首先,我对 Java 和编程都是新手(Matlab 除外),因此非常感谢简单的答案:-)。
我正在尝试创建一个温度转换器(带有 GUI),我需要更新一些标签。一开始效果很好,但现在我必须使用 if 语句中的值。这会导致我尝试更新标签的错误:
tempKelvin 无法解析为变量
单击“转换”按钮时会发生所有操作,代码如下:
// Create and add convert button
JButton fahrenheitButton = new JButton("Convert");
fahrenheitButton.addActionListener(new ActionListener() {
public void actionPerformed(ActionEvent arg0) {
// Check if input is of type double and perform action
if (isNumber(tempTextField.getText())) {
double inputTemp = Double.parseDouble(tempTextField.getText());
// Convert from Kelvin
if (((String) unitDropdown.getSelectedItem()).equals("Kelvin")) {
int tempKelvin = (int) (inputTemp);
int tempCelcius = (int) (inputTemp - 273.15);
int tempFahrenheit = (int) ((inputTemp - 273.15) * (9/5) + 32);
// Convert from Celsius
} else if (((String) unitDropdown.getSelectedItem()).equals("Celsius")) {
int tempKelvin = (int) (inputTemp + 273.15);
int tempCelcius = (int) (inputTemp);
int tempFahrenheit = (int) (inputTemp * (9/5) + 32);
// Convert from Fahrenheit
} else if (((String) unitDropdown.getSelectedItem()).equals("Fahrenheit")) {
int tempKelvin = (int) ((inputTemp - 32) * (5/9) + 273.15);
int tempCelcius = (int) ((inputTemp - 32) * (5/9));
int tempFahrenheit = (int) ((inputTemp - 273.15) * (9/5) + 32);
// If none of the above was selected, it's an error...
} else {
int tempKelvin = 0;
int tempCelcius = 0;
int tempFahrenheit = 0;
warningLabel.setText("Oops, this doesn't look good!");
}
// Update labels
kelvinLabel.setText(tempKelvin + " K");
celsiusLabel.setText(tempCelcius + " C");
fahrenheitLabel.setText(tempFahrenheit + " F");
warningLabel.setText("");
} else {
warningLabel.setText("Input must be numeric!");
}
}
});
fahrenheitButton.setBounds(20, 45, 89, 23);
contentPane.add(fahrenheitButton);
任何帮助将不胜感激,谢谢!!