我的Plex服务器的元数据的日期不正确。我试图将metadata_items.added_at
设置为其关联的media_parts.created_at
,但您只能通过第三个表(media_items
)找到关联。
下面是相关的表模式:
| media_parts |
| ------------------------------- |
| id | media_item_id | created_at |
| media_items |
| --------------------- |
| id | metadata_item_id |
| metadata_items |
| -------------- |
| id | added_at |
我可以使用3个表的连接来获取信息:
SELECT
media_parts.created_at AS media_created_at,
metadata_items.added_at AS metadata_added_at
FROM media_parts
INNER JOIN media_items
ON media_parts.media_item_id = media_items.id
INNER JOIN metadata_items
ON media_items.metadata_item_id = metadata_items.id
但是我很难更新metadata_items
表,因为你不能在UPDATE
中使用JOIN
。
UPDATE metadata_items
SET added_at = (
SELECT
media_parts.created_at
FROM media_parts
INNER JOIN media_items
ON media_parts.media_item_id = media_items.id
INNER JOIN metadata_items
ON media_items.metadata_item_id = metadata_items.id
)
但它错误百出。
我找到的所有示例都是替换2表连接的解决方案,或者设置一个常量值。
如何使用media_items
将metadata_items
与media_parts
相关联,以便使用created_at
中的值更新added_at
?
下面是一个sql转储,用于重新生成最小的数据库:
CREATE TABLE media_parts(
id INTEGER NOT NULL PRIMARY KEY
,media_item_id INTEGER NOT NULL
,created_at VARCHAR(19) NOT NULL
);
INSERT INTO media_parts(id,media_item_id,created_at) VALUES (33,33,'2016-05-13 19:59:06');
INSERT INTO media_parts(id,media_item_id,created_at) VALUES (44,44,'2015-10-12 02:15:21');
INSERT INTO media_parts(id,media_item_id,created_at) VALUES (72,72,'2016-01-01 02:13:58');
INSERT INTO media_parts(id,media_item_id,created_at) VALUES (118,118,'2016-03-05 19:57:32');
CREATE TABLE media_items(
id INTEGER NOT NULL PRIMARY KEY
,metadata_item_id INTEGER NOT NULL
);
INSERT INTO media_items(id,metadata_item_id) VALUES (33,34);
INSERT INTO media_items(id,metadata_item_id) VALUES (44,45);
INSERT INTO media_items(id,metadata_item_id) VALUES (72,73);
INSERT INTO media_items(id,metadata_item_id) VALUES (118,117);
CREATE TABLE metadata_items(
id INTEGER NOT NULL PRIMARY KEY
,added_at VARCHAR(19) NOT NULL
);
INSERT INTO metadata_items(id,added_at) VALUES (34,'2019-05-13 20:00:54');
INSERT INTO metadata_items(id,added_at) VALUES (45,'2018-10-12 02:19:14');
INSERT INTO metadata_items(id,added_at) VALUES (73,'2019-01-01 02:42:43');
INSERT INTO metadata_items(id,added_at) VALUES (117,'2019-03-07 20:58:50');
1条答案
按热度按时间ahy6op9u1#
您可以使用SQLite的UPDATE ... FROM语法:
请参阅demo。