Question

我制作了sqlfiddle示例：http://sqlfiddle.com/#!9/3bbb6d/1/0

    SELECT
        `Image`.`id`,
        `Topic`.`title`
    FROM
        `images` AS `Image`, `topics` AS `Topic`
    WHERE
        `Image`.`vender_id` = 'model' AND `Topic`.`id` = IFNULL((SELECT article_id FROM articles_images WHERE image_id = `Image`.`id` ORDER BY id DESC LIMIT 0, 1), 999)
    ORDER BY
        `Image`.`id` DESC
    LIMIT 10

结果：（正确）

id  title
8   
7   
5   CCC
4   DDD
3   BBB
2   BBB
1

没有IFNULL(..., 999) http://sqlfiddle.com/#!9/3bbb6d/2

结果：

id  title
5   CCC
4   DDD
3   BBB
2   BBB

但是我还需要记录，并且记录中的空“ 标题” => id：8、7和1

是否可以不使用 IFNULL 而拥有所有记录？

+++更新+++

@Used_By_Already

感谢您的帮助。我得到了结果：（它有两个ID为5的记录）

id  title
8   (null)
7   (null)
5   AAA              << X
5   CCC              << correct 
4   DDD
3   BBB
2   BBB
1   (null)

有两个具有相同 ID 的记录：5，我怎么只能获得其中一个 articles_images.id 最高的记录？

我需要得到的结果是：

id  title
8   
7   
5   CCC
4   DDD
3   BBB
2   BBB
1

Answer 1

使用左联接：

SELECT
     I.id
    ,T.title
FROM images AS i
LEFT JOIN articles_images AS ai ON I.id = ai.image_id
LEFT JOIN topics AS t on ai.article_id = t.id
WHERE i.vender_id = 'model' 
ORDER BY
    i.id DESC
LIMIT 10

此结果与您请求的结果略有不同：

+----+-------+
| id | title |
+----+-------+
|  8 | NULL  |
|  7 | NULL  |
|  5 | AAA   |
|  5 | CCC   |
|  4 | DDD   |
|  3 | BBB   |
|  2 | BBB   |
|  1 | NULL  |
+----+-------+

针对RE修改的问题（包括NULL）：

SELECT
     I.id
    ,max(T.title) as title
FROM images AS i
LEFT JOIN articles_images AS ai ON I.id = ai.image_id
LEFT JOIN topics AS t on ai.article_id = t.id
WHERE i.vender_id = 'model' 
GROUP BY
     I.id
;

+----+-------+
| id | title |
+----+-------+
|  1 | NULL  |
|  2 | BBB   |
|  3 | BBB   |
|  4 | DDD   |
|  5 | CCC   |
|  7 | NULL  |
|  8 | NULL  |
+----+-------+

请参阅：https://rextester.com/LCOBW94449

为保证您使用的是最高的article_，如果我建议这样做：

SELECT
     I.id
    ,T.title
FROM ximages AS i
LEFT JOIN (
            SELECT image_id, max(article_id) article_id
            FROM xarticles_images
            GROUP BY image_id
          ) AS ai ON I.id = ai.image_id
LEFT JOIN xtopics AS t on ai.article_id = t.id
WHERE i.vender_id = 'model' 
ORDER BY
    i.id DESC
;

Answer 2

使用左联接

      SELECT
        i.`id`,
        t.`title`
    FROM
        `images` AS i left join `topics` AS  t on  i.`vender_id` = 'model' 
         and t.`id`=i.article_id

如何在没有IFNULL的情况下获得相同的结果-MySQL

2 个答案: