「Oracle」like文で日時型を指定するサンプル

環境
Oracle Database 11g Release 11.2.0.1.0 – 64bit Production
PL/SQL Release 11.2.0.1.0 – Production

構文
select * from テーブル名
where TO_CHAR(項目名, '日付の形式’) like '指定文字列%’
TO_CHARで一旦日付型の項目名の文字列にしてからlikeを使用します。

使用例
1.USER_INFOテーブルの確認

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USERID USERNAME AGE CREATEDATE
11 to01         23 2022/10/07 17:30:37
12 user%02         24 2022/10/14 17:30:46
13 %tokyo         25 2022/09/01 17:30:55
14 user05         35 2022/08/01 17:31:02
15 osa_ka06    40 2022/08/01 17:31:14
USERID USERNAME AGE CREATEDATE 11 to01         23 2022/10/07 17:30:37 12 user%02         24 2022/10/14 17:30:46 13 %tokyo         25 2022/09/01 17:30:55 14 user05         35 2022/08/01 17:31:02 15 osa_ka06    40 2022/08/01 17:31:14
USERID	USERNAME	AGE	CREATEDATE
11	to01	        23	2022/10/07 17:30:37
12	user%02	        24	2022/10/14 17:30:46
13	%tokyo	        25	2022/09/01 17:30:55
14	user05	        35	2022/08/01 17:31:02
15	osa_ka06	   40	2022/08/01 17:31:14

2.日時型の項目名をlikeで指定します
SQL構文

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select * from user_info where to_char(createdate,'YYYY/MM') like '2022/08%'
select * from user_info where to_char(createdate,'YYYY/MM') like '2022/08%'
select * from user_info where to_char(createdate,'YYYY/MM') like '2022/08%'

3.実行結果

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USERID USERNAME AGE CREATEDATE
14     user05     35 2022/08/01 17:31:02
15     osa_ka06 40 2022/08/01 17:31:14
USERID USERNAME AGE CREATEDATE 14     user05     35 2022/08/01 17:31:02 15     osa_ka06 40 2022/08/01 17:31:14
USERID	USERNAME	AGE	CREATEDATE
14	    user05	    35	2022/08/01 17:31:02
15	    osa_ka06	40	2022/08/01 17:31:14

 

Oracle

Posted by arkgame