演示

create table mytable (c1 string,c2 string);
insert into mytable values ('A','A'),('B',null),('C',null);

select count(distinct c1) as count_distinct from mytable;

+----------------+
| count_distinct |
+----------------+
|              3 |
+----------------+

如预期-3个不同的值-“a”、“b”和“c”

select count(distinct concat(c1,c2)) as count_distinct from mytable;

+----------------+
| count_distinct |
+----------------+
|              1 |
+----------------+

一如预期。为什么？-请参阅下一个查询

select c1,c2,concat(c1,c2) as concat_c1_c2 from mytable;

+----+------+--------------+
| c1 |  c2  | concat_c1_c2 |
+----+------+--------------+
| A  | A    | AA           |
| B  | NULL | NULL         |
| C  | NULL | NULL         |
+----+------+--------------+

用null连接产生null

select count(distinct c1,c2) as count_distinct from mytable;

+----------------+
| count_distinct |
+----------------+
|              1 |
+----------------+

缺陷！！
下面是一个解决问题的方法：

select count(distinct struct(c1,c2)) as count_distinct from mytable;

+----------------+
| count_distinct |
+----------------+
|              3 |
+----------------+