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"}
};