熟练掌握sql语句的力量

<?php



/*



燕十八 公益PHP培训 

课堂地址:YY频道88354001 

学习社区:www.zixue.it 



表名:stu



有如下表及数据

+------+---------+-------+

| name | subject | score |

+------+---------+-------+

| 张三 | 数学    |    90 |

| 张三 | 语文    |    50 |

| 张三 | 地理    |    40 |

| 李四 | 语文    |    55 |

| 李四 | 政治    |    45 |

| 王五 | 政治    |    30 |

+------+---------+-------+



要求:查询出2门及2门以上不及格者的平均成绩



*/



//正想思维



//1 先找出挂科超过两门的同学



$sql="select name from stu group by name having sum(score<60)>=2;";

//+------+

//| name |

//+------+

//| 张三 |

//| 李四 |

//+------+

//2 求出所有同学的平均分



$sql="select name,avg(score) from stu group by name;";



//+------+------------+

//| name | avg(score) |

//+------+------------+

//| 张三 |    60.0000 |

//| 李四 |    50.0000 |

//| 王五 |    30.0000 |

//| 赵六 |    88.0000 |

//+------+------------+

//有1和2的结果,我们知道结果只要从张三,李四两个人的平均分就可以了



$sql="select name,avg(score) from stu  where name in ('张三','李四') group by name;"



//+------+------------+

//| name | avg(score) |

//+------+------------+

//| 张三 |    60.0000 |

//| 李四 |    50.0000 |

//+------+------------+



//将语句整合

$sql="select name ,avg(score) from stu where name in (select name from (select name from stu group by name having sum(score<60)>=2)as temp) group by name;";

//+------+------------+

//| name | avg(score) |

//+------+------------+

//| 张三 |    60.0000 |

//| 李四 |    50.0000 |

//+------+------------+



//方法2 方向思维



//1想求出平均值



$sql='select name,avg(score) from stu group by name;';

//+------+------------+

//| name | avg(score) |

//+------+------------+

//| 张三 |    60.0000 |

//| 李四 |    50.0000 |

//| 王五 |    30.0000 |

//| 赵六 |    88.0000 |

//+------+------------+

//再求挂科超过2门的学生



$sql="select name,avg(score),sum(score<60) as gk from stu group by name having gk>=2;";



//+------+------------+------+

//| name | avg(score) | gk   |

//+------+------------+------+

//| 张三 |    60.0000 |    2 |

//| 李四 |    50.0000 |    2 |

//+------+------------+------+



//熟悉sql语句,一句话可以搞定,而且效率比前面的高了好几倍