各位大神，请教一个问题

[枫中叶]

我想在mysql的过程里，设置一个变量，得到的是上个月的第一天，但我这里into不进去，不知道该 怎么解决，求大神们帮忙看下

-- 上个月的第一天
DECLARE V_START_DAY DATETIME;
-- 获取上个月的第一天
SELECT DATE_SUB(DATE_SUB(DATE_FORMAT(NOW(),'%Y-%M-%D'),
INTERVAL EXTRACT(DAY FROM NOW())-1 DAY),INTERVAL 1 MONTH) INTO V_START_DAY FROM DUAL;

[ACMAIN_CHM]

mysql> select curdate();
+------------+
| curdate()  |
+------------+
| 2016-07-16 |
+------------+
1 row in set (0.01 sec)

mysql> select curdate()-interval day(curdate())-1 day - interval 1 month ;
+------------------------------------------------------------+
| curdate()-interval day(curdate())-1 day - interval 1 month |
+------------------------------------------------------------+
| 2016-06-01                                                 |
+------------------------------------------------------------+
1 row in set (0.00 sec)

mysql> select curdate()-interval day(curdate())-1 day - interval 1 month into @V_START_DAY  ;
Query OK, 1 row affected (0.00 sec)

mysql> select @V_START_DAY;
+--------------+
| @V_START_DAY |
+--------------+
| 2016-06-01   |
+--------------+
1 row in set (0.00 sec)

mysql>
mysql>

[枫中叶]

引用 1 楼 ACMAIN_CHM 的回复:

mysql> select curdate();
+------------+
| curdate()  |
+------------+
| 2016-07-16 |
+------------+
1 row in set (0.01 sec)

mysql> select curdate()-interval day(curdate())-1 day - interval 1 month ;
+------------------------------------------------------------+
| curdate()-interval day(curdate())-1 day - interval 1 month |
+------------------------------------------------------------+
| 2016-06-01                                                 |
+------------------------------------------------------------+
1 row in set (0.00 sec)

mysql> select curdate()-interval day(curdate())-1 day - interval 1 month into @V_START_DAY  ;
Query OK, 1 row affected (0.00 sec)

mysql> select @V_START_DAY;
+--------------+
| @V_START_DAY |
+--------------+
| 2016-06-01   |
+--------------+
1 row in set (0.00 sec)

mysql>
mysql>

非常感谢