Question

我正在尝试编写一个有点复杂的MySQL查询，它将计算一列中列出的与其他条件相关的不同项目。

目前，该列中的唯一项目是：MC，SR，MP和EV。

这段代码确实有效，但我的问题是，有更好更有效的方法吗？另外，添加到列中的新项目怎么样？

$sql = "Select ps.id, ps.primary_skill, 
count(*) 
(select COUNT(*)
from items i, item_primary_skill_map ipsm 
where i.id = ipsm.item_id 
and ipsm.skill_id = ps.id 
and i.itemType = 'MC') as mc_count,
(select COUNT(*) 
from items i, item_primary_skill_map ipsm 
where i.id = ipsm.item_id 
and ipsm.skill_id = ps.id 
and i.itemType = 'SR') as sr_count,
(select COUNT(*) 
from items i, item_primary_skill_map ipsm 
where i.id = ipsm.item_id 
and ipsm.skill_id = ps.id 
and i.itemType = 'MP') as mp_count,
(select COUNT(*) 
from items i, item_primary_skill_map ipsm 
where i.id = ipsm.item_id 
and ipsm.skill_id = ps.id 
and i.itemType = 'EV') as ev_count,
(select COUNT(*) 
from items i, item_primary_skill_map ipsm 
where i.id = ipsm.item_id 
and ipsm.skill_id = ps.id 
and i.dok = '1') as dok1,
(select COUNT(*) 
from items i, item_primary_skill_map ipsm 
where i.id = ipsm.item_id 
and ipsm.skill_id = ps.id 
and i.dok = '2') as dok2,
(select COUNT(*) 
from items i, item_primary_skill_map ipsm 
where i.id = ipsm.item_id 
and ipsm.skill_id = ps.id 
and i.dok = '3') as dok3,
(select COUNT(*) 
from items i, item_primary_skill_map ipsm 
where i.id = ipsm.item_id 
and ipsm.skill_id = ps.id 
and i.dok = '4') as dok4
FROM primary_skills ps
where grade =  '$pl' and subject='$subject'
";

Answer 1

你可以试试这个： -

Select ps.id, ps.primary_skill, count(*), 
       CASE i.itemType = 'MC' THEN COUNT(*) END AS as mc_count,
       CASE i.itemType = 'SR' THEN COUNT(*) END AS as sr_count,
       CASE i.itemType = 'MP' THEN COUNT(*) END AS as mp_count,
       CASE i.itemType = 'EV' THEN COUNT(*) END AS as ev_count,
       CASE i.dok = '1' THEN COUNT(*) END,
       CASE i.dok = '2' THEN COUNT(*) END,
       CASE i.dok = '3' THEN COUNT(*) END,
       CASE i.dok = '4' THEN COUNT(*) END
from items i, item_primary_skill_map ipsm
where i.id = ipsm.item_id 
and ipsm.skill_id = ps.id FROM 
and grade =  '$pl' 
and subject='$subject'

我认为这是解决问题的方法。

Answer 2

怎么样

select i.itemType, count(*)
  from items i, item_primary_skill_map ipsm, primary_skills ps
 where i.id = ipsm.item_id 
   and ipsm.skill_id = ps.id 
   and ps.grade      = '$pl' 
   and ps.subject    = '$subject'      
 group by i.itemType;

不同之处在于，此查询每个唯一值返回1行，而不是查询中的1列。对于i.dok你需要做同样的事情。

Answer 3

我提出的解决方案使用循环：

$sql = "Select ps.id, ps.primary_skill ";
foreach ($iTypes as $type){
    $sql .=", (select COUNT(*)
    from items i, item_primary_skill_map ipsm 
    where i.id = ipsm.item_id 
    and ipsm.skill_id = ps.id 
    and i.itemType = '$type') as " . strtolower($type) . "_count ";
}
$sql .= "FROM primary_skills ps
where grade =  '$pl' and subject='$subject'";

感谢您的所有回复。

在单个列内计数以获得不同的结果

3 个答案: