我越來越模糊更新數據option_id
@option_name
在同一時間。我如何更新數據到MySQL
當前分貝
option_id option_name option_content option_status
1 web_url http://localhost.com 1
2 web_name My Website 1
3 web_description Welcome to my website 1
4 web_keywords movies, power, ranger 1
PHP更新數據
$web_name = $_POST['web_name'];
$web_url = $_POST['web_url'];
$web_desc = $_POST['web_desc'];
$web_keyword = $_POST['web_keyword'];
更新從DR anwser
$query = "UPDATE web_options SET option_content=
'{$db->string_escape($web_name, true)}'
WHERE option_name = 'web_name'";
$db->rq($query);
$query = "UPDATE web_options SET option_content=
'{$db->string_escape($web_url, true)}'
WHERE option_name = 'web_url'";
$db->rq($query);
$query = "UPDATE web_options SET option_content=
'{$db->string_escape($web_desc, true)}'
WHERE option_name = 'web_desc'";
$db->rq($query);
$query = "UPDATE web_options SET option_content=
'{$db->string_escape($web_keyword, true)}'
WHERE option_name = 'web_keyword'";
$db->rq($query);
有有辦法讓這個更新查詢更簡單嗎?
又是什麼問題? – 2010-11-29 09:40:06
你能否澄清一下你想用什麼來更新? – deceze 2010-11-29 09:59:44
看起來像對我的問題大聲笑錯誤的查詢.. DR是給了一個線索。但我想短暫查詢。 – Blur 2010-11-29 10:08:48