0

我有这样的查询：

    SELECT 
    `om_chapter`.`manganame` as `link`,
    (SELECT `manganame` FROM `om_manga` WHERE `Active` = '1' AND `om_manga`.`link` = `om_chapter`.`manganame` LIMIT 0,1) AS `manganame`,
    (SELECT `cover` FROM `om_manga` WHERE `Active` = '1' AND `om_manga`.`link` = `om_chapter`.`manganame` LIMIT 0,1) AS `cover`,
    (SELECT `othername` FROM `om_manga` WHERE `Active` = '1' AND `om_manga`.`link` = `om_chapter`.`manganame` LIMIT 0,1) AS `othername`
FROM `om_chapter`
WHERE 
    `Active` = '1' AND 
    (SELECT `Active` From `om_manga` WHERE `om_manga`.`link` = `om_chapter`.`manganame` LIMIT 0,1) AND 
    `id` IN ( SELECT MAX(`id`) FROM `om_chapter` WHERE `Active` = '1' GROUP BY `manganame` )
ORDER BY `id` DESC LIMIT 10

那么我怎样才能使这个查询更快呢？

这是我的表：

om_章节：

id  | manganame     | chapter   | Active
-----------------------------------------
1   | naruto        | 1         | 1
2   | naruto        | 12        | 1
3   | naruto        | 22        | 1
4   | bleach        | 10        | 1
5   | bleach        | 15        | 1
6   | gents         | 1         | 1
7   | naruto        | 21        | 1

om_manga:

id  | othername | manganame     | cover     | Active
-----------------------------------------------------
1   | naruto    | naruto        | n.jpg     | 1
2   | bleach    | bleach        | b.jpg     | 1
4   | gents     | gents         | g.jpg     | 1

我想要形成这个查询的第一件事是通过将 manganame 分组并按 id 排序给我 om_chapter 的最后 10 行。我尝试通过使用 group 甚至 distinct 来使用简单的查询，但它们都没有给我正确的结果......

在使用 group 或 distinct 的简单查询中，结果如下：

id  | manganame     | chapter   | Active
-----------------------------------------
7   | prince        | 21        | 1
5   | gent          | 15        | 1
2   | naruto        | 12        | 1
1   | bleach        | 1         | 1

但我想要这个结果：

id  | manganame     | chapter   | Active
-----------------------------------------
9   | gents         | 21        | 1
8   | bleach        | 21        | 1
7   | prince        | 21        | 1
6   | naruto        | 1         | 1

所以我用这个：

WHERE 
`Active` = '1' AND 
(SELECT `Active` From `om_manga` WHERE `om_manga`.`link` = `om_chapter`.`manganame` LIMIT 0,1) AND 
`id` IN ( SELECT MAX(`id`) FROM `om_chapter` WHERE `Active` = '1' GROUP BY `manganame` )

我在 where 中使用 sub select 因为我希望 om_manga 表中的 Active 字段为 1..

对于子选择的重置，我实际上没有尝试加入，但我会..！

4

4 回答 4

1

我可能误解了你的意图。但这里有一个尝试：

SELECT c.`manganame` AS `link`
     , m.`manganame`
     , m.`cover`
     , m.`othername`
FROM 
     `om_manga` m 
     INNER JOIN `om_chapter` c 
     ON m.`link` = c.`manganame`
     INNER JOIN 
     ( SELECT `manganame`, MAX(`id`) AS `maxid` 
       FROM `om_chapter` 
       WHERE `Active` = '1' 
       GROUP BY `manganame` ) mx
     ON mx.`maxid` = c.`id`
ORDER BY c.`id` DESC LIMIT 10
于 2011-11-22T19:13:28.283 回答
1

我将在 om_chapter 表中引入一个外键约束，以说明从漫画到相应章节的链接。

这就是我将问题概念化的方式。

A manga can have many chapters.  A chapter belongs to one manga.

然后我会改变 om_chapter 表，包括一个外键，用于链接到漫画的章节。

ALTER TABLE om_Chapter (
ADD mangaID int references om_Manga (id)
)

并删除 manganame 列，因为它现在只是多余的

 ALTER TABLE om_Chapter (
 DROP COLUMN manganame
)

你的桌子可能看起来像这样。

om_manga:

id  | othername | manganame     | cover     | Active
-----------------------------------------------------
1   | naruto    | naruto        | n.jpg     | 1
2   | bleach    | bleach        | b.jpg     | 1
4   | gents     | gents         | g.jpg     | 1

om_章节：

id  | chapter   | Active  | mangaID
-----------------------------------------
1   | 1         | 1       |  1
2   | 12        | 1       |  1
3   | 22        | 1       |  1
4   | 10        | 1       |  2
5   | 15        | 1       |  2
6   | 1         | 1       |  4

最后你可以像这样查询表

SELECT TOP 10 m.Manganame as link,
  m.Manganame,
  m.cover,
  m.othername,

FROM om_manga as m INNER JOIN
  om_chapter as c ON m.ID = c.mangaID

WHERE m.active = 1 AND c.active = 1
ORDER BY m.ID DESC
于 2011-11-22T19:30:55.057 回答
0

为什么不简单的加入？

SELECT om_chapter.manganame, cover, othername
FROM om_chapter
JOIN om_manga ON om_chapter.manganame = om_manga=manganame
WHERE om_chapter.Active = 1 AND om_manga.Active = 1

除非我误读了你的版本。

于 2011-11-22T19:03:54.833 回答
0

使用左外连接（并丢失子查询和反引号）：

SELECT c.manganame AS link,
       m.manganame AS manganame,
       m.cover     AS cover,
       m.othername AS `othername
  FROM om_chapter    AS c
  LEFT JOIN om_manga AS m
    ON c.manganame = m.manganame
 WHERE c.Active = '1'
   AND c.id IN (SELECT MAX(o.id)
                  FROM om_chapter AS o
                 WHERE o.active = 1
                 GROUP BY o.manganame)
 ORDER BY c.id DESC LIMIT 10

如果是我的查询，我可能也会选择“ c.id AS id”。

于 2011-11-22T19:09:40.073 回答