【发布时间】:2018-04-18 22:45:40
【问题描述】:
它以空值打印...我该如何摆脱它?它给了我例如:
现在:
狮子座 null null null
如果它没有达到 4,我希望它只打印 Leo。
这是我当前的代码:
import java.util.*;
import java.io.*;
public class Trial {
public static void main(String[] args) {
String [] Student ={"Tom", "Jack", "Gio", "Leo"};
String[][] Attendance = new String[4][4];
System.out.println("STUDENT LIST-TAKE ATTENDANCE");
for(int i = 0; i<4; i++){
System.out.println("Type 1 for Present, 2 for Absent, 3 for Fieldtrip, or 4 for Late " + "for " + Student[i]);
Scanner scan1 = new Scanner(System.in);
double Att = scan1.nextDouble();
int j = 0;
if (Att == 1){
Attendance[0][j] = Student[i];
j++;
}
else if(Att == 2){
Attendance[1][j] = Student[i];
j++;
}
else if(Att == 3){
Attendance[2][j] = Student[i];
j++;
}
else if(Att == 4){
Attendance[3][j] = Student[i];
j++;
}
else
System.out.println("Retry");
}
System.out.println("Present: ");
for (int k = 0; k<=3; k++)
System.out.print(Attendance[0][k] + " ");
System.out.println();
System.out.println("Absent: ");
for (int k = 0; k<=3; k++)
System.out.print(Attendance[1][k] + " ");
System.out.println();
System.out.println("Fieldtrip: ");
for (int k = 0; k<=3; k++)
System.out.print(Attendance[2][k] + " ");
System.out.println();
System.out.println("Late: ");
for (int k = 0; k<=3; k++)
System.out.print(Attendance[3][k] + " ");
System.out.println();
/*
String[][] Attendance = {
{ "Present", },
{ "Absent", },
{ "Fieldtrip", },
{ "Late", }
};
System.out.println();
for (int i = 0; i < Attendance; i++) {
System.out.print(Attendance[i][0] + ": ");
for (int j = 1; j < Attendance[i].length; j++) {
System.out.print(Attendance[i][j] + " ");
}
*/
}
}
【问题讨论】:
-
嗯,打印前可以检查是否为null。
-
A) 阅读有关 java 命名约定 - 变量名采用 camelCase - 总是 B) 阅读有关使用浮点数的信息。您基本上永远不会直接使用 == 比较它们。
标签: java arrays loops oop dynamic-arrays