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.
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:
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:
- 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;
 - +------+----------+ --Vevb.com
 - | 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:
- 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 | --Vevb.com
 - | cat | m | 1 |
 - | dog | f | 1 |
 - | dog | m | 2 |
 - | hamster | f | 1 |
 - | snake | m | 1 |
 - +---------+------+----------+
 
mysql sum,代码如下:
select name,count((score<60)!=0) as a,avg(score) from result group by name having a >=2;
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