MySQL使用具有多个连接的组_CONCAT

我有以下四张桌子。我的查询工作正常，但我需要“AUTHORIZED_VIEWER”和“AUTHORIZED_VIEWER_EMAIL”字段返回所有值，而不仅仅是第一个值。我相信这可以通过使用GROUP_CONCAT来实现，但是，我不确定这部分应该如何实现。注意-当尝试使用GROUP_CONCAT时，我必须使用以下语法，因为它返回一个BLOB：

以下是四张表格：

users_tbl
+-----+------------------+
|id   |email             |  
+-----+------------------+
|10   | scott@co.com     |
|8    | cesar@co.com     |
|11   | kevin@co.com     |
|12   | jake@co.com      |
+-----+------------------+

authorized_viewers_tbl (authorized_viewer linked to id in users_tbl)
+-----+------------+------------------+
|id   |lightbox_id |authorized_viewer |   
+-----+------------+------------------+
|1    | 50         |11                |
|7    | 50         |8                 |
|3    | 31         |11                |
|5    | 30         |8                 |
|6    | 30         |11                |
|8    | 16         |11                |
|9    | 16         |10                |
|10   | 5          |10                |
|11   | 5          |11                |
+-----+------------+------------------+

lightboxes_tbl
+-----+------------------+---------------+
|id   |lightbox_name     |author         |   
+-----+------------------+---------------+
|5    | Test Lightbox #1 |jake@co.com    |
|16   | Test Lightbox #2 |cesar@co.com   |
|30   | Test Lightbox #3 |scott@co.com   |
|31   | Test Lightbox #4 |kevin@co.com   |
|50   | Test Lightbox #5 |cesar@co.com   |
+-----+------------------+---------------+

lightbox_assets_tbl
+-------+-------------+------------------+------------------=---+----------+
|id     |lightbox_id  |asset_name        |asset_path            | asset_id |
+-------+-------------+------------------+----------------------+----------+
|232    |30           |b757.jpg          |SWFs/b757.jpg         | 3810     |
|230    |31           |b757.jpg          |SWFs/b757.jpg         | 3810     |
|233    |16           |a321_takeoff.jpg  |SWFs/a321_takeoff.jpg | 3809     |
|234    |31           |a321_takeoff.jpg  |SWFs/a321_takeoff.jpg | 3809     |
|235    |50           |a330_landing.png  |SWFs/a330_landing.png | 3789     |
+-------+-------------+------------------+-----------------------+---------+

以下是我当前使用的查询：

SELECT lb.id,
   lb.lightbox_name,
   lb.author,
   avt.authorized_viewer,
   u.email AS authorized_viewer_email,
   COUNT(lba.lightbox_id) total_assets
FROM lightboxes_tbl lb
LEFT JOIN lightbox_assets_tbl lba ON lb.id = lba.lightbox_id
LEFT JOIN authorized_viewers_tbl avt ON avt.lightbox_id = lb.id
LEFT JOIN users_tbl u ON u.id = avt.authorized_viewer
WHERE lb.author = 'scott@co.com'
  OR avt.authorized_viewer =
    (SELECT id
     FROM users_tbl
     WHERE email = 'scott@co.com')
GROUP BY lb.id
ORDER BY lb.lightbox_name ASC

谢谢

[编辑] 基于SQL FIDLE的预期结果：

 +-------+----------------+--------------+-------------------+--------------------------+--------------+
 |id     |lightbox_name   |author        |authorized_viewer  | email                    | total_assets |
 +-------+----------------+--------------+-------------------+--------------------------+--------------+
 |5      |Test Lightbox#1 |jake@co.com   |10,11              |scott@co.com,kevin@co.com |0             |             
 |16     |Test Lightbox#2 |cesar@co.com  |10,11              |scott@co.com,kevin@co.com |1             |
 |30     |Test Lightbox#3 |scott@co.com  |11,8               |kevin@co.com,cesar@co.com |1             |
 +-------+-------------+-----------------+-------------------+--------------------------+--------------+

试试这个：-

SELECT lb.id,
   lb.lightbox_name,
   lb.author,
   avt.authorized_viewer,
   u.email AS authorized_viewer_email,
   COUNT(lba.lightbox_id) total_assets
FROM lightboxes_tbl lb
LEFT JOIN lightbox_assets_tbl lba ON lb.id = lba.lightbox_id
LEFT JOIN authorized_viewers_tbl avt ON avt.lightbox_id = lb.id
LEFT JOIN users_tbl u ON u.id = avt.authorized_viewer
WHERE lb.author = 'scott@co.com'
OR avt.authorized_viewer =
(SELECT id
 FROM users_tbl
 WHERE email = 'scott@co.com')
GROUP BY lb.id, lb.lightbox_name, lb.author, avt.authorized_viewer, u.email
ORDER BY lb.lightbox_name

---

有一种更干净的方法可以做到这一点，但我还没有时间去思考

这是一个有趣的问题，非常感谢大家的分享，希望我们能帮上忙

我们将

group\u concat

添加到

avt.authorized\u viewer

和

u.email

我们将

distinct

添加到

组_concat

中，以便仅根据请求提取唯一值

我们为每个非聚合值添加了

groupby

我们修改了where子句，将Scott作为审阅者的所有灯箱都拉进去。通过使用author字段作为限制，我们排除了其他审阅者。通过基于灯箱Id的过滤器，我们保留了所有用户；这允许

组\u concat

按需工作

---

希望这能帮你解决问题！（从基本主题中清除了一些评论，因为我现在已经包含了它们或这里收集的信息。）

@xQbert这很接近。排在哪scott@co.com'不是作者，而是授权的_查看器，它只是列表scott@co.com. 如果可能的话，我需要它列出所有授权的观众，在哪里scott@co.com是其中之一。这有意义吗？谢谢是的，我认为你的第一次尝试更接近。所以我期待的是任何一个灯箱scott@co.com'是否返回作者，包括所有授权的观众（以及列出的所有其他信息-看起来是正确的）。此外，任何灯箱scott@co.com'是可返回的授权\u查看器，包括其他授权\u查看器。用户可以是作者或授权的_查看器，但不能同时是两者。希望这有帮助。谢谢你的持续帮助。我正看着这把小提琴，它列出了30磅的两项资产——我不知道为什么。不管怎么说，我只是看了你最新的小提琴，我相信它是正确的！！！谢谢你的帮助。如果你想创建一个答案，我会将它标记为正确。好的，我可以复制它。看看这个SQL FIDLE，最后一行-LB#31。A&B列有重复的条目。简单的方法是将distinct添加到group by<代码>组密码（不同的avt.授权的\u查看器）a，组密码（不同的u.电子邮件）b，需要不同的密码。这是因为我们需要lightbox\u assets\u tbl中的双重输入。我绝对不是SQL专家，所以如果有更干净的方法来编写此文件，并且您有时间，我很乐意看到它。再次感谢！！！

SELECT lb.id,
   lb.lightbox_name,
   lb.author,
   avt.authorized_viewer,
   u.email AS authorized_viewer_email,
   COUNT(lba.lightbox_id) total_assets
FROM lightboxes_tbl lb
LEFT JOIN lightbox_assets_tbl lba ON lb.id = lba.lightbox_id
LEFT JOIN authorized_viewers_tbl avt ON avt.lightbox_id = lb.id
LEFT JOIN users_tbl u ON u.id = avt.authorized_viewer
WHERE lb.author = 'scott@co.com'
OR avt.authorized_viewer =
(SELECT id
 FROM users_tbl
 WHERE email = 'scott@co.com')
GROUP BY lb.id, lb.lightbox_name, lb.author, avt.authorized_viewer, u.email
ORDER BY lb.lightbox_name

SELECT lb.id,
       lb.lightbox_name,
       lb.author,
       group_concat(distinct avt.authorized_viewer) a,
       group_concat(distinct u.email) b,
       COUNT(distinct lba.id) total_assets
FROM lightboxes_tbl lb
LEFT JOIN lightbox_assets_tbl lba ON lb.id = lba.lightbox_id
LEFT JOIN authorized_viewers_tbl avt ON avt.lightbox_id = lb.id
LEFT JOIN users_tbl u ON u.id = avt.authorized_viewer
where lb.author = 'scott@co.com'
or 
lb.id in (Select lightbox_ID 
          from authorized_Viewers_tbl X
          INNER JOIN users_Tbl U on U.ID = X.authorized_Viewer
          WHERE email = 'scott@co.com')
GROUP BY lb.id, lb.lightbox_name, lb.author
ORDER BY lb.lightbox_name ASC