[英]SELECT just one FROM the LEFT JOINTs +mySQL
Got following Tables in my SQL-Database (simplified):在我的 SQL 数据库中得到以下表格(简化版):
Table Blogs :表博客:
+----+----------------------+----------+
| ID | Date | TitleGer |
+----+----------------------+----------+
| 1 | 2017-04-28 15:09:46 | Huhu |
| 2 | 2017-04-28 15:16:18 | Miau |
| 3 | 2017-04-28 15:17:14 | Kleff |
+----+----------------------+----------+
Table PicturesJoin :表格图片加入:
+-------------+---------+---------------------+
| IDPicture | IDBlog | Date |
+-------------+---------+---------------------+
| 86 | 1 | 2017-06-28 17:41:11 |
| 87 | 1 | 2017-06-28 17:41:11 |
+-------------+---------+---------------------+
Table Pictures :表图片:
+------+-------------------------+---------------------+
| ID | Filename | Date |
+------+-------------------------+---------------------+
| 86 | 20170512200326_320.jpg | 2017-05-12 20:03:26 |
| 87 | 20170512200326_384.jpg | 2017-05-12 20:03:30 |
+------+-------------------------+---------------------+
PictureJoin "joins" the Picture with the Blog-Table. PictureJoin 将图片与博客表“连接起来”。 Now I use following SQL-Command to joine these two Tables (Blog - PictureJoin) / (PictureJoin - Pictures) together.现在我使用以下 SQL-Command 将这两个表 (Blog - PictureJoin) / (PictureJoin - Pictures) 连接在一起。
SELECT
Blogs.ID,
Blogs.Date,
TitleGer,
Pictures.Filename
FROM
Blogs
LEFT JOIN
PicturesJoin ON PicturesJoin.IDBlog = Blogs.ID
LEFT JOIN
Pictures ON Pictures.ID = PicturesJoin.IDPicture
ORDER BY
DATE DESC
The result might look like that:结果可能如下所示:
+------+----------------------+-----------+------------------------+
| ID | Date | TitleGer | Filename |
+------+----------------------+-----------+------------------------+
| 1 | 2017-06-28 15:09:46 | Huhu | 20170512200326_320.jpg |
| 1 | 2017-06-28 15:09:46 | Huhu | 20170512200326_384.jpg |
| 2 | 2017-04-28 15:16:18 | Miau | NULL |
| 3 | 2017-04-28 15:17:14 | Kleff | NULL |
+------+----------------------+-----------+------------------------+
He makes a cross-product out of the available Pictures, which is also logical.他从可用的图片中制作了一个交叉产品,这也是合乎逻辑的。 But I want him to just use the first Picture he finds.但我希望他只使用他找到的第一张图片。 In the end it should look like that:最后它应该是这样的:
+------+----------------------+-----------+------------------------+
| ID | Date | TitleGer | Filename |
+------+----------------------+-----------+------------------------+
| 1 | 2017-06-28 15:09:46 | Huhu | 20170512200326_320.jpg |
| 2 | 2017-04-28 15:16:18 | Miau | NULL |
| 3 | 2017-04-28 15:17:14 | Kleff | NULL |
+------+----------------------+-----------+------------------------+
Tried several hours but couldn't get it to work.尝试了几个小时,但无法让它工作。 Please help!请帮忙!
The easiest approach may be to select one IDPicture per IDBlog from PicturesJoin only:最简单的方法可能是仅从 PicturesJoin 中为每个 IDBlog 选择一个 IDPicture:
SELECT
b.ID,
b.Date,
b.TitleGer,
p.Filename
FROM Blogs b
LEFT JOIN
(
SELECT
IDBlog,
MIN(IDPicture) AS IDPicture
FROM PicturesJoin
GROUP BY IDBlog
) pj ON pj.IDBlog = b.ID
LEFT JOIN Pictures p ON p.ID = pj.IDPicture
ORDER BY b.Date DESC;
SELECT b.ID,b.Date,b.TitleGer,p.Filename
FROM Blogs b
LEFT JOIN
(
SELECT main_table.*
FROM PicturesJoin main_table LEFT JOIN PicturesJoin child_table
ON (main_table.IDBlog= child_table.IDBlog AND main_table.IDPicture> child_table.IDPicture)
WHERE child_table.id IS NULL
)
OUTER_TABLE ON OUTER_TABLE .IDBlog = b.ID
LEFT JOIN Pictures p ON p.ID = pj.IDPicture
ORDER BY b.Date DESC;
Try above query.试试上面的查询。
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