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Mysql下count()和sum()区别

时间:2021-07-01 10:21:17 帮助过:10人阅读

在mysql中有两个函数count()与sum()函数,有很多朋友搞不清楚,从英文的角度我们可以分析出来count是统计个数,sum是求各并且只能是数值型哦

要求:查询出2门及2门以上不及格者的平均成绩。
经常会用两种查询语句有两种:

代码如下

1. select name,sum(score < 60) ,avg(score) from result group by name having sum(score<60) >=2;

再看

算你拥有动物的总数目与“在pet表中有多少行?”是同样的问题,因为每个宠物有一个记录。COUNT(*)函数计算行数,所以计算动物数目的查询应为:

代码如下
mysql> SELECT COUNT(*) FROM pet;
+----------+
| COUNT(*) |
+----------+
| 9 |
+----------+

在前面,你检索了拥有宠物的人的名字。如果你想要知道每个主人有多少宠物,你可以使用COUNT( )函数:

代码如下
mysql> SELECT owner, COUNT(*) FROM pet GROUP BY owner;
+--------+----------+
| owner | COUNT(*) |
+--------+----------+
| Benny | 2 |
| Diane | 2 |
| Gwen | 3 |
| Harold | 2 |
+--------+----------+

注 意,使用GROUP BY对每个owner的所有记录分组,没有它,你会得到错误消息:

代码如下
mysql> SELECT owner, COUNT(*) FROM pet;
ERROR 1140 (42000): Mixing of GROUP columns (MIN(),MAX(),COUNT(),...)
with no GROUP columns is illegal if there is no GROUP BY clause

COUNT( )和GROUP BY以各种方式分类你的数据。下列例子显示出进行动物普查操作的不同方式。

每种动物的数量:

代码如下
mysql> SELECT species, COUNT(*) FROM pet GROUP BY species;
+---------+----------+
| species | COUNT(*) |
+---------+----------+
| bird | 2 |
| cat | 2 |
| dog | 3 |
| hamster | 1 |
| snake | 1 |
+---------+----------+

每种性别的动物数量:

代码如下
mysql> SELECT sex, COUNT(*) FROM pet GROUP BY sex;
+------+----------+
| sex | COUNT(*) |
+------+----------+
| NULL | 1 |
| f | 4 |
| m | 4 |
+------+----------+

(在这个输 出中,NULL表示“未知性别”。)

按种类和性别组合的动物数量:

代码如下
mysql> SELECT species, sex, COUNT(*) FROM pet GROUP BY species, sex;
+---------+------+----------+
| species | sex | COUNT(*) |
+---------+------+----------+
| bird | NULL | 1 |
| bird | f | 1 |
| cat | f | 1 |
| cat | m | 1 |
| dog | f | 1 |
| dog | m | 2 |
| hamster | f | 1 |
| snake | m | 1 |
+---------+------+----------+

若 使用COUNT( ),你不必检索整个表。例如, 前面的查询,当只对狗和猫进行时,应为:

代码如下
mysql> SELECT species, sex, COUNT(*) FROM pet
-> WHERE species = 'dog' OR species = 'cat'
-> GROUP BY species, sex;
+---------+------+----------+
| species | sex | COUNT(*) |
+---------+------+----------+
| cat | f | 1 |
| cat | m | 1 |
| dog | f | 1 |
| dog | m | 2 |
+---------+------+----------+

或, 如果你仅需要知道已知性别的按性别的动物数目:

代码如下

mysql> SELECT species, sex, COUNT(*) FROM pet
-> WHERE sex IS NOT NULL
-> GROUP BY species, sex;
+---------+------+----------+
| species | sex | COUNT(*) |
+---------+------+----------+
| bird | f | 1 |
| cat | f | 1 |
| cat | m | 1 |
| dog | f | 1 |
| dog | m | 2 |
| hamster | f | 1 |
| snake | m | 1 |
+---------+------+----------+

mysql sum

代码如下

2.select name ,count((score<60)!=0) as a,avg(score) from result group by name having a >=2;

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