文件的主要方法：主要问题是double celciustofahrenheit等在数据传入后不返回计算结果，这是一个恒定的流，就像我想返回main并再次提示用户输入和缩放选择，这是一个switch case语句，它根据所选内容输出缩放信息已传递到inputemp，但即使无法输出

public double inputTemp(int scale) {
    Scanner localInput = new Scanner(System.in);
        System.out.println("enter temperature to convert \n" 
+ "please numeric data only");
        double celciusToFahrenheitCall = cToF(inputField.nextDouble());
        double kelvinToCelciusCall = kToC(inputField.nextDouble());
        double fahrenheitToCelciusCall = fToC(inputField.nextDouble());
        double kelvinToFahrenheitCall = kToF(inputField.nextDouble());
        double celciusToKelvinCall = cToK(inputField.nextDouble());
        double fahrenheitToKelvinCall = fToK(inputField.nextDouble());
        return scale;
    }
double cToF(double inputField) {
        inputField = (9/5 * inputField + 32);
        return inputField;
        //converts Celsius to Fahrenheit and returns the answer
    }
    double kToF(double inputField) {
        double kelvinToFahrenheit = (9/5 * inputField - 273 + 32);
        return kelvinToFahrenheit;
        //converts Kelvin to Fahrenheit and returns the answer
    }
    double fToC(double inputField) {
        return (5/9 * (inputField - 32));
        //converts Fahrenheit to Celsius and returns the answer
    }
    double cToK(double inputField) {  
        return (inputField + 273);
        //converts Celsius to Kelvin and returns the answer
    }
    double kToC(double inputField) {
        return (inputField - 273);
            //converts Kelvin to Celsius and returns the answer
    }
    double fToK(double inputField) {
        return ((5/9 * (inputField -32 ) + 273));
          //converts Fahrenheit to Kelvin and returns the answer
    }

}