Java——如何在一定范围内正确验证数字用户输入

Question

只是想寻求一点帮助！我现在正在做一个体重转换项目。我已经让它正常工作了，但现在我正试图让它成为傻瓜式的，也就是说，如果用户输入的数值低于或超过某个范围（在本例中，我寻找的是0到450之间的KG），则会出现一条消息，告知错误，然后会提示用户再次输入其值。我可以用下面的代码实现这一点，但问题是，当用户输入一个有效值时，它不仅会打印有效输入的转换，还会打印之前不正确的值的转换。我附上了一个命令提示符的屏幕截图来说明这个问题。谁能告诉我哪里出了问题吗？谢谢

public void kgToStonesAndPounds()
    {
    double kg = 0;
        
        
        System.out.println("Please enter weight in KG here, range must be between 1 and 450: ");
        kg = input.nextDouble();
        
        
        if ( kg >= 1 && kg <= 450 ) // validate kg
         System.out.printf("\nThe weight you have entered is %.0f KG\n" , kg);
        else
        {System.out.println( "Weight in KG must be in the range of 1 - 450" );
        this.kgToStonesAndPounds();
        }
      
      
        double pounds = kg * 2.204622;

        double stonesWithDecimal = pounds / 14;

        int stone = (int) stonesWithDecimal; // cast int to get rid of the decimal
        
        long poundsWithoutStone = (long)((stonesWithDecimal - stone) * 14); // Take the fractional remainder and multiply by 14
        
        System.out.println("This converts to " + stone + " Stone " + poundsWithoutStone + " Pounds " );
    }//end method kgToStonesAndPounds

命令提示符中的示例

Answer 1

递归（在其内部调用方法）不是处理错误的方法，它只应在逻辑需要时使用。

要再次询问，请使用一个循环，该循环仅在输入有效时退出

double kg;
do {
    System.out.println("Please enter weight in KG here, range must be between 1 and 450: ");
    kg = input.nextDouble();
    if (kg >= 1 && kg <= 450) {
        System.out.printf("\nThe weight you have entered is %.0f KG\n", kg);
        break;
    } else {
        System.out.println("Weight in KG must be in the range of 1 - 450");
    }
} while (true);

你可以用模来简化你的代码

double pounds = kg * 2.204622;
int stone = (int) pounds / 14;
int poundsWithoutStone = (int) pounds % 14;
System.out.println("This converts to " + stone + " Stone " + poundsWithoutStone + " Pounds ");

Answer 2

Java——如何在一定范围内正确验证数字用户输入

只要你得到想要的效果/结果（无不良副作用），一种方法就不比另一种方法更正确。

我可以这样做。只需最初提示，如果输入不正确，则重复提示。

double kg;
String prompt = "Please enter weight in KG here, range must be between 1 and 450: ";
System.out.println(prompt);
while ((kg = input.nextDouble()) < 1 || kg > 450) {
    System.out.println(prompt);
}
System.out.printf("\nThe weight you have entered is %.0f KG\n" , kg);
// now do something with kg

Answer 3

在无效情况下再次调用该方法后，必须添加一个return。这样，当从方法调用返回时，如果从该方法调用它，它将不会移出else语句并执行以下代码。

public void kgToStonesAndPounds() {
    ...

    if ( kg >= 1 && kg <= 450 ) // validate kg
        System.out.printf("\nThe weight you have entered is %.0f KG\n" , kg);
    else {
        System.out.println( "Weight in KG must be in the range of 1 - 450");
        this.kgToStonesAndPounds();
        return; // here
    }
    
    ...
}