根据日期从db中选择行

Question

我在mysql表中有一个行列表

+--------+-----+------------+------------+------------+-----------+
| off_id | uid | leave_from | leave_to   | leave_code | reason    |
+--------+-----+------------+------------+------------+-----------+
|      1 |   1 | 2012-01-01 | 2012-01-05 | OFF        | asdsda    |
|      2 |   1 | 2012-01-15 | 2012-01-16 | OFF        | asdd      |
|      5 |   1 | 2012-02-03 | 2012-02-05 | OFF        | gfjghjhgj |
+--------+-----+------------+------------+------------+-----------+

我必须选择日期2012-01-01到2012-01-05之间的行。 我该怎么办呢，请帮忙。

Answer 1

SELECT * 
FROM tbl
WHERE leave_from >= '2012-01-01'
AND leave_to <= '2012-01-05'

Answer 2

select * from table where leave_from between '2012-01-01' and '2012-01-05'

Answer 3

SELECT Off_Id,uid,Leave_from,leave_to,leave_code
FROM YourTableName
WHERE Leave_From>="2012-01-01" and Leave_to <="2012-01-05"

Answer 4

如果我理解你的问题：

SELECT * FROM `table_name` 
WHERE leave_from >= "2012-01-01" AND leave_to <= "2012-01-05"

Answer 5

第1点：你可以使用> << => =和date数据类型。

以上可以通过实现

SELECT * FROM table WHERE leave_from > '2012-01-01' AND leave_from <'2012-01-05';