2 个表上的 UNION 每次返回一个几乎为空的结果

Question

我有一个叫做画廊的东西，它将媒体组合在一起。该媒体可以是照片或视频。我将照片存储在一个表中，将视频存储在另一个表中，因此我使用 UNION 查询来查找属于图库的照片和视频。

我

的问题似乎是我的结果包含其中一个表的空对象（没有 ID）——改写一下，如果该表中没有结果，它将始终为正在查询的其中一个表返回无用的结果。

首先，查询：

SELECT * from (
    SELECT g.id AS gallery_id,  'photo' AS type, p.id AS id, p.filename, p.caption, null AS title, null AS service, null AS embed, null AS width, null AS height, p.display_order FROM galleries g 
        LEFT OUTER JOIN photos AS p ON p.gallery_id = g.id
        WHERE g.id = {$this->id}
    UNION
    SELECT g.id AS gallery_id, 'video' AS type, v.id AS id, null AS filename, null AS caption, v.title, v.service, v.embed, v.width, v.height, v.display_order FROM galleries g 
        LEFT OUTER JOIN videos AS v ON v.gallery_id = g.id 
        WHERE g.id = {$this->id}
) AS u ORDER BY display_order;

我正在添加type列，以便我可以确定我得到的结果类型。我已经取消了表之间不常见的结果。

就像我说的，它有效，但与预期不尽如人意。如果我有一个只包含照片的画廊，我仍然得到一个（几乎）空的视频结果。示例结果：

[] => Galleries Object
    (
        [id] => 
        [name] => 
        [slug] => 
        [gallery_id] => 32
        [type] => video
        [filename] => 
        [caption] => 
        [title] => 
        [service] => 
        [embed] => 
        [width] => 
        [height] => 
        [display_order] => 
    )
[39] => Galleries Object
    (
        [id] => 39
        [name] => 
        [slug] => 
        [gallery_id] => 32
        [type] => photo
        [filename] => 39-studio-blue-pacific.jpg
        [caption] => 
        [title] => 
        [service] => 
        [embed] => 
        [width] => 
        [height] => 
        [display_order] => 1
    )

第一个结果，标有 [type]=>video 是空的，我称之为，因为它没有视频、标题、嵌入代码等的 ID......它只包含gallery_id和type。

这是我迄今为止整理的最复杂的查询，我确信我缺少一些东西。如果图库只包含视频或仅包含照片，我希望结果能够反映这一点。

作为一个黑客，我可以在回应某些内容之前检查一下这些结果是否有ID，但我知道我的查询可以改进。帮助？

Answer 1

您返回这些结果的原因是您在表上使用外部联接。试试这个：

SELECT * from (
    SELECT g.id AS gallery_id,  'photo' AS type, p.id AS id, p.filename, p.caption, null AS title, null AS service, null AS embed, null AS width, null AS height, p.display_order FROM galleries g 
        JOIN photos AS p ON p.gallery_id = g.id
        WHERE g.id = {$this->id}
    UNION
    SELECT g.id AS gallery_id, 'video' AS type, v.id AS id, null AS filename, null AS caption, v.title, v.service, v.embed, v.width, v.height, v.display_order FROM galleries g 
        JOIN videos AS v ON v.gallery_id = g.id 
        WHERE g.id = {$this->id}
) AS u ORDER BY display_order;

另外，听起来你可能会从阅读我写的这篇关于SQL的文章中获得很多好处，以帮助你更好地理解这些概念。