进一步细化@SQLMeance和@Brannon给出的答案,使用OP提到但@SQLMenace的回答中没有出现的GROUP by子句
CREATE TABLE table1 (
id INT
);
INSERT INTO table1 VALUES
(1),
(2),
(NULL),
(2),
(NULL),
(3),
(1),
(4),
(NULL),
(2);
SELECT * FROM table1;
+------+
| id |
+------+
| 1 |
| 2 |
| NULL |
| 2 |
| NULL |
| 3 |
| 1 |
| 4 |
| NULL |
| 2 |
+------+
10 rows in set (0.00 sec)
SELECT id, COUNT(*) FROM table1 GROUP BY id;
+------+----------+
| id | COUNT(*) |
+------+----------+
| 1 | 2 |
| 2 | 3 |
| NULL | 3 |
| 3 | 1 |
| 4 | 1 |
+------+----------+
5 rows in set (0.00 sec)
这里,COUNT(*)计数每种类型的id(包括NULL)的出现次数
SELECT id, COUNT(id) FROM table1 GROUP BY id;
+------+-----------+
| id | COUNT(id) |
+------+-----------+
| 1 | 2 |
| 2 | 3 |
| NULL | 0 |
| 3 | 1 |
| 4 | 1 |
+------+-----------+
5 rows in set (0.00 sec)
这里,COUNT(id)计算每种类型的id的出现次数,但不计算NULL的出现次数
SELECT id, COUNT(DISTINCT id) FROM table1 GROUP BY id;
+------+--------------------+
| id | COUNT(DISTINCT id) |
+------+--------------------+
| NULL | 0 |
| 1 | 1 |
| 2 | 1 |
| 3 | 1 |
| 4 | 1 |
+------+--------------------+
5 rows in set (0.00 sec)
这里,COUNT(DISTINCT id)只计算每种类型id的出现次数一次(不计算重复次数),也不计算NULL的出现次数