回答編集履歴
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追記
answer
CHANGED
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@@ -10,4 +10,15 @@
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10
10
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FROM hotels_reviews
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11
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GROUP BY hotel_id
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12
12
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) step1
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13
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+
```
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試してないけど、こっちでもいけるかも
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15
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```SQL
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16
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SELECT hotel_id, AVG(rating) AS rate
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, CASE WHEN AVG(rating) > 4.5 THEN '非常に良い'
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WHEN AVG(rating) > 3 THEN '普通'
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WHEN AVG(rating) <= 3 THEN '平均未満'
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ELSE '評価なし'
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END AS ratingresult
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22
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FROM hotels_reviews
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23
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GROUP BY hotel_id
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13
24
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```
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