以下是日期的格式:
XXXXXX001221
这个日期是 12 月 21 日
如果我想使用 CURDATE() 函数获取昨天的条目,我该怎么做?
如果你只是在做月-日(索引基于 CURDATE() 的字符串格式,即 YYYY-MM-DD):
SELECT * FROM table_name
WHERE date_field LIKE
CONCAT('%', SUBSTR(DATE_SUB(CURDATE(), INTERVAL 1 DAY) FROM 6 FOR 2),
SUBSTR(DATE_SUB(CURDATE(), INTERVAL 1 DAY) FROM 9 FOR 2));
或者使用 EXTRACT() 的另一种方法:
SELECT * FROM table_name
WHERE date_field LIKE
CONCAT('%', EXTRACT(MONTH FROM DATE_SUB(CURDATE(), INTERVAL 1 DAY)),
EXTRACT(DAY FROM DATE_SUB(CURDATE(), INTERVAL 1 DAY)));
也适用于年份:
SELECT * FROM table_name
WHERE date_field LIKE
CONCAT('%', SUBSTR(DATE_SUB(CURDATE(), INTERVAL 1 DAY) FROM 3 FOR 2),
SUBSTR(DATE_SUB(CURDATE(), INTERVAL 1 DAY) FROM 6 FOR 2),
SUBSTR(DATE_SUB(CURDATE(), INTERVAL 1 DAY) FROM 9 FOR 2));
SELECT * FROM table_name
WHERE date_field LIKE
CONCAT('%', EXTRACT(YEAR FROM DATE_SUB(CURDATE(), INTERVAL 1 DAY)),
EXTRACT(MONTH FROM DATE_SUB(CURDATE(), INTERVAL 1 DAY)),
EXTRACT(DAY FROM DATE_SUB(CURDATE(), INTERVAL 1 DAY)));
编辑:而不是WHERE date_field LIKE ...
你可以使用WHERE RIGHT(date_field, 6) = ...
codethis'回答中提到的那样:
SELECT * FROM table_name
WHERE RIGHT(date_field, 4) =
CONCAT(EXTRACT(MONTH FROM DATE_SUB(CURDATE(), INTERVAL 1 DAY)),
EXTRACT(DAY FROM DATE_SUB(CURDATE(), INTERVAL 1 DAY)));
SELECT * FROM table_name
WHERE RIGHT(date_field, 6) =
CONCAT(EXTRACT(YEAR FROM DATE_SUB(CURDATE(), INTERVAL 1 DAY)),
EXTRACT(MONTH FROM DATE_SUB(CURDATE(), INTERVAL 1 DAY)),
EXTRACT(DAY FROM DATE_SUB(CURDATE(), INTERVAL 1 DAY)));
注意:您也可以使用变量来保存计算 DATE_SUB(CURDATE(), INTERVAL 1 DAY) 的所有时间。
SET @yesterday = DATE_SUB(CURDATE(), INTERVAL 1 DAY);
SELECT ...
尝试类似:
SELECT * FROM table WHERE RIGHT(date_col, 6) = (SELECT CONVERT(VARCHAR(6), GETDATE()-1, 12) AS [YYMMDD])