java - 尝试打印链接数组时，学生可能没有作业或有很多作业。 printStudentDetails 打印每个学生加上 x 个作业对象

Question

就像在图书馆一样，学生可以借 n 本书。我想从两个数组“students[]”和“currentJobs[]”打印，因此输出针对每个学生 - ID1、姓名、job1、job2。 ID2、姓名(无职位)。 ID3，作业1。

我创建了一个嵌套的 for 循环，但不知道如何对链接进行编码。公共(public)字段是studentID，所以我需要外部循环来获取studentID并比较内部循环(数组)上的每个studentID并打印每个匹配的详细信息

 public void displayStudentsReport()
   {
      System.out.println("***** Display Students Report *****");
      System.out.println();

      for (int i = 0; i < indexCount; i++)
      {
         students[i].printStudentDetails();
         // print job details for each student
         for (int j = 0; j < indexNo; j++)
         {
            currentJobs[j].printJobDetails();
         }
      }
   }

   public void printStudentDetails()
   {
      System.out.println("Printing from Student class");
      System.out.printf("%-25s%-40s\n", "Student ID: ", studentID);
      System.out.printf("%-25s%-40s\n", "Name: ", name);
      System.out.println();
   }

   public void printJobDetails()
   {
      System.out.printf("%-25s%-40s\n", "Job: ", jobTitle);
      System.out.printf("%-25s%-4.1f", "Current Working Hours: ", totalHours);
      System.out.println("hrs");
      System.out.printf("%-25s%-40s\n", "Employer: ", employer);
      System.out.println();
   }

预期结果解释

Answer 1

只是对你所做的事情进行了一些修正。在第二个列表中，您应该检查它是否是列表第一个中的正确学生。

public void displayStudentsReport()
{
  System.out.println("***** Display Students Report *****");
  System.out.println();

  for (int i = 0; i < students.length; i++)
  {
     String comp;
     comp = students[i].getStudentID();
     students[i].printStudentDetails();
     // print job details for each student
     boolean studentHasJob = false;
     for (int j = 0; j < currentJobs.length; j++)
     {
        if(currentJobs[j].getStudentID().equalsIgnoreCase(comp)) {
            studentHasJob = true;
            currentJobs[j].printJobDetails();
        }
     }
     if(!studentHasJob){
         System.out.println("Student " + comp + " has no jobs");
     }
  }


  }

如果您想在学生没有工作的情况下打印特定的内容，我添加了一个小 boolean 值。

希望这就是您正在寻找的!

---编辑

所以我再次检查了代码，因为我认为这里存在性能问题。事实上，在代码中您多次传递了 currentJobs 数组。正好是 Students.length * currentJobs.length 倍。这是很多。因此，这里是可以做得更好的预览:

private void displayStudentsReport() {
    System.out.println("***** Display Students Report *****");
    System.out.println();

// build map of <StudentId : List of job for this student>
// perf is much better here because you pass on the job list only once
// whereas before you passed (students.length * currentJobs.length) times
Map<String, List<Job>> jobByStudent = new HashMap<>();
for(Job currentJob : currentJobs) {
    if(jobByStudent.containsKey(currentJob.getStudentID())) {
        jobByStudent.get(currentJob.getStudentID()).add(currentJob);
    } else {
        List<Job> jobs = new ArrayList<>();
        jobs.add(currentJob);
        jobByStudent.put(currentJob.getStudentID(), jobs);
    }
}

for(Student student : students) {
    String comp = student.getStudentID();
    student.printStudentDetails();

    if(jobByStudent.containsKey(comp)) {
        for(Job job : jobByStudent.get(comp)) {
            job.printJobDetails();
        }
    } else {
        System.out.println("Student " + comp + " has no jobs");
    }
}
}