存储过程中的多案例选择答案

【问题标题】：multi case selection in a stored procedure存储过程中的多案例选择
【发布时间】：2013-11-11 06:01:02
【问题描述】：

SELECT image_id, 
       CASE image_id 
         WHEN 0 THEN (SELECT image_path 
                      FROM   images 
                      WHERE  image_id IN (SELECT default_image 
                                          FROM   registration 
                                          WHERE  reg_id IN (SELECT favorite_id 
                                                            FROM   @favDT))) 
         ELSE (SELECT image_path 
               FROM   images 
               WHERE  image_id = b.image_id 
                      AND active = 1) 
       END AS image_path 
FROM   buddies b 
WHERE  reg_id IN (SELECT favorite_id 
                  FROM   @favDT)

我在这个查询中遇到了一个问题，因为select favorite_id from @favDT 在这种情况下返回了许多最喜欢的 id，但我需要获得在来自FROM buddies b where reg_id in (select favorite_id from @favDT) 部分中选择的相同的最喜欢的 id 并使用它case when 0

有什么帮助吗？

【问题讨论】：

标签： mysql sql sql-server stored-procedures case

【解决方案1】：

也许是这样的：

SELECT image_id, 
       CASE image_id 
         WHEN 0 THEN (SELECT image_path 
                      FROM   images 
                      WHERE  image_id IN (SELECT default_image 
                                          FROM   registration 
                                          WHERE  reg_id IN (SELECT favorite_id 
                                                            FROM   @favDT
                                                            WHERE favorite_id = b.reg_id
                                                           )
                                         )
                     ) 
         ELSE (SELECT image_path 
               FROM   images 
               WHERE  image_id = b.image_id 
                      AND active = 1) 
       END AS image_path 
FROM   buddies b 
WHERE  reg_id IN (SELECT favorite_id 
                  FROM   @favDT)

【讨论】：

如果没有选择行，我如何添加案例？
没有从图像中选择行？在这种情况下你想要哪个 image_path？
没有在好友中选择的行在这种情况下需要在image_id为0时进行相同的操作
但在这种情况下，您将从表图像中获取所有图像，这些图像与@favDT 中具有 reg_id 的任何缓冲区无关。在这种情况下，您可以在我的查询后添加UNION ALL SELECT image_id, image_path FROM images WHERE IMAGE_ID NOT IN (SELECT default_image FROM registration WHERE reg_id IN (SELECT favorite_id FROM @favDT WHERE favorite_id = b.reg_id))

【解决方案2】：

你需要像下面这样写条件：

SELECT image_id, 
       CASE image_id 
         WHEN 0 THEN (SELECT image_path 
                      FROM   images 
                      WHERE  image_id IN (SELECT default_image 
                                          FROM   registration 
                                          WHERE  reg_id  = b.reg_id)) 
         ELSE (SELECT image_path 
               FROM   images 
               WHERE  image_id = b.image_id 
                      AND active = 1) 
       END AS image_path 
FROM   buddies b 
WHERE  reg_id IN (SELECT favorite_id 
                  FROM   @favDT)

【讨论】：

【解决方案3】：

不带 CASE 试试这个：

SELECT b.image_id,
       im.image_path as image_path
FROM buddies b 
JOIN @favDT fdt on b.reg_id=fdt.favorite_id
JOIN images im on (b.image_id<>0 and b.image_id=im.image_id and active=1) 
                  OR
                  (b.image_id=0 and r.default_image=im.image_id)

LEFT JOIN registration r on (b.image_id=0 and fdt.favorite_id=r.reg_id)

【讨论】：