您必须实现一个条件来模拟模式7行为。

SELECT @y := 2013 as `@y`, @w := 1 as `@w`,
       @d := STR_TO_DATE(CONCAT(@y, @w, ' Monday'), '%x%v %W') as `@d`,
       IF (YEAR(@d) < @y, DATE_ADD(@d, INTERVAL 1 WEEK), @d) as first_day;

你会得到：

+------+----+------------+------------+
| @y   | @w | @d         | first_day  |
+------+----+------------+------------+
| 2013 |  1 | 2012-12-31 | 2013-01-07 |
+------+----+------------+------------+

2007年1月1日是一年中的第一个星期一：

SELECT @y := 2007 as `@y`, @w := 1 as `@w`,
       @d := STR_TO_DATE(CONCAT(@y, @w, ' Monday'), '%x%v %W') as `@d`,
       IF (YEAR(@d) < @y, DATE_ADD(@d, INTERVAL 1 WEEK), @d) as first_day;

+------+----+------------+------------+
| @y   | @w | @d         | first_day  |
+------+----+------------+------------+
| 2007 |  1 | 2007-01-01 | 2007-01-01 |
+------+----+------------+------------+

新增2018-10-25：
所以，显然这对2013年第2周不起作用，它在中使用上述查询结果 2013-01-07 实际上是第一周的星期一。事后看来，基于我最初提出的方法，这是很明显的，显而易见的解决方案突然出现在我眼前。

SELECT @y := 2013 as `@y`, @w := 1 as `@w`,
       @d := STR_TO_DATE(CONCAT(@y, @w, ' Monday'), '%x%v %W') as `@d`,
       IF (WEEK(@d, 7) <> @w, DATE_ADD(@d, INTERVAL 1 WEEK), @d) as first_day;

+------+----+------------+------------+
| @y   | @w | @d         | first_day  |
+------+----+------------+------------+
| 2013 |  1 | 2012-12-31 | 2013-01-07 |
+------+----+------------+------------+

SELECT @y := 2013 as `@y`, @w := 2 as `@w`,
       @d := STR_TO_DATE(CONCAT(@y, @w, ' Monday'), '%x%v %W') as `@d`,
       IF (WEEK(@d, 7) <> @w, DATE_ADD(@d, INTERVAL 1 WEEK), @d) as first_day;

+------+----+------------+------------+
| @y   | @w | @d         | first_day  |
+------+----+------------+------------+
| 2013 |  2 | 2013-01-07 | 2013-01-14 |
+------+----+------------+------------+

嘿。