Sql 连接三个表,产生重复的记录
下面的查询几乎是正确的,只不过它会导致list_taxonomies表中的重复行。我只需要list_taxonomies表中唯一的行。我一直在努力寻找一些解决方案,但似乎没有找到。我在items表上尝试了左外部和内部联接 谢谢你的帮助 查询:Sql 连接三个表,产生重复的记录,sql,postgresql,join,Sql,Postgresql,Join,下面的查询几乎是正确的,只不过它会导致list_taxonomies表中的重复行。我只需要list_taxonomies表中唯一的行。我一直在努力寻找一些解决方案,但似乎没有找到。我在items表上尝试了左外部和内部联接 谢谢你的帮助 查询: SELECT lists.*, json_agg(items ORDER BY items.id) AS _items, json_agg(list_taxonomies ORDER BY list_taxonomies.ty
SELECT
lists.*,
json_agg(items ORDER BY items.id) AS _items,
json_agg(list_taxonomies ORDER BY list_taxonomies.type) AS taxonomy
FROM
lists
JOIN
list_taxonomies ON list_taxonomies.list_id = lists.id
JOIN
items ON items.list_id = lists.id
WHERE
lists.id = 3
GROUP BY
lists.id
当前结果:
{
"status": "success",
"data": [{
"id": 3,
"name": "tincidunt pede ac urna. Ut",
"description": "Lorem ipsum dolor sit amet, consectetuer adipiscing",
"created": "2016-08-24T12:00:00.000Z",
"updated": "2016-08-24T12:00:00.000Z",
"owner": 9,
"likes": 3,
"private": 0,
"location": "United States",
"nsfw": 0,
"_items": [{
"id": 2,
"name": "sem semper",
"description": "sollicitudin commodo",
"list_id": 3,
"type": 2,
"image": "http://fillmurray.com/",
"list_order": 6,
"created": "2016-08-24T05:00:00-07:00",
"link": "http://amazon.com"
}, {
"id": 14,
"name": "magna sed",
"description": "bibendum. Donec felis",
"list_id": 3,
"type": 2,
"image": "http://fillmurray.com/",
"list_order": 1,
"created": "2016-08-24T05:00:00-07:00",
"link": "http://google.com"
}],
"taxonomy": [{
"list_id": 3,
"taxonomy": "Art",
"type": 1
}, {
"list_id": 3,
"taxonomy": "Art",
"type": 1
}, {
"list_id": 3,
"taxonomy": "Art",
"type": 1
}, {
"list_id": 3,
"taxonomy": "Art",
"type": 1
}, {
"list_id": 3,
"taxonomy": "Art",
"type": 1
}, {
"list_id": 3,
"taxonomy": "Art",
"type": 1
}, {
"list_id": 3,
"taxonomy": "Art",
"type": 1
}]
}],
"message": "Retrieved list 1"
}
从多个表聚合时,请在加入前聚合:
SELECT
l.*,
i._items,
lt.taxonomy
FROM lists l
JOIN
(
select list_id, json_agg(list_taxonomies.* order by type) AS taxonomy
from list_taxonomies
group by list_id
) lt ON lt.list_id = l.id
JOIN
(
select list_id, json_agg(items.* order by id) AS _items
from items
group by list_id
) i ON i.list_id = l.id
WHERE l.id = 3;
从多个表聚合时,请在加入前聚合:
SELECT
l.*,
i._items,
lt.taxonomy
FROM lists l
JOIN
(
select list_id, json_agg(list_taxonomies.* order by type) AS taxonomy
from list_taxonomies
group by list_id
) lt ON lt.list_id = l.id
JOIN
(
select list_id, json_agg(items.* order by id) AS _items
from items
group by list_id
) i ON i.list_id = l.id
WHERE l.id = 3;
用左手试试JOIN@McNets我试过了,同样的结果。应该是分类内的项目?因为双左连接应该有效。@McNets分类法应该在数据内部,而不是在项目内部。返回的数据的结构如上所述是正确的。请使用LEFT重试JOIN@McNets我试过了,同样的结果。应该是分类内的项目?因为双左连接应该有效。@McNets分类法应该在数据内部,而不是在项目内部。返回数据的结构如上所述是正确的。这对我来说是有意义的,但是查询失败了,因为_items实际上不是items中的列。不,它不是。它是项目上json聚合的别名。我看不出有什么瑕疵。也许它需要是
items
或`items.*`而不是json\u agg(*order by id)
中的*
?您收到的确切错误消息是什么?我收到的\u项不存在。你说得对,是*本身造成了错误。一旦我改变了这一点,它就是对的。谢谢这对我来说是有意义的,但是查询失败了,因为_items实际上不是items中的列。不,它不是。它是项目上json聚合的别名。我看不出有什么瑕疵。也许它需要是items
或`items.*`而不是json\u agg(*order by id)
中的*
?您收到的确切错误消息是什么?我收到的\u项不存在。你说得对,是*本身造成了错误。一旦我改变了这一点,它就是对的。谢谢