Php 如何从多索引查询中获取[id]索引
下面是我的sql查询示例Php 如何从多索引查询中获取[id]索引,php,mysql,sql,Php,Mysql,Sql,下面是我的sql查询示例 $query = mysql_query("SELECT COUNT(*) as `box.id` from boxes as box left join page_boxes as pbox on box.id=pbox.bid left join page_subcribers as pages on pages.page_id=pbox.page_id
$query = mysql_query("SELECT COUNT(*) as `box.id`
from boxes as box
left join page_boxes as pbox
on box.id=pbox.bid
left join page_subcribers as pages
on pages.page_id=pbox.page_id
left join category_boxes as cbox
on box.id=cbox.bid
left join subcribers as catsb
on cbox.category_id=catsb.cid
where pages.uid='".$session_id."' or catsb.uid='".$session_id."'
and box.status='".$approval."'")or die (mysql_error());
$row = mysql_fetch_array($query);
$total = $row['id'];
我需要box.id
作为索引$total=$row['id']
来自此查询,但当我使用类似于$total=$row['id']的方法时代码>我收到错误消息
Notice: Undefined index: id in C:\xampp\htdocs\media\ctrx.php on line 12
如何获取此索引id值?只需删除反勾号即可
$query = mysql_query("SELECT COUNT(*) as box.id
或者你可以像那样使用反勾号
$query = mysql_query("SELECT COUNT(*) as `box`.`id`
但不包括表和列以及反勾号
`box.id` // <---this wrong
`box.id`/我想应该是$total=$row['box.id']
将选择COUNT(*)作为box.id
更改为选择COUNT(*)作为id
您已将计数别名为box.id
,因此这就是您在PHP中得到的,$row['box.id']
。