配文件名，但无济于事。

知道我在做什么错吗？

提前感谢！

我面临很大的挑战。我有我的功能where（），它跟踪用户在网站上的访问位置，并获取脚本文件名，并提供描述并插入数据库中。 ...

Answer 1

Answer 2