第一个查询的输出是来自当前用户后面的用户的iduser编号。$sql = mysql_query("SELECT * FROM followrelations Where iduser2='$uid' "); //followers from relationstable$userid1 = $row['iduser1
以下查询起作用: FROM `employee` (SELECT `department_manager(SELECT DISTINCT `manager` WHERE `manager` IS NOT NULL) WHERE `manager` IS NOT NULL))ERROR 1064 (42000
我是mysql的新手,所以请多关照。我的脚本出现了以下错误,我不确定它有什么问题。uoidWHERE category='Health, Fitness' 给出
#1064 - You have an error in your SQLsyntax; check the manual that corresponds to your MySQL server version for the right syntax to use near
我试图查询数据库的唯一结果,句子(从SQL控制台请求)是ok的,我发送到数据库的参数也是ok的,所以必须抛出一个结果。这是我的问题:
$cliente = mysql_fetch_assoc($mySQL->query("SELECT nombre FROM clientes WHERE rfc=(SELECT rfc