MySQL - Tip

Database 2015.12.02 11:09 Posted by 파란크리스마스

MySQL - Update 질의문 사용시 조회된 결과로 반영

출처 : MySQL - UPDATE query based on SELECT Query

update item_code a
       left join item_type b on a.type_name like ( concat('%', b.type_name , '%') )
   set a.type_id = b.id
UPDATE code_type a
       LEFT OUTER JOIN code_group b 
       ON a.g_id = b.id AND b.category_id in ('M', 'E')
   SET a.t_code = CONCAT(b.g_code, SUBSTR(a.t_code, 4, 2))
 WHERE a.g_id = b.id AND b.category_id in ('M', 'E')

MySQL - 날짜 연산

출처 : 12.7 Date and Time Functions

mysql> select now(), date_sub(now(), interval 1 DAY);
+---------------------+---------------------------------+
| now()               | date_sub(now(), interval 1 DAY) |
+---------------------+---------------------------------+
| 2015-12-03 23:44:52 | 2015-12-02 23:44:52             |
+---------------------+---------------------------------+
1 row in set (0.00 sec)

MySQL - DATETIME 타입을 DATE 타입으로 형변환

select *
  from memo1
 where survey_date = cast(now() as DATE)

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