+----+---------------+--------------------+------------+----------+-----------------+
| id | restaurant_id | filename | is_profile | priority | show_in_profile |
+----+---------------+--------------------+------------+----------+-----------------+
| 40 | 20 | 1320849687_390.jpg | | | 1 |
| 60 | 24 | 1320853501_121.png | 1 | | 1 |
| 61 | 24 | 1320853504_847.png | | | 1 |
| 62 | 24 | 1320853505_732.png | | | 1 |
| 63 | 24 | 1320853505_865.png | | | 1 |
| 64 | 29 | 1320854617_311.png | 1 | | 1 |
| 65 | 29 | 1320854617_669.png | | | 1 |
| 66 | 29 | 1320854618_636.png | | | 1 |
| 67 | 29 | 1320854619_791.png | | | 1 |
| 74 | 154 | 1320922653_259.png | | | 1 |
| 76 | 154 | 1320922656_332.png | | | 1 |
| 77 | 154 | 1320922657_106.png | | | 1 |
| 84 | 130 | 1321269380_960.jpg | 1 | | 1 |
| 85 | 130 | 1321269383_555.jpg | | | 1 |
| 86 | 130 | 1321269384_251.jpg | | | 1 |
| 89 | 28 | 1321269714_303.jpg | | | 1 |
| 90 | 28 | 1321269716_938.jpg | 1 | | 1 |
| 91 | 28 | 1321269717_147.jpg | | | 1 |
| 92 | 28 | 1321269717_774.jpg | | | 1 |
| 93 | 28 | 1321269717_250.jpg | | | 1 |
| 94 | 28 | 1321269718_964.jpg | | | 1 |
| 95 | 28 | 1321269719_830.jpg | | | 1 |
| 96 | 43 | 1321270013_629.jpg | 1 | | 1 |
+----+---------------+--------------------+------------+----------+-----------------+
我有这个表，我想为给定的餐馆ID列表选择文件名。 例如24,29,154：
+----+---------------
| filename |
+----+---------------
1320853501_121.png (has is_profile 1)
1320854617_311.png (has is_profile 1)
1320922653_259.png (chosen as profile picture because restaurant doesn't have a profile pic but has pictures)
我尝试了分组和案例陈述，但我无处可去。如果你使用分组，它应该是一个完整的分组。
答案 0 :(得分：2)
你可以通过聚合和一些逻辑来做到这一点：
select restaurant_id,
coalesce(max(case when is_profile = 1 then filename end),
max(filename)
) as filename
from t
where restaurant_id in (24, 29, 154)
group by restaurant_id;
首先查找/ a个人资料文件名。接下来只选择任意一个。