在MySQL中需要查询表中不重复的记录时，可以使用distinct关键字过滤重复记录。
语法：
select distinct <字段名1>[,<字段名2>...,<字段名n>] from <表名>;
数据表如下：
mysql> select * from emp;
+-------+--------+------------+------+------------+------+------+--------+
| empno | ename | job | mgr | hiredate | sal | comm | deptno |
+-------+--------+------------+------+------------+------+------+--------+
| 7369 | smith | clerk | 7902 | 1980-12-17 | 800 | NULL | 20 |
| 7499 | allen | salesman | 7698 | 1981-02-20 | 1600 | 300 | 30 |
| 7521 | ward | salesman | 7698 | 1981-02-22 | 1250 | 500 | 30 |
| 7566 | jones | manager | 7839 | 1981-04-02 | 2975 | NULL | 20 |
| 7654 | martin | salesman | 7698 | 1981-09-28 | 1250 | 1400 | 30 |
| 7698 | blake | manager | 7839 | 1981-05-01 | 2850 | NULL | 30 |
| 7782 | clark | manager | 7839 | 1981-06-09 | 2450 | NULL | 10 |
| 7788 | scott | analyst | 7566 | 1987-04-19 | 3000 | NULL | 20 |
| 7839 | king | persident | NULL | 1981-11-17 | 5000 | NULL | 10 |
| 7844 | turner | salesman | 7698 | 1981-09-08 | 1500 | 0 | 30 |
| 7876 | adams | clerk | 7788 | 1987-05-23 | 1100 | NULL | 20 |
| 7900 | james | clerk | 7698 | 1981-12-03 | 950 | NULL | 30 |
| 7902 | ford | analyst | 7566 | 1981-12-03 | 3000 | NULL | 20 |
| 7934 | miller | clerk | 7782 | 1982-01-23 | 1300 | NULL | 10 |
+-------+--------+------------+------+------------+------+------+--------+
示例1：单个字段去重
mysql> select distinct deptno from emp;
+--------+
| deptno |
+--------+
| 20 |
| 30 |
| 10 |
+--------+
mysql> select distinct job from emp;
+------------+
| job |
+------------+
| clerk |
| salesman |
| manager |
| analyst |
| persident |
+------------+
示例2：多个字段去重
mysql> select distinct deptno,job from emp;
+--------+------------+
| deptno | job |
+--------+------------+
| 20 | clerk |
| 30 | salesman |
| 20 | manager |
| 30 | manager |
| 10 | manager |
| 20 | analyst |
| 10 | persident |
| 30 | clerk |
| 10 | clerk |
+--------+------------+
多个字段去重时，distinct关键字必须位于第一个字段前，多个字段完全一样的情况下，才会过滤。
非重复计数：
select count(distinct <字段名1>[,<字段名2>...,<字段名n>]) from <表名>;
示例：
mysql> select count(distinct deptno,job) from emp;
+----------------------------+
| count(distinct deptno,job) |
+----------------------------+
| 9 |
+----------------------------+
