創建條件（陣列 - JAVA）

Question

我想創建一個變量「INT EDAD」的條件，但我很困惑...

我需要一個條件，這條線尤其

Integer.parseInt(JOptionPane.showInputDialog("Ingrese edad")), 

這是一個int變量，我想驗證它是在0到100之間，然後給用戶一個消息並再次詢問這個變量。

package pruebas; 

import javax.swing.JOptionPane; 

public class Persona { 

public static void main(String[] args) { 
    Estudiante [] misEstudiantes2 = new Estudiante[1]; 

    for(int i = 0; i< misEstudiantes2.length; i++){ 

     misEstudiantes2[i] = new Estudiante(JOptionPane.showInputDialog("Ingrese nombre").toUpperCase(), 
      Integer.parseInt(JOptionPane.showInputDialog("Ingrese edad")), 
      JOptionPane.showInputDialog("Ingrese carrera"), 
      Integer.parseInt(JOptionPane.showInputDialog("Ingrese año")), 
      Integer.parseInt(JOptionPane.showInputDialog("Ingrese mes")), 
      Integer.parseInt(JOptionPane.showInputDialog("Ingrese dia")), 
      Double.parseDouble(JOptionPane.showInputDialog("Ingrese sueldo"))); 

     misEstudiantes2[i].estableceSueldo(1); 

     System.out.println("Objetos del array \n" + misEstudiantes2[i].dameNombre() + "\n" + misEstudiantes2[i].dameSueldo() + "\n" + 
     misEstudiantes2[i].dameCarrera() + "\n" + misEstudiantes2[i].dameFecha() + "\n" + "La edad es " + misEstudiantes2[i].dameEdad()); 
    } 
} 
} 

Answer 1

如果您在構建新學生對象之前將所有用戶輸入存儲到變量中，這是可能的。

String name = JOptionPane.showInputDialog("Ingrese nombre").toUpperCase(); 
int age = Integer.parseInt(JOptionPane.showInputDialog("Ingrese edad")); 
String career = JOptionPane.showInputDialog("Ingrese carrera"); 
int year = Integer.parseInt(JOptionPane.showInputDialog("Ingrese año")); 
int month = Integer.parseInt(JOptionPane.showInputDialog("Ingrese mes")); 
int day = Integer.parseInt(JOptionPane.showInputDialog("Ingrese dia")); 
double salary = Double.parseDouble(JOptionPane.showInputDialog("Ingrese sueldo"))); 

misEstudiantes2[i] = new Estudiante(name, age, career, year, month, day, salary); 

從這裏，你可以在你的變量周圍進行必要的檢查。例如：

int age = Integer.parseInt(JOptionPage.showInputDialog("Ingrese edad")); 
while (age < 0 || age > 100) { // in case they put in an invalid number 
           // try again, with a notice about the range 
    age = Integer.parseInt(JOptionPage.showInputDialog("Ingrese edad, mas de 0 y menos de 100")); 
} 

Answer 2

試着寫一個方法，爲你做。例如：

public static Integer promptForInteger(int min, int max, String message, String message2) { 
    Integer result = Integer.parseInt(JOptionPane.showInputDialog(message)); 
    while(result < min || result > max) { 
    result = Integer.parseInt(JOptionPane.showInputDialog(message2)); 
    } 
    return result; 
} 

當您創建對象，然後調用它：

misEstudiantes2[i] = new Estudiante(JOptionPane.showInputDialog("Ingrese nombre").toUpperCase(), 
            promptForInteger(0, 100, "Ingrese edad", "Ingrese edad, mas de 0 y menos de 100"), 
            JOptionPane.showInputDialog("Ingrese carrera"), 
            Integer.parseInt(JOptionPane.showInputDialog("Ingrese año")), 
            Integer.parseInt(JOptionPane.showInputDialog("Ingrese mes")), 
            Integer.parseInt(JOptionPane.showInputDialog("Ingrese dia")), 
            Double.parseDouble(JOptionPane.showInputDialog("Ingrese sueldo"))); 

我不知道你的母語，所以我使用的字符串從johnhopkins ES答案。

Answer 3

做用戶一個忙，並提供他們一個微調選擇號碼。

import javax.swing.*; 

class VerifiedIntegerInRange { 

    public VerifiedIntegerInRange() { 
     SpinnerNumberModel numberRange = new SpinnerNumberModel(-1,-1,100,1); 
     JSpinner spinner = new JSpinner(numberRange); 
     int i = numberRange.getNumber().intValue(); 
     while (i<0) { 
      JOptionPane.showMessageDialog(
       null, 
       spinner, 
       "Select a number between 0 & 100", 
       JOptionPane.QUESTION_MESSAGE); 
      i = numberRange.getNumber().intValue(); 
     } 
     System.out.println("User chose " + i); 
    } 

    public static void main(String[] args) { 
     Runnable r = new Runnable() { 
      public void run() { 
       new VerifiedIntegerInRange(); 
      } 
     }; 
     SwingUtilities.invokeLater(r); 
    } 
}