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 | 复制代码 |
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1. select name,sum(score < 60) ,avg(score) from result group by name having sum(score<60) >=2; |
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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 | 复制代码 |
| mysql> SELECT COUNT(*) FROM pet; +----------+ | COUNT(*) | +----------+ | 9 | +----------+ |
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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 | 复制代码 |
| mysql> SELECT owner, COUNT(*) FROM pet GROUP BY owner; +--------+----------+ | owner | COUNT(*) | +--------+----------+ | Benny | 2 | | Diane | 2 | | Gwen | 3 | | Harold | 2 | +--------+----------+ |
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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 | 复制代码 |
| 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 |
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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 | 复制代码 |
| mysql> SELECT species, COUNT(*) FROM pet GROUP BY species; +---------+----------+ | species | COUNT(*) | +---------+----------+ | bird | 2 | | cat | 2 | | dog | 3 | | hamster | 1 | | snake | 1 | +---------+----------+ |
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每种性别的动物数量：
| 代码如下 | 复制代码 |
| mysql> SELECT sex, COUNT(*) FROM pet GROUP BY sex; +------+----------+ | sex | COUNT(*) | +------+----------+ | NULL | 1 | | f | 4 | | m | 4 | +------+----------+ |
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（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 | 复制代码 |
| 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 | +---------+------+----------+ |
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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 | 复制代码 |
| 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 | +---------+------+----------+ |
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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 | 复制代码 |
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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 | +---------+------+----------+ |
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mysql sum
| 代码如下 | 复制代码 |
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2.select name ,count((score<60)!=0) as a,avg(score) from result group by name having a >=2; |
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时间： 2024-08-01 00:07:29