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2

指摘の修正

2016/04/27 06:28

投稿

ttyp03
ttyp03

スコア17002

answer CHANGED
@@ -2,5 +2,5 @@
2
2
  SELECT ID, COUNT(ID) AS CNT FROM TABLE
3
3
  WHERE 回数=1
4
4
  GROUP BY ID
5
- HAVING COUNT(ID) > 1
5
+ HAVING COUNT(ID) = 1
6
6
  ```

1

質問に沿った回答に修正

2016/04/27 06:28

投稿

ttyp03
ttyp03

スコア17002

answer CHANGED
@@ -1,3 +1,6 @@
1
1
  ```SQL
2
- SELECT DISTINCT * FROM テーブル WHERE 回数=1
2
+ SELECT ID, COUNT(ID) AS CNT FROM TABLE
3
+ WHERE 回数=1
4
+ GROUP BY ID
5
+ HAVING COUNT(ID) > 1
3
6
  ```