SQL查询重复

我的查询是这样设置（不能正常工作）SQL查询重复

SELECT RentACar.dbo.Car.car_id as clm_CAR_ID, 
RentACar.dbo.Car_Image.car_image_url as clm_PICTURE_URL, 
RentACar.dbo.Users.user_agent_name as clm_USERNAME 

from RentACar.dbo.Car 

left join RentACar.dbo.Users 
on RentACar.dbo.Car.car_user_id = RentACar.dbo.Users.user_id 

left join RentACar.dbo.Car_Image 
on RentACar.dbo.Car.car_id = RentACar.dbo.Car_Image.car_image_car_id

我的目标是，在显示所有car_table汽车蒙山的USER_TABLE user_name和 picture_url image_table的。

但在image_table我有一个问题.. 一辆车可以有多个图片。

此查询结果;

clm_CAR_ID  clm_PICTURE_URL         clm_USERNAME   
1  NULL           ac ad 
2  NULL           ac ad 
3  NULL           ac ad 
4  NULL           ac ad 
5  /upload/10CeskUntitled23-11.png         ac ad 
5  /upload/10CveloperDesktopdeskUntitl4ed-11.png       ac ad 
6  NULL           ac ad 
7  NULL           ac ad 
8  NULL           ac ad 
9  /upload/9ngsdeveloperDesktopmelekkım.jpg         ac ad 
9  /upload/9eskt1opmelekkktc.jpg         ac ad 
10  /upload/10CeveloperDesktopd2eskxm.jpg         ac ad 
10  /upload/10ngsdeveloperDesktopdeskxm.jpg         ac ad 
10  /upload/10sdeveloperDesktopdeskU5ntitled-11.png       ac ad

但我expaction结果

clm_CAR_ID  clm_PICTURE_URL         clm_USERNAME   
1  NULL           ac ad 
2  NULL           ac ad 
3  NULL           ac ad 
4  NULL           ac ad 
5  /upload/10CeskUntitled23-11.png         ac ad 
6  NULL           ac ad 
7  NULL           ac ad 
8  NULL           ac ad 
9  /upload/9ngsdeveloperDesktopmelekkım.jpg         ac ad 
10  /upload/10CeveloperDesktopd2eskxm.jpg         ac ad

来源

2012-04-03 Mehmet

如何处理拥有多个图像的汽车 - 每辆汽车每张图像一排，每辆汽车一行，汽车首张（或最后一张）图像，每辆汽车一行，或连接成一个单一的领域......？另外，这是SQLServer？ – 2012-04-03 12:31:31

你可以在你的问题中发布你期望的输出吗？ – GarethD 2012-04-03 14:09:35

@GarethD编辑我的职务 – Mehmet 2012-04-03 15:24:05

既然你没有指定哪个RDBMS您使用我张贴的最通用的答案我可以想出：

SELECT Car_ID AS clm_CAR_ID, 
     Car_Image_URL AS clm_PICTURE_URL, 
     user_agent_name AS clm_USERNAME 
FROM Car 
     INNER JOIN Users 
      ON Car_User_ID = User_ID 
     LEFT JOIN 
     ( SELECT Img.* 
      FROM Car_Image Img 
        INNER JOIN 
        ( SELECT MIN(Image_ID) AS Image_ID 
         FROM Car_Image 
         GROUP BY Car_Image_Car_ID 
        ) AS MaxImg 
         ON MaxImg.Image_ID = Img.Image_ID 
     ) AS Img 
      ON Img.Car_Image_Car_ID = Car_ID;

我已经在This SQL Fiddle上测试过它，它适用于MySQL，MSSQL，PostgreSQL和SQLLite。我几乎没有使用Oracle的经验，因此无法在使用Oracle进行测试时修复语法错误。

来源

2012-04-03 13:14:14 GarethD

感谢GarethD我尝试这回每一辆汽车，但结果是错误的:( '选择car_id，USER_NAME，car_image_url FROM RentACar.dbo.Car INNER JOIN RentACar.dbo.Users ON car_user_id = USER_ID LEFT JOIN （SELECT图。* FROM 图RentACar.dbo.Car_Image INNER JOIN （SELECT MAX（car_image_id）AS Image_ID FROM RentACar.dbo.Car_Image GROUP BY car_image_car_id）AS MaxImg ON MaxImg .Image_ID = Img.car_image_id）AS Img ON Img.car_image_id = car_id' – Mehmet 2012-04-03 13:41:05

你能否给我一个线索，看看结果出了什么问题？当我不知道问题是什么时，纠正某些事情是相当困难的。 – GarethD 2012-04-03 13:43:59

RentACar.dbo.Car上有5张车图片。当查询运行时，程序只返回两个车号为6和9的图像，但车号为6的车没有图像。 – Mehmet 2012-04-03 13:55:28

你应该加入Car_Table，USER_TABLE。
然后加入Image_Table，其中包含每辆车一张照片。

SELECT car_id, user_name, image_url 
FROM 
    (Car_Table JOIN User_Table AS cars_and_photos ON car_user_id = user_id) JOIN 
    (SELECT * FROM Image_Table GROUP_BY image_car_id) AS one_photo_per_car 
ON one_photo_per_car.image_car_id = cars_and_photos.user_id

我没有测试此查询

来源

2012-04-03 12:34:28

喜雅罗斯瓦夫我没有运行此， '选择RentACar.dbo.Car.car_id，RentACar.dbo.Users.user_agent_name，RentACar.dbo.Car_Image.car_image_url FROM （RentACar.dbo.Car JOIN RentACar.dbo.Users ON RentACar.dbo.Car.car_user_id = USER_ID）AS cars_and_photos JOIN （SELECT * FROM RentACar.dbo.Car_Image GROUP_BY RentACar.dbo.Car_Image.car_image_car_id）AS one_photo_per_car ON one_photo_per_car.image_car_id = cars_and_photos.user_id '结果 **消息156，级别15，状态1，行3 关键字'AS'附近的语法不正确。 Msg 102，Level 15，State 1，Line 5 'RentACar'附近的语法不正确。** – Mehmet 2012-04-03 13:18:30

尝试'SELECT RentACar.dbo.Car.car_id，RentACar.dbo.Users.user_agent_name，RentACar.dbo.Car_Image.car_image_url FROM （RentACar.dbo.Car JOIN RentACar.dbo.Users AS cars_and_photos ON RentACar.dbo.Car.car_user_id = user_id）JOIN（SELECT * FROM RentACar.dbo.Car_Image GROUP BY RentACar.dbo.Car_Image.car_image_car_id）AS one_photo_per_car ON one_photo_per_car。 image_car_id = cars_and_photos.user_id' – 2012-04-03 14:01:56

SELECT rentacar.dbo.car.car_id, 
     rentacar.dbo.car_image.car_image_url AS image, 
     rentacar.dbo.users.user_agent_name AS USER 
from rentacar.dbo.car 
left join rentacar.dbo.car_image 
on rentacar.dbo.car.car_id = rentacar.dbo.car_image.image_car_id 
left join rentacar.dbo.users 
on rentacar.dbo.car.car_user_id = rentacar.dbo.users.user_id

来源

2012-04-03 12:36:52

感谢Hagbart席琳这也跑不过， 'SELECT DISTINCT RentACar.dbo.Car.car_id， RentACar.dbo.Car_Image.car_image_url为Resim， RentACar.dbo.Users.user_agent_name为Sahip 来自RentACar.dbo的。汽车左加入RentACar.dbo.Car_Image 上RentACar.dbo.Car.car_id = RentACar.dbo.Car_Image.car_image_car_id 左连接上RentACar.dbo.Car.car_user_id = RentACar.dbo.Users RentACar.dbo.Users 。 user_id' 程序返回Image_Table中有多少辆车的图片数量。我从这项计划希望只有一次 – Mehmet 2012-04-03 13:10:58

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