MySQL 复杂查询未产生正确结果

发布于 2024-08-01 14:04:42 字数 871 浏览 2 评论 0原文

我有两个表:车辆(Id,VIN)和图像(Id,VehicleId,名称,默认)。 我需要选择要在表格中显示的车辆 VIN 及其默认图片。 我遇到的问题是,如果未设置默认图片,我仍然想选择要显示的图像(如果存在)。 如果不存在图像,则车辆信息显然仍必须显示。 这是我到目前为止所得到的:

SELECT
    Vehicles.Id, Vehicles.VIN, Images.Name AS Image,
    (SELECT COUNT(*) FROM Images WHERE VehicleId = Vehicles.Id) AS ImageCount
FROM
    Vehicles
LEFT JOIN
    Images ON Images.VehicleId = Vehicles.Id
WHERE
    Images.Default = 1

此语句将显示具有默认图像的车辆,但如果未设置默认图像,则不会显示任何内容

为了更好地描述我的问题,这里有一些测试数据:

VEHICLES:
ID VIN
1  12341234123412341
2  23452345234523452
3  34534534534534534

IMAGES:
ID VEHICLEID NAME DEFAULT
1  1         a    1
2  1         b    0
3  2         c    0
4  2         d    0

尽管车辆 2 没有默认值,但我希望它选择要显示的图像。 此外,车辆 3 根本没有图像,但我仍然需要它在没有图像的情况下显示在表格中。 车辆 1 将显示其默认图像,因为它已设置。 我希望这能解决问题。

I have two table: Vehicles(Id, VIN) and Images(Id, VehicleId, Name, Default). I need to select the vehicles VIN and its default picture to display in a table. The problem I am having is that if a default picture is not set I still would like to select an image to display if it exists. If no images exist the vehicle information obviously must also still display. Here is what I have so far:

SELECT
    Vehicles.Id, Vehicles.VIN, Images.Name AS Image,
    (SELECT COUNT(*) FROM Images WHERE VehicleId = Vehicles.Id) AS ImageCount
FROM
    Vehicles
LEFT JOIN
    Images ON Images.VehicleId = Vehicles.Id
WHERE
    Images.Default = 1

This statement will display vehicles which have a default image but will not display anything if no default is set.

To describe my problem better here is some test data:

VEHICLES:
ID VIN
1  12341234123412341
2  23452345234523452
3  34534534534534534

IMAGES:
ID VEHICLEID NAME DEFAULT
1  1         a    1
2  1         b    0
3  2         c    0
4  2         d    0

Even though vehicle 2 has no default I want it to select an image to display. Also vehicle 3 has no images at all but I still need it to show up in the table with no image. Vehicle 1 will display its default image because it is set. I hope this clears things up.

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评论(2

囚你心 2024-08-08 14:04:42

好的,设置如下:

mysql> create table VEHICLES ( ID INT PRIMARY KEY, VIN CHAR( 17 ) );
mysql> INSERT INTO VEHICLES ( ID, VIN ) VALUES( 1, '12341234123412341' );
mysql> INSERT INTO VEHICLES ( ID, VIN ) VALUES( 2, '23452345234523452' );
mysql> INSERT INTO VEHICLES ( ID, VIN ) VALUES( 3, '34534534534534534' );

注意,我必须将 IMAGES 表中的 DEFAULT 列重命名为 DEF

mysql> CREATE TABLE IMAGES ( ID INT PRIMARY KEY, VEHICLEID INT, NAME VARCHAR(20), DEF INT );
mysql> INSERT INTO IMAGES( ID, VEHICLEID, NAME, DEF ) VALUES( 1, 1, 'a', 1 );
mysql> INSERT INTO IMAGES( ID, VEHICLEID, NAME, DEF ) VALUES( 2, 1, 'b', 0 );
mysql> INSERT INTO IMAGES( ID, VEHICLEID, NAME, DEF ) VALUES( 3, 2, 'c', 0 );
mysql> INSERT INTO IMAGES( ID, VEHICLEID, NAME, DEF ) VALUES( 4, 2, 'd', 0 );

这是解决方案。

首先,我们需要一个查询,为每辆车获取一个图像行,如果有则选择默认图像行。

为此,我们按 DEF 降序对图像进行排序(因此 1 位于顶部),然后按 VEHICLEID 分组以确保每辆车只有一行。

mysql> SELECT * FROM ( SELECT * FROM IMAGES ORDER BY DEF DESC ) sortedimages GROUP BY VEHICLEID;
+----+-----------+------+------+
| ID | VEHICLEID | NAME | DEF  |
+----+-----------+------+------+
|  1 |         1 | a    |    1 | 
|  3 |         2 | c    |    0 | 
+----+-----------+------+------+

现在,我们从 VEHICLES 表中进行选择,并使用 LEFT OUTER JOIN 进行上述查询,以确保我们始终为每辆车获取一行:

mysql> SELECT * FROM VEHICLES LEFT OUTER JOIN ( SELECT * FROM ( SELECT * FROM IMAGES ORDER BY DEF DESC ) sortedimages GROUP BY VEHICLEID ) defaultimages ON VEHICLES.ID = defaultimages.VEHICLEID;
+----+-------------------+------+-----------+------+------+
| ID | VIN               | ID   | VEHICLEID | NAME | DEF  |
+----+-------------------+------+-----------+------+------+
|  1 | 12341234123412341 |    1 |         1 | a    |    1 | 
|  2 | 23452345234523452 |    3 |         2 | c    |    0 | 
|  3 | 34534534534534534 | NULL |      NULL | NULL | NULL | 
+----+-------------------+------+-----------+------+------+

OK, here's the setup:

mysql> create table VEHICLES ( ID INT PRIMARY KEY, VIN CHAR( 17 ) );
mysql> INSERT INTO VEHICLES ( ID, VIN ) VALUES( 1, '12341234123412341' );
mysql> INSERT INTO VEHICLES ( ID, VIN ) VALUES( 2, '23452345234523452' );
mysql> INSERT INTO VEHICLES ( ID, VIN ) VALUES( 3, '34534534534534534' );

Note I had to rename the column DEFAULT to DEF in the IMAGES table:

mysql> CREATE TABLE IMAGES ( ID INT PRIMARY KEY, VEHICLEID INT, NAME VARCHAR(20), DEF INT );
mysql> INSERT INTO IMAGES( ID, VEHICLEID, NAME, DEF ) VALUES( 1, 1, 'a', 1 );
mysql> INSERT INTO IMAGES( ID, VEHICLEID, NAME, DEF ) VALUES( 2, 1, 'b', 0 );
mysql> INSERT INTO IMAGES( ID, VEHICLEID, NAME, DEF ) VALUES( 3, 2, 'c', 0 );
mysql> INSERT INTO IMAGES( ID, VEHICLEID, NAME, DEF ) VALUES( 4, 2, 'd', 0 );

And here's the solution.

First we need a query that gets one image row per vehicle, choosing the default one if there is one.

We do this by sorting the images in descending order of DEF (so the 1s are at the top), and then grouping by VEHICLEID to make sure there is only one row per vehicle.

mysql> SELECT * FROM ( SELECT * FROM IMAGES ORDER BY DEF DESC ) sortedimages GROUP BY VEHICLEID;
+----+-----------+------+------+
| ID | VEHICLEID | NAME | DEF  |
+----+-----------+------+------+
|  1 |         1 | a    |    1 | 
|  3 |         2 | c    |    0 | 
+----+-----------+------+------+

Now we select from the VEHICLES table, and LEFT OUTER JOIN to the above query to make sure we always get one row per vehicle:

mysql> SELECT * FROM VEHICLES LEFT OUTER JOIN ( SELECT * FROM ( SELECT * FROM IMAGES ORDER BY DEF DESC ) sortedimages GROUP BY VEHICLEID ) defaultimages ON VEHICLES.ID = defaultimages.VEHICLEID;
+----+-------------------+------+-----------+------+------+
| ID | VIN               | ID   | VEHICLEID | NAME | DEF  |
+----+-------------------+------+-----------+------+------+
|  1 | 12341234123412341 |    1 |         1 | a    |    1 | 
|  2 | 23452345234523452 |    3 |         2 | c    |    0 | 
|  3 | 34534534534534534 | NULL |      NULL | NULL | NULL | 
+----+-------------------+------+-----------+------+------+
云淡风轻 2024-08-08 14:04:42
SELECT
    Vehicles.Id, Vehicles.VIN, Images.Name AS Image,
    (SELECT COUNT(*) FROM Images WHERE VehicleId = Vehicles.Id) AS ImageCount
FROM
    Vehicles
LEFT JOIN
    Images ON Images.VehicleId = Vehicles.Id
WHERE
    1

应该可以解决这个问题,因为您受到默认设置为 1 的事实的限制。我认为这回答了您的问题,但可能需要针对您想要的实际结果进行细化 - (您是否想要获取图像的名称, 或者是什么?)

SELECT
    Vehicles.Id, Vehicles.VIN, Images.Name AS Image,
    (SELECT COUNT(*) FROM Images WHERE VehicleId = Vehicles.Id) AS ImageCount
FROM
    Vehicles
LEFT JOIN
    Images ON Images.VehicleId = Vehicles.Id
WHERE
    1

should do the trick, as you were limiting by the fact that default was set to 1. I think this answers your question, but may need to be refined for the actual result you want- (are you looking to get the name of the image, or what?)

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