2021/11/27大数据1+XJava答案

2021/11/27

步骤二

//已提供Student类的属性，补充完成该类的有参（五个参数）及无参构造方法  public Student(){}  public Student(String name,String pwd,int age,int grade,int rate){   this.name=name;   this.pwd=pwd;   this.age=age;   this.grade=grade;   this.rate=rate;  }

步骤三

//请修改此方法，赋值时，课程难度只能是高、中、低三种之一，其余值不予处理。  public void setDifficulty(String difficulty) {   switch(difficulty){    case "高":this.difficulty = difficulty;break;    case "中":this.difficulty = difficulty;break;    case "低":this.difficulty = difficulty;break;    default :break;   }     }

步骤四

//请修改此方法，以确保输出格式如下：(sname=zs;cname=语文;score=80)  @Override  public String toString() {   return "(sname=" sname ";cname=" cname ";score=" score ")";  }

步骤五

/**   * 插入学生   * @param s 学生对象   * @return   */  public int add(Student s) {                 // 请补全sql语句                 String sql = "insert into `student` values (?,?,?,?,?,?)";                 return studentUtil.add(sql, s.getName(),s.getPwd(),s.getAge(),s.getGrade(),s.getRate());         }

步骤六

/**   * 查询学生总数   * @return 返回总人数   */  public int queryNum() {                 String sql = "select * from student";                 List<Student> list = studentUtil.getList(sql, Student.class);                 // 请修改以下代码，确保返回值为总人数。假设所有学生的名字都不一样                 int num = list.size();                 return num;         }

步骤七

/**   * 查询最小年龄学生姓名   * @return 返回学生姓名   */  public String queryMinAge() {                 // 请补全sql语句                 String sql = "select `name` from `student` where age = (select min(age) from `student`)";                 Student g = studentUtil.getOne(sql, Student.class);                 return g.getName();         }

步骤八

/**   * 根据课程名称查询课程   * @return 返回课程对象   */  public Course queryCourse(String name) {                 // 请补全sql语句                 String sql = "select * from `course` where `name` = ?";                 return courseUtil.getOne(sql, Course.class, name);         }

步骤九

/**   * 根据课程名称更新课程难度   * @return 更新成功返回true，未更新成功返回false   */  public boolean updateDifficultyByName(String name,String difficulty){                 // 请补全sql语句                 String sql = "update `course` set `difficulty` = ? where `name` = ?";                 int a = courseUtil.update(sql, difficulty, name);                 if(a>0){                         return true;                 }else{                         return false;                   }         }

步骤十

/**   * 查询平均分数最高的学生姓名   * @return 返回学生姓名   */  public String queryAvgMax() {                 // 请补全sql语句   select `sname` from (select `sname`,avg(`score`) as aa from `score` group by `sname`)tab1 where aa = (select max(aa) from (select `sname`,avg(`score`) as aa from `score` group by `sname`)tab1)                 //select max(aa) from (select `sname`,avg(`score`) as aa from `score` group by `sname`)tab1                 String sql = "select `sname` from (select `sname`,avg(`score`) as aa from `score` group by `sname`)tab1 where aa = (select max(aa) from (select `sname`,avg(`score`) as aa from `score` group by `sname`)tab1)";                 Score s = scoreUtil.getOne(sql, Score.class);                 return s.getSname();         }

步骤十一

// 把集合 li 取出每个成绩对象的名称，放入集合中 s 中,并返回                 // 请补充以下代码                 for (int i = 0; i < li.size(); i  ) {                              s.add(li.get(i).getSname());                 }                                  return s;