小编oso*_*ejo的帖子

function Foo(){...}
Foo.bar = function (){...};

这是将静态方法添加到构造函数的唯一模式吗？特别是,是否无法在Foo()本身的定义中创建静态方法bar()？

+----+-------+-------+
| id | style | color |
+----+-------+-------+
|  1 |     1 | red   |
|  2 |     1 | blue  |
|  3 |     2 | red   |
|  4 |     2 | blue  |
|  5 |     2 | green |
|  6 |     3 | blue  |
+----+-------+-------+

查询：

SELECT style, COUNT(*) as count from t GROUP BY style WITH ROLLUP HAVING count > 1;

产生：

+-------+-------+
| style | count |
+-------+-------+
|     1 | …