Question

我有以下查询：

SELECT m.*, COUNT(c.chapter_nr) as chapters, MAX(c.chapter_nr) as latest_chapter
FROM comic_series AS m
JOIN chapters as c on c.comic_id = m.id
WHERE m.title_en = 'test'

这允许我通过给出漫画的名称来从漫画书中找到章节。该查询选择comic_series表中的所有列，并添加chapters（该漫画的章节数）和latest_chapter（最新章节）列。

现在，这个问题困扰了我很长一段时间：

我有第三列，其中包含链接到类别ID的漫画ID（类别详细信息位于第四个表格中，但这里不相关），我想通过搜索来选择漫画书及其章节类别ID。

这是我到目前为止所提出的：

SELECT `m`.*, COUNT(c.chapter_nr) as chapters, MAX(c.chapter_nr) as latest_chapter
FROM `comic_series` AS `m`
JOIN `comic_categories` AS `mc` ON mc.comic_id = m.id 
JOIN chapters as c on c.comic_id = m.id
where mc.category_id = 5
GROUP BY `m`.`id`

这会返回正确的COUNT(c.chapter_nr)但是当我添加更多类别ID时，它会返回错误的数量：

SELECT `m`.*, COUNT(c.chapter_nr) as chapters, MAX(c.chapter_nr) as latest_chapter
FROM `comic_series` AS `m`
JOIN `comic_categories` AS `mc` ON mc.comic_id = m.id 
JOIN `chapters` as c on c.comic_id = m.id
where mc.category_id = 5 OR mc.category_id = 1
GROUP BY `m`.`id`

上面的查询应该为COUNT(c.chapter_nr)返回1，但它返回2.这可能是因为comic_categories表中有2条记录用于该漫画，而chapters中只有1条记录} table。

Answer 1

根据您的查询，您选择的所有漫画书的category_id为5 AND 1.由于此处无法实现，因此您无法获得任何结果。将其更改为OR或更好的IN子句：

SELECT `m`.*, COUNT(c.chapter_nr) as chapters, MAX(c.chapter_nr) as latest_chapter
FROM `comic_series` AS `m`
JOIN `comic_categories` AS `mc` ON mc.comic_id = m.id 
JOIN chapters as c on c.comic_id = m.id
where mc.category_id IN (1, 5) -- <<<<< HERE
GROUP BY `m`.`id`

编辑（修改后的问题之后）：

使用JOIN，您的结果会爆炸。我假设，漫画有多个类别，因此您可以获得每个类别的结果。

在不进行分组的情况下尝试使用该语句，以便在总结之前掌握实际使用的内容：

SELECT `m`.*
FROM `comic_series` AS `m`
JOIN `comic_categories` AS `mc` ON mc.comic_id = m.id
JOIN chapters as c on c.comic_id = m.id
where mc.category_id IN (1, 5);

这将为您提供两行，但结果非常相似（最可能完全相同的两行）。

您可以使用子选择解决此问题，如

SELECT `m`.*, COUNT(c.chapter_nr) as chapters, MAX(c.chapter_nr) as latest_chapter
FROM `comic_series` AS `m`
JOIN chapters as c on c.comic_id = m.id
where m.id IN (SELECT comic_id FROM comic_categories WHERE category_id IN (1, 5))
GROUP BY `m`.`id`;

这将返回您的期望。这有一个陷阱：一旦数据库显着增长，Subselect与IN语法配对可能会非常慢。因此，请确保在您的网站经常访问后立即将其保留在内存中！

Answer 2

您应该使用OR代替AND：

SELECT `m`.*, COUNT(c.chapter_nr) as chapters, MAX(c.chapter_nr) as latest_chapter
FROM `comic_series` AS `m`
JOIN `comic_categories` AS `mc` ON mc.comic_id = m.id 
JOIN chapters as c on c.comic_id = m.id
where mc.category_id IN (1, 5)
GROUP BY `m`.`id`

编辑：下面的子查询将解决错误的计数。它们是在comic_series属于多个类别时引起的。然后，连接将导致重复的行。子查询没有这个问题。

SELECT m.*, COUNT(c.chapter_nr) as chapters, MAX(c.chapter_nr) as latest_chapter
FROM comic_series AS m
JOIN chapters as c on c.comic_id = m.id
WHERE m.id IN 
(
  SELECT comic_id 
  FROM comic_categories 
  WHERE category_id IN (1, 5)
)

Answer 3

当您加入类别表时，每个匹配类别会获得一行。如果漫画分为两类，您将获得重复的行数。

请尝试使用COUNT(DISTINCT c.chapter_nr)：

SELECT `m`.*, COUNT(DISTINCT c.chapter_nr) as chapters, MAX(c.chapter_nr) as latest_chapter
FROM `comic_series` AS `m`
JOIN `comic_categories` AS `mc` ON mc.comic_id = m.id 
JOIN `chapters` as c on c.comic_id = m.id
where mc.category_id = 5 OR mc.category_id = 1
GROUP BY `m`.`id`

如何为以下内容创建mysql查询？

3 个答案: