我想为专辑的封面选择一个随机图片。如果相册包含标记为“盖子”的图像,那么它应该只从这些图像中进行选择。如果一个子相册有“封面”标记的图像,那么它就不应该选择没有标记“覆盖”的图像。
这是一个视觉结构。
Album 1
├───Image 1
├───Album 2
│ ├───Image 2
│ ├───Image 3 (cover)
│ └───Image 4 (cover)
└───Album 3
├───Image 5
└───Image 6这是我的桌子
表标签
| id | name |
|----|-------|
| 1 | cover |表image_tag
| id | image_id | tag_id |
|----|----------|--------|
| 1 | 3 | 1 |
| 2 | 4 | 1 |表album_image
| id | album_id | image_id |
|----|----------|----------|
| 1 | 1 | 1 |
| 2 | 2 | 2 |
| 3 | 2 | 3 |
| 4 | 2 | 4 |
| 5 | 3 | 5 |
| 6 | 3 | 6 |表相册
| id | name | parent_id |
|----|---------|-----------|
| 1 | album 1 | null |
| 2 | album 2 | 1, |
| 3 | album 3 | 1, |这就是我提出的SQL。
SELECT ai.image_id AS image_id FROM albums AS a
INNER JOIN album_image AS ai ON ai.album_id = a.id
LEFT JOIN image_tag AS it ON it.image_id = ai.image_id AND it.tag_id = 1[1]
WHERE a.id = 1[2] OR parent_id LIKE '1,%'[3]
ORDER BY CASE
WHEN it.tag_id IS NOT NULL AND a.id = 1[4] THEN RAND()
ELSE RAND() + 1
END ASC LIMIT 1;1:这是PHP中另一个SQL当前的get。如果可以合并,那就太好了。是否可以左加入一个内部连接表?如果标签不存在怎么办?
2,4:这是来自PHP的专辑id。
3:这是来自PHP,以确保子相册包括在内。
该解决方案将生成以下结果。
是否有一种方法可以修改SQL以便生成预期的结果?
我在运行MySQL。
编辑用于测试的模式
CREATE TABLE tags (
id INT AUTO_INCREMENT PRIMARY KEY,
name TEXT NOT NULL
);
INSERT INTO tags (name) VALUES ('cover');
CREATE TABLE image_tag (
id INT AUTO_INCREMENT PRIMARY KEY,
image_id INT NOT NULL,
tag_id INT NOT NULL
);
INSERT INTO image_tag (image_id, tag_id) VALUES (3, 1);
INSERT INTO image_tag (image_id, tag_id) VALUES (4, 1);
CREATE TABLE album_image (
id INT AUTO_INCREMENT PRIMARY KEY,
album_id INT NOT NULL,
image_id INT NOT NULL
);
INSERT INTO album_image (album_id, image_id) VALUES (1, 1);
INSERT INTO album_image (album_id, image_id) VALUES (2, 2);
INSERT INTO album_image (album_id, image_id) VALUES (2, 3);
INSERT INTO album_image (album_id, image_id) VALUES (2, 4);
INSERT INTO album_image (album_id, image_id) VALUES (3, 5);
INSERT INTO album_image (album_id, image_id) VALUES (3, 6);
CREATE TABLE albums (
id INT AUTO_INCREMENT PRIMARY KEY,
name TEXT NOT NULL,
parent_id TEXT
);
INSERT INTO albums (name, parent_id) VALUES ('album 1', NULL);
INSERT INTO albums (name, parent_id) VALUES ('album 2', '1,');
INSERT INTO albums (name, parent_id) VALUES ('album 3', '1,');发布于 2017-12-08 02:01:41
这是我的解决办法。
SELECT ai.image_id AS image_id, (
SELECT COUNT(*) FROM albums AS a2
INNER JOIN album_image AS ai2 ON ai2.album_id=a2.id
LEFT JOIN image_tag AS it2
INNER JOIN tags AS t2 ON t2.id = it2.tag_id
ON it2.image_id = ai2.image_id AND t2.name = 'cover'
WHERE a2.id = a.id AND t2.name IS NOT NULL AND it.tag_id IS NULL
) AS count FROM albums AS a
INNER JOIN album_image AS ai ON ai.album_id=a.id
LEFT JOIN image_tag AS it
INNER JOIN tags AS t ON t.id = it.tag_id
ON it.image_id = ai.image_id AND t.name = 'cover'
WHERE a.id = 1 OR parent_id LIKE '1,%'
ORDER BY CASE
WHEN it.tag_id IS NOT NULL AND a.id = 1 THEN RAND()
WHEN count > 0 THEN RAND() + 2
ELSE RAND() + 1
END ASC LIMIT 1;https://stackoverflow.com/questions/47687471
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