下面是一个链接表的示例:+----+-------+---------++----+----1 | 3 |5 rows in set (0.00 sec)SELECT v.id, GROUP_CONCAT(l.tagid) as tags FROM videos v LEFT JOIN links l ON l.vid
我从mysql获取数据就像..。SELECT legal_country, count(*) FROM accounts GROUP BY legal_country ORDER BY count(*) DESC"; while ( $row = mysql_fetch_assoc( $result ) ) {
$data[] = $": "US","value&qu
例如,我正在尝试创建一个模板,该模板根据url变量?id=1从数据库中获取一行数据。当访问时,我希望显示特定行中的所有信息,例如标题。目前,我通过查看php.net并在某种程度上对其进行了修改而得到了这一点。我对如何修复这个问题知之甚少,因为我是PHP新手,没有太多经验:S。$result = mysql_query("SELECT id,title FROM table WHERE id = urldecode( $_GET['id']