[英]Joining Tables for COUNT Query
我正在使用以下脚本来显示页面,其中$ URL与页面URL匹配(例如MySite/People/Carl_Sagan
)...
$sql= "SELECT COUNT(URL) AS num FROM people WHERE URL = :url";
$stmt = $pdo->prepare($sql);
$stmt->bindParam(':url',$MyURL,PDO::PARAM_STR);
$stmt->execute();
$Total = $stmt->fetch();
switch($Total['num'])
{
case 1:
break;
case 2:
break;
default:
break;
}
在另一个站点上,我想将多个表连接在一起,形成一个小型百科全书。 我知道如何使用UNION
命令,但不适用于此查询。 请注意,表gz_life中的目标字段名为Taxon,而不是URL。 我以为可以以某种方式为它命名-Taxon AS URL-但这似乎也不起作用。
$sql= "SELECT COUNT(URL) AS num FROM pox_topics WHERE URL = :url
UNION ALL
SELECT COUNT(URL) AS num FROM people WHERE URL = :url
UNION ALL
SELECT COUNT(Taxon) AS num FROM gz_life WHERE Taxon = :url";
谁能告诉我在PDO查询中将表连接在一起的最佳方法?
有几种方法可以做到这一点(如果我了解您要实现的目标)。 一种方法是使用您已有的方法,但执行最后一步来累加计数:
SELECT SUM(num) FROM (
SELECT COUNT(URL) AS num FROM pox_topics WHERE URL = :url
UNION ALL
SELECT COUNT(URL) AS num FROM people WHERE URL = :url
UNION ALL
SELECT COUNT(Taxon) AS num FROM gz_life WHERE Taxon = :url
) as subquery
请注意,您将需要子查询的别名以使查询正确。
如果您可以将计数保持在一行上,则可以编写一个仅使用一次: :url
的查询:
select tp.num_topic, tp.num_people, count(*) as num_taxon
from (select t.url, t.num_topic, count(*) as num_people
from (SELECT t.url, COUNT(URL) AS num_topic
from pox_topics t
where t.URL = :url
) t join
people p
on t.url = p.url
) tp join
gz_life gl
on gl.Taxon = tp.url;
这会将:url
变成一列,然后使用连续的子查询层来计算计数。
编辑:
某些计数可能为0。要处理这种情况:
select tp.num_topic, tp.num_people, count(gl.taxon) as num_taxon
from (select t.url, t.num_topic, count(p.url) as num_people
from (SELECT const.url, COUNT(t.URL) AS num_topic
from (select :url as url) const left outer join
pox_topics t
on t.url = const.url
) t left outer join
people p
on t.url = p.url
) tp left outer join
gz_life gl
on gl.Taxon = tp.url;
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