sql的优势之一是使用单一的命令能够访问到一个记录集，其重要的特点就是非过程化。
        在很多企业流行着360评分体系。比如，有A、B、C三个员工竞选某要职，HR组织吃瓜群众位对其三位进行评分，出于不清楚的原因，HR在拿到群众们的评分结果后，会先对某侯选人的得分结果记录集进行排序，再按比例去掉记录两头（最高、最低）的N%评分记录，最后再进行后续计算。
        本文是直接解决以上的“去掉记录两头的N%评分记录”的问题的。最少有以下几种方案：
      （本文代码基于SQL Server2005，关于@tb_eval表变量的定义见本文最后的附录）

-- 去掉最高20%、最低20%评分

select *
from
@tb_eval a01
where a01.id not in (select top 20 percent id from @tb_eval a02 where a01.empid=a02.empid order by a02.score desc)
and a01.id not in (select top 20 percent id from @tb_eval a02 where a01.empid=a02.empid order by a02.score asc)
order by empid, score desc


 

-- 去掉最高20%、最低20%评分

select * 
from 
(
	select row_number() over (partition by empid order by score desc) rowno, *
	from @tb_eval
) a01
where rowno > (select cast(round(count(*)*0.2,0) as int) padcnt from @tb_eval a02 where a01.empid=a02.empid)
and rowno <= (select (count(*) - cast(round(count(*)*0.2,0) as int)) padcnt from @tb_eval a02 where a01.empid=a02.empid)


 

-- 去掉最高20%、最低20%评分

select a01.*
from 
(
	select row_number() over (partition by empid order by score desc) rowno, *
	from @tb_eval
) a01
left join 
(
	select empid, cast(round(count(*)*0.2,0) as int) topcnt, cast(round(count(*)*0.2,0) as int) lastcnt, count(*) as totalcnt 
	from @tb_eval 
	group by empid
) a02
on a01.empid=a02.empid
where a01.rowno>a02.topcnt and a01.rowno <= (a02.totalcnt - a02. lastcnt)


  
本问题其实像是大学时代的课后练习作业，写作本文的原因主要是想说明sql的非过程化，你不应该使用游标去完成本任务。
 附录：
1、完整的代码

-- 人员表（本例不使用）
declare @tb_emp table table
(
    id    int,				-- 人员ID
    name    varchar(64)		-- 人员名称
);

insert into @tb_emp (id, name) values(1, 'emp01');
insert into @tb_emp (id, name) values(2, 'emp02');
insert into @tb_emp (id, name) values(3, 'emp03');
insert into @tb_emp (id, name) values(4, 'emp04');
insert into @tb_emp (id, name) values(5, 'emp05');
insert into @tb_emp (id, name) values(6, 'emp06');
insert into @tb_emp (id, name) values(7, 'emp07');
insert into @tb_emp (id, name) values(8, 'emp08');
insert into @tb_emp (id, name) values(9, 'emp09');
insert into @tb_emp (id, name) values(10, 'emp10');
insert into @tb_emp (id, name) values(11, 'emp11');
insert into @tb_emp (id, name) values(12, 'emp12');
insert into @tb_emp (id, name) values(13, 'emp13');
insert into @tb_emp (id, name) values(14, 'emp14');
insert into @tb_emp (id, name) values(15, 'emp15');
insert into @tb_emp (id, name) values(16, 'emp16');
insert into @tb_emp (id, name) values(17, 'emp17');
insert into @tb_emp (id, name) values(18, 'emp18');
insert into @tb_emp (id, name) values(19, 'emp19');
insert into @tb_emp (id, name) values(20, 'emp20');
insert into @tb_emp (id, name) values(21, 'emp21');
insert into @tb_emp (id, name) values(22, 'emp22');
insert into @tb_emp (id, name) values(23, 'emp23');
insert into @tb_emp (id, name) values(24, 'emp24');
insert into @tb_emp (id, name) values(25, 'emp25');


-- 评分表
declare @tb_eval table
(
	    id	int not null identity(1,1) primary key,		-- 评分表自动编号
	    empid int,						-- 被评分人员ID
	    score decimal(10, 2),				-- 评分
            examinerid int					-- 评分人员ID
);

-- 给empid为1, 3, 5的评分
insert into @tb_eval (empid, score, examinerid) values (1,9, 4);
insert into @tb_eval (empid, score, examinerid) values (1,76, 19);
insert into @tb_eval (empid, score, examinerid) values (1,37, 10);
insert into @tb_eval (empid, score, examinerid) values (1,90, 13);
insert into @tb_eval (empid, score, examinerid) values (1,29, 7);
insert into @tb_eval (empid, score, examinerid) values (1,6, 2);
insert into @tb_eval (empid, score, examinerid) values (1,69, 17);
insert into @tb_eval (empid, score, examinerid) values (1,76, 5);
insert into @tb_eval (empid, score, examinerid) values (1,60, 15);
insert into @tb_eval (empid, score, examinerid) values (3,13, 20);
insert into @tb_eval (empid, score, examinerid) values (3,49, 6);
insert into @tb_eval (empid, score, examinerid) values (3,72, 25);
insert into @tb_eval (empid, score, examinerid) values (3,84, 2);
insert into @tb_eval (empid, score, examinerid) values (3,4, 22);
insert into @tb_eval (empid, score, examinerid) values (3,17, 11);
insert into @tb_eval (empid, score, examinerid) values (3,74, 18);
insert into @tb_eval (empid, score, examinerid) values (3,66, 21);
insert into @tb_eval (empid, score, examinerid) values (3,62, 8);
insert into @tb_eval (empid, score, examinerid) values (3,4, 4);
insert into @tb_eval (empid, score, examinerid) values (3,66, 13);
insert into @tb_eval (empid, score, examinerid) values (3,41, 3);
insert into @tb_eval (empid, score, examinerid) values (3,94, 16);
insert into @tb_eval (empid, score, examinerid) values (5,87, 24);
insert into @tb_eval (empid, score, examinerid) values (5,71, 14);
insert into @tb_eval (empid, score, examinerid) values (5,39, 22);
insert into @tb_eval (empid, score, examinerid) values (5,55, 12);
insert into @tb_eval (empid, score, examinerid) values (5,13, 19);
insert into @tb_eval (empid, score, examinerid) values (5,73, 16);
insert into @tb_eval (empid, score, examinerid) values (5,45, 10);
insert into @tb_eval (empid, score, examinerid) values (5,9, 15);
insert into @tb_eval (empid, score, examinerid) values (5,4, 8);
insert into @tb_eval (empid, score, examinerid) values (5,31, 11);

-- 全部评分按empid分组后，每组从高到低评分
select row_number() over (partition by empid order by score desc) rowno, * 
from @tb_eval

-- 去掉最高20%、最低20%评分
select *
from 
@tb_eval a01 
where a01.id not in (select top 20 percent id from @tb_eval a02 where a01.empid=a02.empid order by a02.score desc) 
and a01.id not in (select top 20 percent id from @tb_eval a02 where a01.empid=a02.empid order by a02.score asc) 
order by empid, score desc 

-- 去掉最高20%、最低20%评分
select a01.*
from 
(
	select row_number() over (partition by empid order by score desc) rowno, * 
	from @tb_eval 
) a01 
left join 
(
	select empid, cast(round(count(*)*0.2,0) as int) topcnt, cast(round(count(*)*0.2,0) as int) endcnt, count(*) as totalcnt 
	from @tb_eval 
	group by empid 
) a02 
on a01.empid=a02.empid 
where a01.rowno>a02.topcnt and a01.rowno <= (a02.totalcnt - a02.endcnt) 

-- 去掉最高20%、最低20%评分
select * 
from 
( 
	select row_number() over (partition by empid order by score desc) rowno, * 
	from @tb_eval 
) a01 
where rowno > (select cast(round(count(*)*0.2,0) as int) padcnt from @tb_eval a02 where a01.empid=a02.empid) 
and rowno <= (select (count(*) - cast(round(count(*)*0.2,0) as int)) padcnt from @tb_eval a02 where a01.empid=a02.empid) 


2、未进行去头、去尾的结果： 


rowno


id


empid


score


examinerid


1


4


1


90


13


2


2


1


76


19


3


8


1


76


5


4


7


1


69


17


5


9


1


60


15


6


3


1


37


10


7


5


1


29


7


8


1


1


9


4


9


6


1


6


2


1


22


3


94


16


2


13


3


84


2


3


16


3


74


18


4


12


3


72


25


5


17


3


66


21


6


20


3


66


13


7


18


3


62


8


8


11


3


49


6


9


21


3


41


3


10


15


3


17


11


11


10


3


13


20


12


14


3


4


22


13


19


3


4


4


1


23


5


87


24


2


28


5


73


16


3


24


5


71


14


4


26


5


55


12


5


29


5


45


10


6


25


5


39


22


7


32


5


31


11


8


27


5


13


19


9


30


5


9


15


10


31


5


4


8



3、去头、去尾后的结果：

rowno


id


empid


score


examinerid


3


8


1


76


5


4


7


1


69


17


5


9


1


60


15


6


3


1


37


10


7


5


1


29


7


4


12


3


72


25


5


17


3


66


21


6


20


3


66


13


7


18


3


62


8


8


11


3


49


6


9


21


3


41


3


10


15


3


17


11


3


24


5


71


14


4


26


5


55


12


5


29


5


45


10


6


25


5


39


22


7


32


5


31


11


8


27


5


13


19

<pre name="code" class="cpp"><p><span style="font-family:Microsoft YaHei;font-size:18px;">①   创建一个Teacher数组，利用上面的sortTeachers(Teacher  teacher[ ] ,int count,CompareFunctionPointer  cfp)函数，调用函数(使用switch…case，通过控制台输入的数字与枚举值匹配实现调用不同的函数)来分别实现按姓名、员工编号、评分排序，其中每种排序的升、降方式都要实现。比如：分别实现对年龄按照升序、降序排列。</span></p><p><span style="font-family:Microsoft YaHei;font-size:18px;">②   输出教师数组中全部的男老师。</span></p><p><span style="font-family:Microsoft YaHei;font-size:18px;">③   输出教师数组中的所有女老师。 </span></p>


/<span style="font-size:24px;">/  Teacher.h</span>
//  Exame
//
//  Created by lanou3g on 15-3-25.
//  Copyright (c) 2015年 lanou3g. All rights reserved.
//

#import <Foundation/Foundation.h>
// 第二种定义
// 性别
// typedef enum sex
// {
//    Man,
//    Woman
// } Sex;

// typedef struct teacher
// {
//    char name[20];
//     Sex sex;
//    int age;
//    int num;
//    float score;
//    
// } Teacher;

typedef struct teacher
{
    char name[20];
    char sex;
    int age;
    int num;
    float score;
    
} Teacher;


enum FunctionName
{
    CompareNameByAscending = 1,
    CompareNameByDescending,
    CompareNumberByAscending,
    CompareNumberByDescending,
    CompareScoreByAscending,
    CompareScoreByDescending,
    PrintFemaleTeacher,
    PrintMaleTeacher
    
    
};
void printTeacher(Teacher *teacher);

typedef BOOL (*CompareFunctionPointer)(Teacher teacher1,Teacher teacher2);

void sortTeachers(Teacher teacher[],int count,CompareFunctionPointer cfp);//<span style="font-size:24px;"><strong>函数指针的利用</strong></span>

void printTeachers(Teacher *teachers,int count);

BOOL compareNumberAscending(Teacher teacher1,Teacher teacher2);
BOOL compareNumberDescending(Teacher teacher1,Teacher teacher2);

BOOL compareNameAscending(Teacher teacher1,Teacher teacher2);
BOOL compareNameDescending(Teacher teacher1,Teacher teacher2);

BOOL compareScoreDescending(Teacher teacher1,Teacher teacher2);
BOOL compareScoreAscending(Teacher teacher1,Teacher teacher2);


void printMaleTeacher(Teacher *teacher,int count);
void printFemaleTeacher(Teacher *teacher,int count);







<pre name="code" class="cpp">//
// <strong><span style="font-size:18px;"> Teacher.m</span></strong>
//  Exame
//
//  Created by lanou3g on 15-3-25.
//  Copyright (c) 2015年 lanou3g. All rights reserved.
//

#import "Teacher.h"

void printTeacher(Teacher *teacher)
{
    printf("姓名：%s,性别：%c,年龄：%d,编号：%d,评分：%f\n",teacher->name,teacher->sex,teacher->age,teacher->num,teacher->score);
}

typedef BOOL (*CompareFunctionPointer)(Teacher teacher1,Teacher teacher2);//<span style="font-size:24px;">定义函数指针</span>

void sortTeachers(Teacher teacher[],int count,CompareFunctionPointer cfp)//<span style="font-size:24px;">函数回调</span>
{
    for (int i = 0; i < count - 1; i ++) {
        for (int j = 0; j < count - i - 1; j ++) {
            if (cfp(teacher[j],teacher[j+1])) {
                Teacher tea = teacher[j];
                teacher[j] = teacher[j+1];
                teacher[j+1] = tea;
            }
        }
    }
}

void printTeachers(Teacher *teachers,int count)
{
    for (int i = 0; i < count; i ++) {
        printTeacher(&teachers[i]);
    }
}

BOOL compareNumberAscending(Teacher teacher1,Teacher teacher2)
{
    return teacher1.num > teacher2.num;
}
BOOL compareNumberDescending(Teacher teacher1,Teacher teacher2)
{
    return teacher1.num < teacher2.num;
}
BOOL compareNameAscending(Teacher teacher1,Teacher teacher2)
{
    //return strcmp(teacher1.name,teacher2.name);
    if (strcmp(teacher1.name,teacher2.name) > 0) {
        return YES;
    }
    else
        return NO;
}
BOOL compareNameDescending(Teacher teacher1,Teacher teacher2)
{
    //return strcmp(teacher2.name, teacher1.name);
    if (strcmp(teacher1.name,teacher2.name) < 0) {
        return YES;
    }
    else
        return NO;

}

BOOL compareScoreDescending(Teacher teacher1,Teacher teacher2)
{
    return teacher1.score < teacher2.score;
}
BOOL compareScoreAscending(Teacher teacher1,Teacher teacher2)
{
    return teacher1.score > teacher2.score;
}


void printMaleTeacher(Teacher *teacher,int count)
{
    for (int i = 0; i < count; i ++) {
        if (teacher[i].sex == 'm') {
            printTeacher(&teacher[i]);
        }
    }
}
void printFemaleTeacher(Teacher *teacher,int count)
{
    for (int i = 0; i < count; i ++) {
        if (teacher[i].sex == 'f') {
            printTeacher(&teacher[i]);
        }
    }
}

<strong><span style="font-size:24px;">主函数的实现</span></strong>
<pre name="code" class="cpp">//
//  main.m
//  Exame
//
//  Created by lanou3g on 15-3-25.
//  Copyright (c) 2015年 lanou3g. All rights reserved.
//

#import <Foundation/Foundation.h>
#import "Teacher.h"

int main(int argc, const char * argv[])
{
    printf("输入1：实现按照姓名进行升序排列\n输入2：实现按照姓名进行降序排列\n输入3：实现按照员工编号进行升序排列\n输入4：实现按照员工编号进行降序排列\n输入5：实现按照评分进行升序排列\n输入6：实现按照评分进行降序排列\n输入7：实现输出所有的女老师\n输入8：实现输出所有的男老师\n");
    printf("=====================\n");
    printf("请输入实现功能对应的数字：\n");

 
    Teacher teacher[4] = {{"zhangsan",'m',23,34,5.5},{"lishi",'f',45,56,3.5},{"wangwu",'m',22,12,1.3},{"songliu",'f',45,12,3.2}};
    
    int n = 0;
    while (scanf("%d",&n)!=EOF) {
       
        if (n >= 1 && n <= 8) {
            switch (n) {
                case CompareNameByAscending:
                    sortTeachers(teacher,4,compareNameAscending);
                    printTeachers(teacher,4);
                    break;
                case CompareNameByDescending:
                    sortTeachers(teacher,4,compareNameDescending);
                    printTeachers(teacher,4);
                    break;
                case CompareNumberByAscending:
                    sortTeachers(teacher,4,compareNumberAscending);
                    printTeachers(teacher,4);
                    break;
                case CompareNumberByDescending:
                    sortTeachers(teacher,4, compareNumberDescending);
                    printTeachers(teacher,4);
                    break;
                case CompareScoreByAscending:
                    sortTeachers(teacher,4, compareScoreAscending);
                    printTeachers(teacher,4);
                     break;
                case CompareScoreByDescending:
                    sortTeachers(teacher,4, compareScoreDescending);
                    printTeachers(teacher,4);
                     break;
                case PrintFemaleTeacher:
                    printFemaleTeacher(teacher,4);
                    break;
                case PrintMaleTeacher :
                    printMaleTeacher(teacher,4);
                    break;
               

                default:
                    break;
            }
        }
        
        else
            printf("请重新输入：\n");
    }
        

    return 0;
}

	

这是一个很有趣的案例。任务要求：部门每个人绩效考核评分后，表格根据分数，对所有员工进行自动排名并排序，不是显示名次，而是自动按名次重新排列员工顺序。说的多，不如看看动图化的要求吧，就像这样：看明白了么？其实一开始觉得这个任务是个悖论，一方面我算出来结果，然后根据结果又原位排列顺序，是不合逻辑的。不过我们可以换个思路，我们可以改为2张表，表1用来计算结果，表2接收表1的结果并自动排列，这样就打破了悖论。这段话没看懂也不要紧，文章看完再回来看这一段。第一步：完成表格版式，及表1数据录入为演示方便，将表1设为静态表，实际工作中加公式就可以改造为动态的。表1的排名其实可以不设置，这里加上是为了看的清楚。将员工姓名复制粘贴到表2，并用VLOOKUP从表1取得分。第二步：手动排序，并录制宏这步很关键。自动排列，首先想到这是代码干的，所以要用到宏或VBA。我们先告诉EXCEL要干什么，即录个片段，让EXCEL自动执行就好。点击录制宏，名字随意。然后选择“得分”和“员工姓名”列，记住这里一定不要反哦，因为我们是要按分数排序。然后点击“数据”选项卡——“降序”，再点击停止录制宏，可以查看代码。
至此，静态的排序操作已经完成了，我们加个排名吧。
第三步：让表格动起来这步思想是，每次我操作后，让EXCEL自动做一次自动排序工作，即调用一次“宏2”。点击“开发工具”选项卡——“Visual Basic”，就打开了VBE环境。双击左侧的Sheet1，然后右侧界面第一个下拉框选择“Worksheet”，第二个下拉框选择“SelectionChange”。然后在第一个SUB过程中写上“宏2”，关掉就好。其实前面步骤可以给“宏2”起个好点的名字，比如“自动排名排序”，这里引用的时候就很清楚。
下面就是验证操作结果的时候了！再看一遍动图：完成任务！注意保存文件！因为是有代码的，所以需要保存为启用宏的工作簿xlsm格式。你以为这就结束了？？等等，发现排名有点问题。好几个并列分，排名怎么不一样？这里我们写个公式就好，E3=COUNTIF($G$3:$G$20,">"&G3)+1，在右侧得分已经排好序的情况下进行计算。公式具体什么意思呢？简单说，就是计算这个区域数字比G3大的次数，因为牛人7已经最高分了，所以COUNTIF的结果就是0，然后加个1，就是排名；同样啊，牛人9也是并列最高分，所以COUNTIF的结果也是0，然后加个1，就实现了并列第一的目的。再到牛人3，有2个分比他高，所以排名就是2+1=3。
保存文件，打完收工。
学有用的知识，做有用的表

需求分析

1.定义一个Teacher结构体包括：姓名，性别，年龄，员工编号，评分 

2.定义一个对结构体数组排序的函数，包括一个返回值为BOOl类型的回调函数，实现按照不同的条件对结构体成员排序 

3.定义一个打印所有老师信息的函数 

4.实现多个返回值为BOOL类型的函数，用来判断排序中的if交换变量条件 

5.定义一个判断并打印所有男老师和女老师的函数 

6.定义一个枚举类型，和控制台输入功能的数字相对应 

7.使用while循环实现重新输入功能，若输入功能以外的数字，则提示重新输入

函数声明

Teacher.h

//定义一个Teacher结构体
typedef struct teacher {
    char name[20];
    char gender;
    int age;
    int number;
    float score;
}Teacher;
//声明一个枚举和用户从控制台输入的功能数字对应
enum FunctionName {
    CompareNameByAscending = 1,
    CompareNameByDescending,
    CompareNumberByAscending,
    CompareNumberByDescending,
    CompareScoreByAscending,
    CompareScoreByDescending,
    PrintMaleTeacher,
    PrintFemaleTeacher
};
//打印一个Teacher的函数
void printTeacher(Teacher *teacher);
//声明一个函数指针类型，判断条件真假
typedef BOOL(*CompareFunctionPointer)(Teacher teacher1, Teacher teacher2);
//Teacher结构体数组排序函数
void sortTeachers(Teacher teacher[], int count, CompareFunctionPointer cfp);
//打印Teach结构体数组中所有老师信息
void printTeachers(Teacher *teachers, int count);
//按照名字升序排序
BOOL compareNameByAscending(Teacher teacher1, Teacher teacher2);
//按照名字降序排序
BOOL compareNameByDescending(Teacher teacher1, Teacher teacher2);
//按编号升序排序
BOOL compareNumberByAscending(Teacher teacher1, Teacher teacher2);
//按编号降序排序
BOOL compareNumberByDescending(Teacher teacher1, Teacher teacher2);
//按评分升序排序
BOOL compareScoreByAscending(Teacher teacher1, Teacher teacher2);
//按评分降序排序
BOOL compareScoreByDescending(Teacher teacher1, Teacher teacher2);
//输出全部男老师
void printMaleTeacher(Teacher *teachers, int count);
//输出全部女老师
void printFemaleTeacher(Teacher *teachers, int count);
//打印操作说明
void printInstructions();
//重命名输出男女老师函数指针
typedef void (*PrintByGender)(Teacher *teachers, int count);



函数实现

Teacher.m

//打印Teacher的函数
void printTeacher(Teacher *teacher) {
    printf("Name:%-20sGender:%-20cAge:%-20dNumber:%-20dScore:%-20f\n", teacher->name, teacher->gender, teacher->age, teacher->number, teacher->score);
}
//Teacher结构体数组排序函数
void sortTeachers(Teacher teacher[], int count, CompareFunctionPointer cfp) {
    for (int i = 0; i < count - 1; i++) {
        for (int j = 0; j < count - 1 - i; j++) {
            if (cfp(teacher[j], teacher[j + 1])) {
                Teacher temp = teacher[j];
                teacher[j] = teacher[j + 1];
                teacher[j + 1] = temp;
            }
        }
    }
}
//打印Teach结构体数组中所有老师信息
void printTeachers(Teacher *teachers, int count) {
    for (int i = 0; i < count; i++) {
        printf("Name:%-20sGender:%-20cAge:%-20dNumber:%-20dScore:%-20f\n", teachers[i].name, teachers[i].gender, teachers[i].age, teachers[i].number, teachers[i].score);
    }
}
//按照名字升序排序
BOOL compareNameByAscending(Teacher teacher1, Teacher teacher2) {
    if (strcmp(teacher1.name, teacher2.name) > 0) {
        return YES;
    }
    return NO;
}
//按照名字降序排序
BOOL compareNameByDescending(Teacher teacher1, Teacher teacher2) {
    if (strcmp(teacher1.name, teacher2.name) < 0) {
        return YES;
    }
    return NO;
}
//按编号升序排序
BOOL compareNumberByAscending(Teacher teacher1, Teacher teacher2) {
    if (teacher1.number > teacher2.number) {
        return YES;
    }
    return NO;
}
//按编号降序排序
BOOL compareNumberByDescending(Teacher teacher1, Teacher teacher2) {
    if (teacher1.number < teacher2.number) {
        return YES;
    }
    return NO;
}
//按评分升序排序
BOOL compareScoreByAscending(Teacher teacher1, Teacher teacher2) {
    if (teacher1.score > teacher2.score) {
        return YES;
    }
    return NO;
}
//按评分降序排序
BOOL compareScoreByDescending(Teacher teacher1, Teacher teacher2) {
    if (teacher1.score < teacher2.score) {
        return YES;
    }
    return NO;
}
//输出全部男老师
void printMaleTeacher(Teacher *teachers, int count) {
    for (int i = 0; i < count; i++) {
        if (teachers[i].gender == 'm') {
            printf("Name:%-20sGender:%-20cAge:%-20dNumber:%-20dScore:%-20f\n", teachers[i].name, teachers[i].gender, teachers[i].age, teachers[i].number, teachers[i].score);
        }
    }
}
//输出全部女老师
void printFemaleTeacher(Teacher *teachers, int count) {
    for (int i = 0; i < count; i++) {
        if (teachers[i].gender == 'f') {
            printf("Name:%-20sGender:%-20cAge:%-20dNumber:%-20dScore:%-20f\n", teachers[i].name, teachers[i].gender, teachers[i].age, teachers[i].number, teachers[i].score);
        }
    }
}
//声明一个枚举和用户从控制台输入的功能数字对应
typedef enum FunctionName List;
//打印操作说明
void printInstructions() {
    printf("输入1：实现按照姓名进行升序排序\n");
    printf("输入2：实现按照姓名进行降序排序\n");
    printf("输入3：实现按照员工编号进行升序排序\n");
    printf("输入4：实现按照员工编号进行降序排序\n");
    printf("输入5：实现按照评分进行升序排序\n");
    printf("输入6：实现按照评分进行降序排序\n");
    printf("输入7：实现输出所有的男老师\n");
    printf("输入8：实现输出所有的女老师\n");
    printf("=================================\n");
    printf("请输入实现功能对应的数字：");
}



函数调用

main.m

//创建Teacher类型的结构体数组
Teacher teachers[] = {
    {"CLS", 'f', 35, 120, 100.0},
    {"XQC", 'f', 32, 100, 98.5},
    {"CGX", 'm', 36, 110, 99.5},
    {"XDL", 'f', 30, 90, 59.5},
    {"ZSY", 'm', 24, 190, 60.0}
};
int main(int argc, const char * argv[]) {
    //定义变量
    CompareFunctionPointer function = NULL;
    int n = 0;
    BOOL isAgain = YES;
    int count = sizeof(teachers) / sizeof(teachers[0]);
    //输出用户操作说明
    printInstructions();
    while (isAgain) {
        function = NULL;
        scanf("%d", &n);
        switch (n) {
            case CompareNameByAscending:
                function = compareNameByAscending;
                break;
            case CompareNameByDescending:
                function = compareNameByDescending;
                break;
            case CompareNumberByAscending:
                function = compareNumberByAscending;
                break;
            case CompareNumberByDescending:
                function = compareNumberByDescending;
                break;
            case CompareScoreByAscending:
                function = compareScoreByAscending;
                break;
            case CompareScoreByDescending:
                function = compareScoreByDescending;
                break;
            case PrintFemaleTeacher:
                printFemaleTeacher(teachers, count);
                break;
            case PrintMaleTeacher:
                printMaleTeacher(teachers, count);
                break;
            default: {
                printf("对不起您输入的数字没有对应的函数，请重新输入:");
                break;
            }
        }
        if (function != NULL) {
            sortTeachers(teachers, count, function);
            printTeachers(teachers, count);
        }
    }
    return 0;
}




案例分析

博主手把手教你玩转回调函数

拿到一个案例，博主首先想到的是通过需求分析来设计一个程序的整体架构，尽可能将程序拆分为几个独立的模块，并提高程序的内聚性，减少程序的耦合性。接下来博主将通过对本案例尽可能详细的分析来说明这一点。
首先我们先放下回调函数不管，按照由整体到局部来对这一案例进行分析，题目要求是创建一个结构体数组并且按照一定的顺序输出，那我们就可以先声明一个结构体，并且写一个函数，用来输出结构体数组的每一个成员的变量。这里楼主使用了typedef对结构体重命名，不懂的童鞋自行脑补，内事问度娘。
第一步我们做完了，题目的要求是输出排序后的信息，我们可以写一个函数对数组进行排序。我们就先做一个按照评分对老湿们进行排序吧，之前都是老湿们给我们排序，现在换做我们给老湿们排序了，想想都暗爽那！闲话少叙，使用冒泡排序，对结构体数组按照比较部分交换整体的思想进行交换。
     for (int i = 0; i < count - 1; i++) {
        for (int j = 0; j < count - 1 - i; j++) {
            if ((teacher[j].score - teacher[j + 1].score) > 0) {
                Teacher temp = teacher[j];
                teacher[j] = teacher[j + 1];
                teacher[j + 1] = temp;
            }
        }
    }

看到这里有人要说了，既然这都已经可以实现了，为什么还要使用函数回调来做？这个问题问得好！如果只按照一种规则对结构体数组排序，确实这样就已经实现了，但题目是要求按照几种方式对老湿门进行排序，这时使用函数回调就能体现其优越性了。通过分析题目，改变排序规则改变的就是冒泡排序中的if判断条件，那么如果我们能将判断条件作为一个参数传入函数的话，就可以通过在主调函数中改变排序的条件。 
(teacher[j].score - teacher[j + 1].score) > 0
按照这个思路继续想，怎样能将一个判断条件作为函数的参数传入到该函数中那？博主告诉你，回调函数就可以实现这个神奇的功能。有的看客们又要问，为什么回调函数可以作为函数参数那？这个问题问得妙，首先函数是存储在内存中一块叫做代码区的空间的，函数就存储在这块区域，并且每个函数都有一个地址，也就是指针，如果能得到这个地址的话，就可以通过调用定义指针变量指向该地址并调用该函数。 

函数名代表函数首地址，这例子代表定义一个返回值是int型并且有2个int型参数的函数指针p指向fun函数。 
int (*p)(int a, int b) = fun;
首先需要定义一个返回值是BOOL类型并且有两个Teacher结构体变量的函数 
BOOL isTrue(Teacher teacher1, Teacher teacher2); 

为了让函数指针作为参数，需要定义一个相同类型的函数指针类型 
typedef BOOL(*CompareFunctionPointer)(Teacher teacher1, Teacher teacher2); 

看到了吧，有了两个结构体参数teacher1和teacher2就可以尽情的对老湿们进行各种方式的排序了。
接下来定义几个CompareFunctionPointer类型的函数用来实现不同的排序，博主写了按照6中方式排序，看客们可以再自行编写对年龄的排序函数。 
void sortTeachers(Teacher teacher[], int count, CompareFunctionPointer cfp); 

根据博主的提示，相信大家都能完成这样一个排序的函数，其中的cfp就是我们讲到的函数指针，在函数实现部分也会很简单，还记得上面提到的冒泡排序嘛，只要在if的括号中调用函数cfp(teacher[j], teacher[j + 1])
好了，万事俱备，接下来就要见证奇迹了，回到主调函数main.m中，先传入一个按照名字升序排序的函数并打印。注意函数指针参数，博主传入的是一个函数名字，而且没有括号和参数哦，原因上面已经说过了，函数名就是函数的地址，也叫做函数指针。输出结果就是按照名字升序排序的老湿信息。
int count = sizeof(teachers) / sizeof(teachers[0]);
printInstructions();
sortTeachers(teachers, count, compareNameByAscending);
printTeachers(teachers, count);



亲们看到这里不知道你们理解了没有，小编答应说要尽量详尽的带着大家学习函数回调，那就再带着大家来总结一下吧。 

1.函数回调适用范围：这个东西是一个比普通函数更有逼格的函数，但亲们也要分析具体案例决定使用与否。一般情况下回调函数可以用来替换在函数中的表达式中，达到动态改变条件的作用。 

2.函数回调的使用步骤：首先找到要替换的表达式，写出替换的函数并实现，并定义一个相同类型的函数指针类型，在原函数中加入函数指针类型参数，用回调函数替换原函数表达式部分。
博主讲到这里函数回调部分的内容就差不多结束了，本题目中要求输入功能对应数字并打印，那能不能通过输入一个字符串来调用对应的函数那？博主先卖个关子，看客们可以自行实现输入一个运算单词，例如：sum，max，minGB实现和函数的匹配，博主将在下一篇博客中和亲们分享：返回值是函数指针的函数。亲们记得关注哦。