Question

我在mysql中有一个包含多个表的数据库，并希望将多个表连接到一个视图中，以免我不得不在PHP中构建3或4个sql语句甚至一个大的连接语句来获取相同的信息。

以下是我要加入的所有表格

track_title               
+----+------------------+
| ID |      TITLE       |
+----+------------------+
|  1 | Title Here       |
|  2 | Another Title    |
|  3 | Some Other Title |
+----+------------------+

track_artist
+----+----------------+-----------+-----------+
| ID | TRACK_TITLE_ID | ARTIST_ID | SYMBOL_ID |
+----+----------------+-----------+-----------+
| 1  | 1              | 1         | 2         |
| 2  | 1              | 2         | 1         |
| 3  | 3              | 1         | 1         |
+----+----------------+-----------+-----------+

artist

+----+-------------+
| ID | ARTIST      |
+----+-------------+
| 1  | Linkin Park |
| 2  | Metallica   |
+----+-------------+

symbol
+----+--------+
| ID | SYMBOL |
+----+--------+
| 1  |        |
| 2  | Feat.  |
+----+--------+

tracklisting
+----+----------+----------+---------------+---------+
| ID | TRACK NO | TITLE_ID | VERSION       | DISC NO |
+----+----------+----------+---------------+---------+
| 1  | 1        | 1        |               | 1       |
| 2  | 1        | 2        | Album Version | 1       |
| 3  | 1        | 3        | Live Version  | 1       |
+----+----------+----------+---------------+---------+

这是我正在寻找的最终视图

+----+----------+------------------+---------------+-----------------------------+---------+
| ID | TRACK NO | TITLE            | VERSION       | ARTIST                      | DISC NO |
+----+----------+------------------+---------------+-----------------------------+---------+
| 1  | 1        | Title Here       |               | Linkin Park Feat. Metallica | 1       |
| 2  | 1        | Another Title    | Album Version |                             | 1       |
| 3  | 1        | Some Other Title | Live Version  | Linkin Park                 | 1       |
+----+----------+------------------+---------------+-----------------------------+---------+

过去3天我一直在用左，右，加入和完全加入来抨击我的头脑，似乎无法让这个工作。

基本上我想要发生的是track_artist表将艺术家和符号组成各自的表并将它们连接成一列。然后加入title和concat列以获得此视图。

full_artist_view
+----------+------------------+-----------------------------+
| TITLE_ID | TITLE            | FULL_ARTIST                 |
+----------+------------------+-----------------------------+
| 1        | Title Here       | Linkin Park Feat. Metallica |
| 2        | Another Title    |                             |
| 3        | Some Other Title | Linkin Park                 |
+----------+------------------+-----------------------------+

我已经做到这一点，但是当我尝试将其加入到跟踪列表中时，我似乎崩溃了我的服务器，这变得非常痛苦。没有mysql错误，所以我猜我正在使用错误的连接，或者这是不可能的。（虽然我看不出这是不可能的）

跟踪列表每周持续增长1000个记录，并且位于+ - 75000条记录。

对我来说，这是应该有效的sql，但不是

FROM full_artist_view LEFT JOIN tracklisting ON 
full_artist_view.TITLE_ID = tracklisting.TITLE_ID

Answer 1

虽然我只有您的小数据样本，但我无法看到您的full_artist_view代码的样子。您应该能够使用以下内容获得结果：

select tt.id,
  tl.`track no`,
  tt.title,
  coalesce(tl.version, '') version,
  group_concat(concat(coalesce(a.artist, ''), ' ', coalesce(s.symbol, '')) order by a.artist SEPARATOR ' ') artist,
  tl.`disc no`
from track_title tt
inner join tracklisting tl
  on tt.id = tl.TITLE_ID
left join track_artist ta
  on tt.id = ta.TRACK_TITLE_ID
left join artist a
  on ta.artist_id = a.id
left join symbol s
  on ta.symbol_id = s.id
group by tt.id, tl.`track no`, tt.title, tl.version, tl.`disc no`

见SQL Fiddle with Demo。这将返回：

| ID | TRACK NO |            TITLE |       VERSION |                       ARTIST | DISC NO |
---------------------------------------------------------------------------------------------
|  1 |        1 |       Title Here |               | Linkin Park Feat. Metallica  |       1 |
|  2 |        1 |    Another Title | Album Version |                              |       1 |
|  3 |        1 | Some Other Title |  Live Version |                 Linkin Park  |       1 |

Answer 2

快速播放，我认为（但不确定）它是您需要的： -

SELECT a.Id, a.Track_No, b.Title, a.Version, GROUP_CONCAT(CONCAT(d.Artists, e.Symbol)), a.Disc_No
FROM tracklisting a
INNER JOIN track_title b ON a.Title_id = b.Id
LEFT OUTER JOIN track_artist c ON b.Id = c.Track_Title_Id
LEFT OUTER JOIN artist d ON c.Artist_Id = d.Id
LEFT OUTER JOIN symbol e ON d.SymbolId = e.Id
GROUP BY a.Id, a.Track_No, b.Title, a.Version, a.Disc_No

但我可能错过了一些让我知道的事情！

连接用于将多个表连接到一个视图中的连接

2 个答案: