Question

以下查询几乎是正确的，但它会导致list_taxonomies表中的重复行。我只需要list_taxonomies表中的唯一行。我一直在努力解决一些问题，但似乎无法解决问题。我已尝试LEFT OUTER和INNER加入项目表。

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查询：

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

目前的结果：

{
    "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"
}

Answer 1

从多个表聚合时，在加入之前进行聚合：

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;

加入三个表，导致重复记录

1 个答案: