我列出了一周内完成任务的个人的开始和停止时间。
当他们更改任务时，会创建一个新条目。
此信息存储在2D数组中。 - 我想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，我想要一行来显示特定日期的某个人。
因此，如果第一个数组存储:(人，日，进，出，总）
{{"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"}};
我想要存储第二个数组：
{{"John","Mon","08:00","17:00","9.00"},
{"John","Tue","08:00","17:00","9.00"},
{"Mike","Tue","08:00","17:00","9.00"}};
到目前为止，这是我的代码：
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
答案 0 :(得分：0)
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）你不应该为完成的数组分配一个5乘5的数组。当然，列数可以是5，但如果您的示例中有重复项，该怎么办？
要解决第二个问题，我会使用 java.lang.UnsatisfiedLinkError: No implementation found for boolean com.esri.core.runtime.LicenseImpl.nativeIsClientIdValid(java.lang.String) (tried Java_com_esri_core_runtime_LicenseImpl_nativeIsClientIdValid and Java_com_esri_core_runtime_LicenseImpl_nativeIsClientIdValid__Ljava_lang_String_2)
at com.esri.core.runtime.LicenseImpl.nativeIsClientIdValid(Native Method)
at com.esri.core.runtime.LicenseImpl.b(SourceFile:103)
at com.esri.android.runtime.ArcGISRuntime$License.b(SourceFile:133)
at com.esri.android.runtime.ArcGISRuntime$License.a(SourceFile:72)
at com.esri.android.runtime.ArcGISRuntime.setClientId(SourceFile:51)
at com.ihs.connect.App.configureArcGIS(App.java:89)
at com.ihs.connect.App.onCreate(App.java:82)
at android.app.Instrumentation.callApplicationOnCreate(Instrumentation.java:1036)
at android.app.ActivityThread.handleBindApplication(ActivityThread.java:6316)
at android.app.ActivityThread.access$1800(ActivityThread.java:221)
at android.app.ActivityThread$H.handleMessage(ActivityThread.java:1860)
at android.os.Handler.dispatchMessage(Handler.java:102)
at android.os.Looper.loop(Looper.java:158)
at android.app.ActivityThread.main(ActivityThread.java:7224)
at java.lang.reflect.Method.invoke(Native Method)
at com.android.internal.os.ZygoteInit$MethodAndArgsCaller.run(ZygoteInit.java:1230)
at com.android.internal.os.ZygoteInit.main(ZygoteInit.java:1120)
循环和计数器变量来遍历每个人的标识变量。在这种情况下，我将假设每个“名称”都不同，因此您可以检查名称和日期是否相同，如果是，则不要递增计数器。最后，你有一个完美尺寸的2D阵列。
接下来，创建一个算法来计算进出时间的工作小时数。
最后，使用另一个for循环，检查名称和日期是否相同。如果是这样，请添加这些小时数并创建包含该数据的新行。
答案 1 :(得分：0)
我建议你使用一个聚合Session集合的对象Employee。 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;
}
}
对象员工：
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));
}
}
但是仍然存在实施员工行为的问题。在下面的例子中，迈克不连续会发生什么：
{{"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"}
};
答案 2 :(得分：0)
感谢您的帮助，但我设法找到一个解决方案，在循环启动数组时解决了我的问题。
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。但是感谢有关制作员工对象的建议，我将来会使用它。
启动数组
John Mon 08:00 12:00 4.0
John Mon 12:00 17:00 5.0
John Tue 08:00 17:00 9.0
Mike Tue 08:00 11:00 3.0
Mike Tue 11:00 17:00 6.0
结束数组
John Mon 08:00 17:00 9.0
John Tue 08:00 17:00 9.0
Mike Tue 08:00 17:00 9.0
null null null null null
null null null null null