java - 使用 if else 语句进行 JOptionPane 输入验证
问题描述
我想当用户输入一个不超过 100 的非正整数时,它会显示一条错误消息,如果输入正确,用户将再次输入,然后继续执行 if/else 语句。
import java.util.Scanner;
import javax.swing.JOptionPane;
public class Culminating {
public static void main(String[] args) {
Scanner in = new Scanner (System.in);
JOptionPane.showMessageDialog(null, "WELCOME");
String userInput = JOptionPane.showInputDialog("KINDLY ENTER YOUR GRADES");
int num = Integer.parseInt(userInput);
if (num>100)
{
JOptionPane.showMessageDialog(null, "ERROR: PLEASE ENTER UP TO 100 ONLY" );
}
else if (num>93)
{
JOptionPane.showMessageDialog(null, "Letter Grade: " + " A " + "\n\nPoints: " + 4.0 );
}
else if (num > 90 )
{
JOptionPane.showMessageDialog(null, "Letter Grade: " +" A- " + "\n\nPoints: " + 3.7 );
}
else if (num> 87){
JOptionPane.showMessageDialog(null, "Letter Grade: " + " B+ " + "\n\nPoints: " + 3.3 );
}
else if (num> 83){
JOptionPane.showMessageDialog(null, "Letter Grade: " + " B " + "\n\nPoints: " + 3.0 );
}
else if (num>80){
JOptionPane.showMessageDialog(null, "Letter Grade: " + " B- " + "\n\nPoints: " + 2.7 );
}
else if (num>77){
JOptionPane.showMessageDialog(null, "Letter Grade: " + " C+ " + "\n\nPoints: " + 2.3 );
}
else if (num>73){
JOptionPane.showMessageDialog(null, "Letter Grade: " + " C " + "\n\nPoints: " + 2.0 );
}
else if (num> 70){
JOptionPane.showMessageDialog(null, "Letter Grade: " + " C- " + "\n\nPoints: " + 1.7 );
}
else if (num> 67){
JOptionPane.showMessageDialog(null, "Letter Grade: " + " D+ " + "\n\nPoints: " + 1.3 );
}
else if (num>64){
JOptionPane.showMessageDialog(null, "Letter Grade: " + " D " + "\n\nPoints: " + 1.0 );
}
else if (num<64){
JOptionPane.showMessageDialog(null, "Letter Grade: " + " E " + "\n\nPoints: " + 0.0 );
}
}
}
解决方案
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