Проблема с SQL запросом

[igramnet 1]

Есть запрос вида

		$sql_update = "UPDATE `" . PREFIX . "_" . $modul_dbtitle . "` SET `autor` = '{$new_author}', `title` = '{$new_title}', `keywords` = '{$keywords}', `url` = '{$urlfile}', `screen_url` = '{$scrurl}', `url1` = '{$url1}', `desc_url1` = '{$desc_url1}', `url2` = '{$url2}', `desc_url2` = '{$desc_url2}', " . $where_ver . " " . $where_platform . "`description` = '{$short_story}', `description_full` = '{$full_story}', `title` = '{$new_title}', `category` = '{$cat_title}', `size` = '{$filesize}', `approve` = '{$approve}', `access` = '{$group_regel}', `alt_name` = '{$alt_name}', `allow_main` = '{$allow_main}', `allow_rate` = '{$allow_rating}', `allow_comm` = '{$allow_comm}', `xfields` = '{$filecontents}', `allow_br` = '{$allow_br}', `reason` = '{$editreason}', `view_edit` = '{$view_edit}', `editdate` = '{$added_time}', `editor` = '{$member_db[2]}', `to_ftp` = '{$to_ftp}', `allow_block` = '{$allow_block}', `news_fixed` = '{$news_fixed}' WHERE `id` = '{$id}'";

Он выдает при сохранении следующий текст

UPDATE `dle_downloads` SET `autor` = 'SErg', `title` = 'Драйвер для AM-1B023 под Windows XP', `keywords` = 'VID_13D3&PID_5097, VID_13D3&PID_5095, VID_13D3&PID_5098, VID_13D3&PID_5099, VID_13D3&PID_5100, VID_13D3&PID_5076, VID_05E3&PID_0507, Azurewave, вебкамеры, Windows, AM1B023, КодыVID_05E3&PID_0503, VID_05E3&PID_0505, VID_13D3&PID_507B, VID_13D3&PID_507B, Драйвер', `url` = '', `screen_url` = '', `url1` = '', `desc_url1` = '', `url2` = '', `desc_url2` = '', `version` = '9.3.23.1', `version1` = '0', `platform1` = '0', `platform` = '', `description` = '<strong>123</strong>', `description_full` = '123qwa', `title` = 'Драйвер для AM-1B023 под Windows XP', `category` = '2', `size` = '0', `approve` = '0', `access` = '', `alt_name` = 'am-1b023_winxp', `allow_main` = '1', `allow_rate` = '1', `allow_comm` = '1', `xfields` = '', `allow_br` = '0', `reason` = '', `view_edit` = '0', `editdate` = '1289920937', `editor` = 'snake', `to_ftp` = '0', `allow_block` = '0', `news_fixed` = '0' WHERE `id` = '3'

но в базу идет не

<strong>123</strong>

а

<strong>123</strong>

Если выполнить запрос через phpmyadmin, то проблем нет. Из-за чего может быть данная ошибка? Код вывода и сохранения следующий

		$sql_update = "UPDATE `" . PREFIX . "_" . $modul_dbtitle . "` SET `autor` = '{$new_author}', `title` = '{$new_title}', `keywords` = '{$keywords}', `url` = '{$urlfile}', `screen_url` = '{$scrurl}', `url1` = '{$url1}', `desc_url1` = '{$desc_url1}', `url2` = '{$url2}', `desc_url2` = '{$desc_url2}', " . $where_ver . " " . $where_platform . "`description` = '{$short_story}', `description_full` = '{$full_story}', `title` = '{$new_title}', `category` = '{$cat_title}', `size` = '{$filesize}', `approve` = '{$approve}', `access` = '{$group_regel}', `alt_name` = '{$alt_name}', `allow_main` = '{$allow_main}', `allow_rate` = '{$allow_rating}', `allow_comm` = '{$allow_comm}', `xfields` = '{$filecontents}', `allow_br` = '{$allow_br}', `reason` = '{$editreason}', `view_edit` = '{$view_edit}', `editdate` = '{$added_time}', `editor` = '{$member_db[2]}', `to_ftp` = '{$to_ftp}', `allow_block` = '{$allow_block}', `news_fixed` = '{$news_fixed}' WHERE `id` = '{$id}'";


echo $sql_update;

		$result = $db->query ( $sql_update );

[zgr 72]

В помощь:

http://www.php.net/manual/en/function.htmlspecialchars-decode.php

http://php.net/manual/en/function.htmlspecialchars.php

[igramnet 1]

Помог) Спасибо.

PS. Решение - функция htmlspecialchars_decode