sql - MySQL:联合、计数和分组依据

Question

(
SELECT root_tags.tag_id, root_tags.tag_name, COUNT( root_tagged.pg_id )
FROM root_tags

LEFT JOIN root_tagged ON ( root_tagged.tag_id = root_tags.tag_id )
LEFT JOIN root_pages ON ( root_pages.pg_id =  root_tagged.pg_id )
LEFT JOIN root_granted ON ( root_granted.pg_id =  root_tagged.pg_id )

WHERE root_pages.parent_id = '5'
AND root_granted.mem_id = '3'

GROUP BY root_tags.tag_id
ORDER BY 3 DESC
)

UNION
(
SELECT root_tags.tag_id, root_tags.tag_name, COUNT( root_tagged.pg_id )
FROM root_tags

LEFT JOIN root_tagged ON ( root_tagged.tag_id = root_tags.tag_id )
LEFT JOIN root_pages ON ( root_pages.pg_id =  root_tagged.pg_id )

WHERE root_pages.parent_id = '5'
AND NOT EXISTS (
    SELECT *
    FROM root_granted
    WHERE root_granted.pg_id =  root_pages.pg_id )

GROUP BY root_tags.tag_id
ORDER BY 3 DESC
)

上面的查询返回如下结果，

tag_id  tag_name                COUNT(root_tags.tag_id)
16      expert-category-c       2
14      expert-category-a       1
15      expert-category-b       1
16      expert-category-c       1

正如您所看到的 tag_id 16 是重复的，我如何重写查询以使 tag_id 16 的计数为 3，我的意思是我希望查询返回这样的结果，

tag_id  tag_name                COUNT(root_tags.tag_id)
16      expert-category-c       3
14      expert-category-a       1
15      expert-category-b       1

我尝试使用此查询，但它返回错误...

(
SELECT root_tags.tag_id, root_tags.tag_name, COUNT( root_tagged.pg_id )
FROM root_tags

LEFT JOIN root_tagged ON ( root_tagged.tag_id = root_tags.tag_id )
LEFT JOIN root_pages ON ( root_pages.pg_id =  root_tagged.pg_id )
LEFT JOIN root_granted ON ( root_granted.pg_id =  root_tagged.pg_id )

WHERE root_pages.parent_id = '5'
AND root_granted.mem_id = '3'

)

UNION
(
SELECT root_tags.tag_id, root_tags.tag_name, COUNT( root_tagged.pg_id )
FROM root_tags

LEFT JOIN root_tagged ON ( root_tagged.tag_id = root_tags.tag_id )
LEFT JOIN root_pages ON ( root_pages.pg_id =  root_tagged.pg_id )

WHERE root_pages.parent_id = '5'
AND NOT EXISTS (
    SELECT *
    FROM root_granted
    WHERE root_granted.pg_id =  root_pages.pg_id )
)

GROUP BY root_tags.tag_id
ORDER BY 3 DESC

您能告诉我如何完成这项工作吗？

谢谢。

Answer 1

在分析您的实际查询后，下面进一步给出了更好的查询。

您可以使用 UNION ALL 而不是 UNION 合并两个查询(以保留重复项)，然后在整个集合中运行 GROUP BY。

SELECT tag_id, tag_name, SUM(CountTags) as CountTags
FROM
(
SELECT root_tags.tag_id, root_tags.tag_name, COUNT( root_tagged.pg_id ) CountTags
FROM root_tags

LEFT JOIN root_tagged ON ( root_tagged.tag_id = root_tags.tag_id )
LEFT JOIN root_pages ON ( root_pages.pg_id =  root_tagged.pg_id )
LEFT JOIN root_granted ON ( root_granted.pg_id =  root_tagged.pg_id )

WHERE root_pages.parent_id = '5'
AND root_granted.mem_id = '3'

GROUP BY root_tags.tag_id

UNION ALL

SELECT root_tags.tag_id, root_tags.tag_name, COUNT( root_tagged.pg_id ) CountTags
FROM root_tags

LEFT JOIN root_tagged ON ( root_tagged.tag_id = root_tags.tag_id )
LEFT JOIN root_pages ON ( root_pages.pg_id =  root_tagged.pg_id )

WHERE root_pages.parent_id = '5'
AND NOT EXISTS (
    SELECT *
    FROM root_granted
    WHERE root_granted.pg_id =  root_pages.pg_id )

GROUP BY root_tags.tag_id
) SQ
GROUP BY tag_id, tag_name
ORDER BY CountTags DESC

由于您的 WHERE 子句针对 root_granted 和 root_pages 进行过滤，因此这些实际上是 INNER JOIN。您还可以使用 EXISTS 测试来模拟 UNION 的第一部分，假设每个 root_pages 记录永远不能拥有超过 1 个 root_granted 记录。

SELECT root_tags.tag_id, root_tags.tag_name, COUNT( root_tagged.pg_id ) CountTags
FROM root_tags
INNER JOIN root_tagged ON ( root_tagged.tag_id = root_tags.tag_id )
INNER JOIN root_pages ON ( root_pages.pg_id =  root_tagged.pg_id )
WHERE root_pages.parent_id = '5'
AND (NOT EXISTS (
    SELECT *
    FROM root_granted
    WHERE root_granted.pg_id =  root_pages.pg_id )
OR EXISTS (
    SELECT *
    FROM root_granted
    WHERE root_granted.pg_id =  root_pages.pg_id AND root_granted.mem_id = '3'))
GROUP BY root_tags.tag_id, root_tags.tag_name
ORDER BY CountTags DESC

由于 not contains 和 exists 是互斥的，因此您可以使用 OR 将它们组合起来进行单个查询。