我使用了以下方法将字符串更改为double,但不幸的是,这关闭了应用程序。EditText输入类型为"NumberDecimal"
numA = (EditText) findViewById(R.id.numA);
numB = (EditText) findViewById(R.id.numB);
//App forceclose here. Not sure why.
final Double a = Double.parseDouble(numA.getText().toString());
final Double b = Double.parseDouble(numB.getText().toString());
calculate.setOnClickListener(new View.OnClickListener() {
@Override
public void onClick(View v) {
numLS.setText("" + ( (- (Double) b) /(2 * (Double) a)));
}
});
试试这个;
String s = b.getText().toString();
final double a = Double.valueOf(s.trim()).doubleValue();
执行此检查:
if (!numA.getText().toString().equals("")) {
final Double a = Double.parseDouble(numA.getText().toString());
}
if (!numB.getText().toString().equals("")) {
final Double b = Double.parseDouble(numB.getText().toString());
}
Double.parseDouble()的空字符串参数会生成NumberFormatException
。
作为一个建议,如果你正在制作一个计算器(或转换器),你应该添加更多的无效输入检查。例如,当用户只输入小数点(.)或输入表单(3.)时,您应该添加一个检查。
您可能希望使用try-catch,因为其他不可分析的数据会引发异常,并且依赖UI仅强制使用有效数字可能不是最好的选择。
numA = (EditText) findViewById(R.id.numA);
numB = (EditText) findViewById(R.id.numB);
Double a;
Double b;
try {
a = Double.parseDouble(numA.getText().toString());
b = Double.parseDouble(numB.getText().toString());
} catch (NumberFormatException e) {
e.printStackTrace();
a = 0.0;
b = 0.0;
}
final double aFin = a;
final double bFin = b;
calculate.setOnClickListener(new View.OnClickListener() {
//Also, you used your class as an onClickListener you would have to make your doubles final.
@Override
public void onClick(View v) {
numLS.setText("" + ( (- (Double) b) /(2 * (Double) a)));
//Division by zero will produce a NaN you should probably check user input data sanity
}
});