如何在Java中操作2D数组数据



我有一个每个人在一周内完成任务的开始和结束时间列表。

当他们改变任务时,一个新的条目被创建。

该信息存储在2D数组中。-我想循环遍历这个数组并将数据存储在另一个数组中。

在新的2D数组中,我想用一行显示某一天的某个人。

如果第一个数组存储:(Person,Day,In,Out,Total)

{{"John","Mon","08:00","12:00","4.00"},
 {"John","Mon","12:00","17:00","5.00"},
 {"John","Tue","08:00","17:00","9.00"},
 {"Mike","Tue","08:00","11:00","3.00"}
 {"Mike","Tue","11:00","17:00","6.00"}};

我希望第二个数组存储:

    {{"John","Mon","08:00","17:00","9.00"},
     {"John","Tue","08:00","17:00","9.00"},
     {"Mike","Tue","08:00","17:00","9.00"}};
下面是目前为止的代码:

 public class CompArrayTest {
public static void main(String args[]){
    String[][] End = new String [5][5];
    String[][] Start = {{"John","Mon","08:00","12:00","4.00"},
                        {"John","Mon","12:00","17:00","5.00"},
                        {"John","Tue","08:00","17:00","9.00"},
                        {"Mike","Tue","08:00","11:00","3.00"},
                        {"Mike","Tue","11:00","17:00","6.00"}};

    //print start 
    for(int i = 0; i<Start.length; i++){ 
      for(int j = 0; j<Start.length; j++){
          System.out.print(Start[i][j]+" ");
      }//j end
     System.out.print("n");
    }//i end
    //change end 
    for(int i = 0; i<Start.length; i++){ 
        String name = Start[i][0];
        String day = Start[i][1];
        String In = Start[i][2];
        String Out = Start[i][3];
        String Total = Start[i][4];
       //look through End
      for(int j = 0; j<5; j++){
      String eN= End[j][0];
      String eD= End[j][1];
      if(eN==name && eD==day){
      //change end time
      End[j][3]=Start[i][3];
      //parse and add times
      double TS = Double.parseDouble(Start[i][4]);
      double TE = Double.parseDouble(End[i][4]);
      double ans = TS + TE;
      String ANS = ans+"";
      End[j][4]= ANS;

      } else {

      End[j][0] = name;
      End[j][1] = day;
      End[j][2] = In;
      End[j][3] = Out;
      End[j][4] = Total;

      }//else end

      }//j end
     System.out.print("n");
    }//i end
 //print end 
    for(int i = 0; i<Start.length; i++){ 
      for(int j = 0; j<Start.length; j++){
          System.out.print(End[i][j]+" ");
      }//j end 
     System.out.print("n");
    }//i end
}//main end

1)您可以为Employee创建一个类,但我的答案将处理您选择2D数组的风格。

2)你不应该为你完成的数组分配一个5 × 5的数组。当然,列的数量可以是5,但是如果像示例中那样有重复的列呢?

要解决第二个问题,我将使用for循环和计数器变量来遍历每个人的标识变量。在本例中,我将假设每个"Name"都是不同的,因此您可以检查Name和the Day是否相同,如果相同,则不增加计数器。最后,你有一个完美大小的二维数组。

接下来,创建一个算法来计算给定进出时间的工作小时数。

最后,使用另一个for循环,遍历并检查Name和Day是否相同。如果是,把这些小时数加起来,用这些数据创建一个新行。

感谢您的帮助,但我设法找到了一个解决方案,解决了我的问题时,通过开始数组循环。

 for (int i = 0; i < Start.length; i++) {
        String name = Start[i][0];
        String day = Start[i][1];
        String In = Start[i][2];
        String Out = Start[i][3];
        String Total = Start[i][4];
        int emptyLine = 0;
        int lastLine = 0;
        //gets first emptyline, lastfilled line
        for (int j = 0; j < 5; j++) {
            if (End[j][0] == null) {
                emptyLine = j;
                if (j > 0) {
                    lastLine = j - 1;
                }
                break;
            }
        }//get Empty
        String eN = End[lastLine][0];
        String eD = End[lastLine][1];
        if (eN == name && eD == day) {
            //change end time
            End[lastLine][3] = Start[i][3];
            //parse and add times
            double TS = Double.parseDouble(Start[i][4]);
            double TE = Double.parseDouble(End[lastLine][4]);
            double ans = TS + TE;
            String ANS = ans + "";
            End[lastLine][4] = ANS;
           // System.out.println("Test If " + name + " " + day + " " + In + " " + Out + " " + Total);
        } else {
            End[emptyLine][0] = name;
            End[emptyLine][1] = day;
            End[emptyLine][2] = In;
            End[emptyLine][3] = Out;
            End[emptyLine][4] = Total;
           // System.out.println("Test Else " + name + " " + day + " " + In + " " + Out + " " + Total);
        }//else end

    }//i end

这给了我想要的输出。但是,感谢您关于制作员工对象的建议,我将在将来使用它。

<

开始数组/strong>

约翰星期一08:00 12:00 4.0
约翰星期一12:00 17:00 5.0
约翰星期二08:00 17:00 9.0
Mike Tue 08:00 11:00 3.0
Mike周二11:00 17:00 6.0

<

结束数组/strong>

约翰星期一08:00 17:00 9.0
约翰星期二08:00 17:00 9.0
迈克星期二08:00 17:00 9.0
Null Null Null Null

我建议你使用一个Employee对象来聚合Session集合。对象会话:

public class Session {
    private String day;
    private String in;
    private String out;
    private String total;
    Session(String day, String in, String out, String total){
        this.setDay(day);
        this.setIn(in);
        this.setOut(out);
        this.setTotal(total);
}
    public String getDay() {
        return day;
    }
    public void setDay(String day) {
        this.day = day;
    }
    public String getIn() {
        return in;
    }
    public void setIn(String in) {
        this.in = in;
    }
    public String getOut() {
        return out;
    }
    public void setOut(String out) {
        this.out = out;
    }
    public String getTotal() {
        return total;
    }
    public void setTotal(String total) {
        this.total = total;
    }   
}

和对象Employee:

public class Employee {
    private String name;
    private List<Session> sessions;
    Employee(String name){
        this.name = name;
        this.sessions = new ArrayList<Session>();
    }
    public boolean addSession(String day, String in, String out, String total){
        return sessions.add(new Session(day, in, out, total));
    }
}

但是如何实施员工的行为仍然是一个问题。在我下面的例子中,如果Mike不连续会发生什么?

{{"John","Mon","08:00","12:00","4.00"},
 {"John","Mon","12:00","17:00","5.00"},
 {"John","Tue","08:00","17:00","9.00"},
 {"Mike","Tue","08:00","11:00","3.00"}
 {"Mike","Tue","12:00","17:00","6.00"}
};

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