回答編集履歴

1

追記

2019/02/21 08:28

投稿

sazi
sazi

スコア25195

test CHANGED
@@ -23,3 +23,25 @@
23
23
  ) step1
24
24
 
25
25
  ```
26
+
27
+ 試してないけど、こっちでもいけるかも
28
+
29
+ ```SQL
30
+
31
+ SELECT hotel_id, AVG(rating) AS rate
32
+
33
+ , CASE WHEN AVG(rating) > 4.5 THEN '非常に良い'
34
+
35
+ WHEN AVG(rating) > 3 THEN '普通'
36
+
37
+ WHEN AVG(rating) <= 3 THEN '平均未満'
38
+
39
+ ELSE '評価なし'
40
+
41
+ END AS ratingresult
42
+
43
+ FROM hotels_reviews
44
+
45
+ GROUP BY hotel_id
46
+
47
+ ```