Question

TL; DR：有没有更好的方法呢？代码有点乱，但它适用于我的应用程序

id是主题ID，id2是回复ID

$dn1 = mysql_query(
    'select c.id, c.name, c.description,c.position,c.bild,
    (
        select count(t.id)
        from topics as t
       where t.parent=c.id and t.id2=1
    ) as topics, 
    (
        select count(t2.id)
        from topics as t2
        where t2.parent=c.id and t2.id2!=1
    ) as replies 
    from categories as c
    group by c.id
    order by c.position asc'
);
$nb_cats = mysql_num_rows($dn1);
while($dnn1 = mysql_fetch_array($dn1)

Answer 1

对我而言，您的查询似乎与此相同：

select c.id, c.name, c.description, c.position, c.bild,
    count(t1.id) as topics, count(t2.id) as replies
from categories as c
    inner join topics as t1 on t1.parent = c.id and t1.id2 = 1
    inner join topics as t2 on t2.parent = c.id and t2.id2 != 1
group by c.id
order by c.position asc

我认为MySQL查询规划器对它有类似的看法。

如果您认为这种情况更具可读性，可以将涉及id2字段的条件移到WHERE子句中。

此查询将仅返回至少包含一个主题和至少一个响应的类别。如果您还需要没有主题或回复的类别，那么您必须将INNER JOIN替换为LEFT JOIN s，将保留替换为id2上的条件

PHP：检查主题和帖子 - 更干净的代码

1 个答案: