#4、查询成绩表中最近一次考试的学号、科目、成绩(使用Max函数)
时间: 2024-01-29 12:02:15 浏览: 135
假设成绩表的表名为Scores,包含字段有:学号(StuID)、科目(Subject)和成绩(Grade),可以使用如下SQL语句查询最近一次考试的学号、科目、成绩:
```
SELECT StuID, Subject, MAX(Grade) as MaxGrade
FROM Scores
GROUP BY StuID, Subject
ORDER BY MaxGrade DESC
LIMIT 1
```
说明:
1. 使用MAX函数获取每个学生每个科目的最高成绩;
2. 使用GROUP BY对学号和科目进行分组;
3. 使用ORDER BY按照最高成绩降序排序;
4. 使用LIMIT 1限制结果只显示第一行,即最高成绩的那一行。
相关问题
使用C语言结构体来写一个程序某班有最多不超过30人(具体人数由键盘输入)参加期末考试,考试科目最多不超过6门(具体门数由键盘输入),定义结构体类型,用结构体数组作函数参数,编程实现如下学生成绩管理: (1)录入每个学生的学号、姓名和各科考试成绩;(2)计算每门课程的总分和平均分; (3)计算每个学生的总分和平均分; (4)按每个学生的总分由高到低排出名次表;(5)按每个学生的总分由低到高排出名次表;(6)按学号由小到大排出成绩表; (7)按姓名的字典顺序排出成绩表; (8)按学号查询学生排名及其各科考试成绩;(9)按姓名查询学生排名及其各科考试成绩; (10)输出每个学生的学号、姓名、各科考试成绩、总分、平均分,以及每门课程的总分和平均分。
以下是使用C语言结构体实现学生成绩管理的程序:
```c
#include <stdio.h>
#include <string.h>
#define MAX_STUDENTS 30
#define MAX_SUBJECTS 6
#define MAX_NAME_LENGTH 20
#define MAX_ID_LENGTH 10
struct Subject {
char name[MAX_NAME_LENGTH];
int total_score;
int num_students;
};
struct Student {
char name[MAX_NAME_LENGTH];
char id[MAX_ID_LENGTH];
int scores[MAX_SUBJECTS];
int total_score;
float avg_score;
};
void input_students(struct Student students[], int num_students, int num_subjects) {
printf("Enter student information:\n");
for (int i = 0; i < num_students; i++) {
printf("Student %d:\n", i+1);
printf("Name: ");
scanf("%s", students[i].name);
printf("ID: ");
scanf("%s", students[i].id);
for (int j = 0; j < num_subjects; j++) {
printf("Score for subject %d: ", j+1);
scanf("%d", &students[i].scores[j]);
students[i].total_score += students[i].scores[j];
}
students[i].avg_score = (float)students[i].total_score / num_subjects;
}
}
void calculate_subject_totals(struct Subject subjects[], struct Student students[], int num_students, int num_subjects) {
for (int j = 0; j < num_subjects; j++) {
strcpy(subjects[j].name, "");
subjects[j].total_score = 0;
subjects[j].num_students = 0;
for (int i = 0; i < num_students; i++) {
subjects[j].total_score += students[i].scores[j];
subjects[j].num_students++;
}
strcpy(subjects[j].name, "Subject ");
char num[2];
sprintf(num, "%d", j+1);
strcat(subjects[j].name, num);
}
}
void calculate_student_totals(struct Student students[], int num_students, int num_subjects) {
for (int i = 0; i < num_students; i++) {
students[i].total_score = 0;
for (int j = 0; j < num_subjects; j++) {
students[i].total_score += students[i].scores[j];
}
students[i].avg_score = (float)students[i].total_score / num_subjects;
}
}
void sort_students_by_total_score_desc(struct Student students[], int num_students) {
for (int i = 0; i < num_students-1; i++) {
for (int j = i+1; j < num_students; j++) {
if (students[i].total_score < students[j].total_score) {
struct Student temp = students[i];
students[i] = students[j];
students[j] = temp;
}
}
}
}
void sort_students_by_total_score_asc(struct Student students[], int num_students) {
for (int i = 0; i < num_students-1; i++) {
for (int j = i+1; j < num_students; j++) {
if (students[i].total_score > students[j].total_score) {
struct Student temp = students[i];
students[i] = students[j];
students[j] = temp;
}
}
}
}
void sort_students_by_id(struct Student students[], int num_students) {
for (int i = 0; i < num_students-1; i++) {
for (int j = i+1; j < num_students; j++) {
if (strcmp(students[i].id, students[j].id) > 0) {
struct Student temp = students[i];
students[i] = students[j];
students[j] = temp;
}
}
}
}
void sort_students_by_name(struct Student students[], int num_students) {
for (int i = 0; i < num_students-1; i++) {
for (int j = i+1; j < num_students; j++) {
if (strcmp(students[i].name, students[j].name) > 0) {
struct Student temp = students[i];
students[i] = students[j];
students[j] = temp;
}
}
}
}
void print_student_ranking(struct Student students[], int num_students) {
printf("Ranking by total score (descending):\n");
printf("%-10s%-20s%-10s%-10s%-10s%-10s%-10s%-10s\n", "Rank", "Name", "ID", "Subject1", "Subject2", "Subject3", "Subject4", "Total");
int rank = 1;
for (int i = 0; i < num_students; i++) {
if (i == 0 || students[i].total_score < students[i-1].total_score) {
rank = i+1;
}
printf("%-10d%-20s%-10s", rank, students[i].name, students[i].id);
for (int j = 0; j < MAX_SUBJECTS; j++) {
printf("%-10d", students[i].scores[j]);
}
printf("%-10d%-10.2f\n", students[i].total_score, students[i].avg_score);
}
}
void print_student_scores(struct Student students[], int num_students) {
printf("Student scores:\n");
printf("%-20s%-10s%-10s%-10s%-10s%-10s%-10s%-10s\n", "Name", "ID", "Subject1", "Subject2", "Subject3", "Subject4", "Total", "Average");
for (int i = 0; i < num_students; i++) {
printf("%-20s%-10s", students[i].name, students[i].id);
for (int j = 0; j < MAX_SUBJECTS; j++) {
printf("%-10d", students[i].scores[j]);
}
printf("%-10d%-10.2f\n", students[i].total_score, students[i].avg_score);
}
}
void print_subject_scores(struct Subject subjects[], int num_subjects) {
printf("Subject scores:\n");
printf("%-20s%-10s%-10s\n", "Name", "Total", "Average");
for (int j = 0; j < num_subjects; j++) {
printf("%-20s%-10d%-10.2f\n", subjects[j].name, subjects[j].total_score, (float)subjects[j].total_score / subjects[j].num_students);
}
}
void print_student_by_id(struct Student students[], int num_students, char id[]) {
int rank = 0;
for (int i = 0; i < num_students; i++) {
if (strcmp(students[i].id, id) == 0) {
rank = i+1;
printf("Ranking for student %s:\n", id);
printf("Rank: %d\n", rank);
printf("Scores:\n");
printf("%-10s%-10s%-10s%-10s%-10s%-10s%-10s%-10s\n", "Name", "ID", "Subject1", "Subject2", "Subject3", "Subject4", "Total", "Average");
printf("%-10s%-10s", students[i].name, students[i].id);
for (int j = 0; j < MAX_SUBJECTS; j++) {
printf("%-10d", students[i].scores[j]);
}
printf("%-10d%-10.2f\n", students[i].total_score, students[i].avg_score);
return;
}
}
if (rank == 0) {
printf("No student with ID %s found.\n", id);
}
}
void print_student_by_name(struct Student students[], int num_students, char name[]) {
int rank = 0;
for (int i = 0; i < num_students; i++) {
if (strcmp(students[i].name, name) == 0) {
rank = i+1;
printf("Ranking for student %s:\n", name);
printf("Rank: %d\n", rank);
printf("Scores:\n");
printf("%-10s%-10s%-10s%-10s%-10s%-10s%-10s%-10s\n", "Name", "ID", "Subject1", "Subject2", "Subject3", "Subject4", "Total", "Average");
printf("%-10s%-10s", students[i].name, students[i].id);
for (int j = 0; j < MAX_SUBJECTS; j++) {
printf("%-10d", students[i].scores[j]);
}
printf("%-10d%-10.2f\n", students[i].total_score, students[i].avg_score);
return;
}
}
if (rank == 0) {
printf("No student with name %s found.\n", name);
}
}
int main() {
int num_students, num_subjects;
struct Student students[MAX_STUDENTS];
struct Subject subjects[MAX_SUBJECTS];
printf("Enter number of students (max %d): ", MAX_STUDENTS);
scanf("%d", &num_students);
printf("Enter number of subjects (max %d): ", MAX_SUBJECTS);
scanf("%d", &num_subjects);
input_students(students, num_students, num_subjects);
calculate_subject_totals(subjects, students, num_students, num_subjects);
calculate_student_totals(students, num_students, num_subjects);
sort_students_by_total_score_desc(students, num_students);
print_student_ranking(students, num_students);
sort_students_by_total_score_asc(students, num_students);
print_student_ranking(students, num_students);
sort_students_by_id(students, num_students);
print_student_scores(students, num_students);
sort_students_by_name(students, num_students);
print_student_scores(students, num_students);
print_subject_scores(subjects, num_subjects);
char id[MAX_ID_LENGTH];
printf("Enter student ID to search: ");
scanf("%s", id);
print_student_by_id(students, num_students, id);
char name[MAX_NAME_LENGTH];
printf("Enter student name to search: ");
scanf("%s", name);
print_student_by_name(students, num_students, name);
return 0;
}
```
以上程序实现了以下功能:
- 输入学生信息
- 计算每门课程的总分和平均分
- 计算每个学生的总分和平均分
- 将学生按总分从高到低排序并输出排名表
- 将学生按总分从低到高排序并输出排名表
- 将学生按学号从小到大排序并输出成绩表
- 将学生按姓名的字典顺序排序并输出成绩表
- 输出每个学生的学号、姓名、各科考试成绩、总分、平均分,以及每门课程的总分和平均分
- 按学号查询学生排名及其各科考试成绩
- 按姓名查询学生排名及其各科考试成绩。
SQL 语句 题目所需学号 第一次考试成绩 ,第二次考试成绩 ,第三次考试成绩 考试成绩超过60分即视为考试通过,查询那一次考核的最高分的分值在三次考核中最高
### 回答1:
可以使用以下 SQL 语句实现:
```
SELECT MAX(score) as max_score
FROM (
SELECT MAX(score) as score
FROM exam
WHERE 学号 = 'xxx' AND 第一次考试成绩 > 60
UNION ALL
SELECT MAX(score) as score
FROM exam
WHERE 学号 = 'xxx' AND 第二次考试成绩 > 60
UNION ALL
SELECT MAX(score) as score
FROM exam
WHERE 学号 = 'xxx' AND 第三次考试成绩 > 60
) as temp
```
其中,`学号` 为需要查询的学号,`exam` 为考试成绩表,`第一次考试成绩`、`第二次考试成绩`、`第三次考试成绩` 为三次考核的成绩字段。
### 回答2:
假设有一个名为"考试成绩"的表格,其中包含学号、第一次考试成绩、第二次考试成绩和第三次考试成绩的数据。那么根据题目要求,我们需要查询在三次考核中哪一次考核的最高分的分值最高。
我们可以按照以下步骤实现:
1. 使用SELECT语句从表格中选择学号和每次考试的最高分值,结合MAX函数和CASE语句进行条件判断。例如:
```sql
SELECT 学号,
CASE
WHEN 第一次考试成绩 >= 60 AND 第一次考试成绩 >= 第二次考试成绩 AND 第一次考试成绩 >= 第三次考试成绩 THEN 第一次考试成绩
WHEN 第二次考试成绩 >= 60 AND 第二次考试成绩 >= 第一次考试成绩 AND 第二次考试成绩 >= 第三次考试成绩 THEN 第二次考试成绩
WHEN 第三次考试成绩 >= 60 AND 第三次考试成绩 >= 第一次考试成绩 AND 第三次考试成绩 >= 第二次考试成绩 THEN 第三次考试成绩
END AS 最高分值
FROM 考试成绩;
```
2. 使用ORDER BY语句按最高分值降序排列,以便找到最高分值最高的考核。例如:
```sql
SELECT 学号,
CASE
WHEN 第一次考试成绩 >= 60 AND 第一次考试成绩 >= 第二次考试成绩 AND 第一次考试成绩 >= 第三次考试成绩 THEN 第一次考试成绩
WHEN 第二次考试成绩 >= 60 AND 第二次考试成绩 >= 第一次考试成绩 AND 第二次考试成绩 >= 第三次考试成绩 THEN 第二次考试成绩
WHEN 第三次考试成绩 >= 60 AND 第三次考试成绩 >= 第一次考试成绩 AND 第三次考试成绩 >= 第二次考试成绩 THEN 第三次考试成绩
END AS 最高分值
FROM 考试成绩
ORDER BY 最高分值 DESC;
```
这样,我们就能够查询到在三次考核中最高分值最高的考核。查询结果中的学号和最高分值即为答案。
### 回答3:
要查询三次考核中最高的那一次考核的最高分值,可以使用以下 SQL 语句:
```
SELECT MAX(考试成绩) AS 最高分值
FROM (
SELECT 学号, MAX(第一次考试成绩) AS 考试成绩
FROM 表名
GROUP BY 学号
UNION
SELECT 学号, MAX(第二次考试成绩) AS 考试成绩
FROM 表名
GROUP BY 学号
UNION
SELECT 学号, MAX(第三次考试成绩) AS 考试成绩
FROM 表名
GROUP BY 学号
) AS 考试分数
WHERE 考试分数.考试成绩 > 60;
```
以上 SQL 语句的步骤是:
1. 先针对每个学生分别计算出三次考核中的最高分,使用 UNION 将结果合并为一张虚拟表;
2. 在虚拟表中筛选出考试成绩超过60分的记录;
3. 在筛选结果中找出最高分值。
请注意将表名替换为实际的表名,并根据实际需求修改字段名。
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