进一步详细说明@SQLMeance 和@Brannon 使用GROUP BY
OP 提到但@SQLMenace 的回答中没有出现的子句给出的答案
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)
在这里,COUNT(*)
计算每种类型id
包括的出现次数NULL
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)
计算每种类型的出现次数,id
但不计算出现的次数NULL
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,只COUNT(DISTINCT id)
统计每种类型的出现次数id
(不计算重复),也不统计出现的次数NULL