当数字格式异常时，如何使用有问题的输入字符串打印自定义消息

Question

线程"main"中的异常java.lang.NumberFormatException：对于输入字符串："a"

如何访问导致NumberFormatException以这种形式打印自定义错误消息的输入字符串：

try {
    /* some code which includes many Integer.parseInt(some_string); */
}
catch (NumberFormatException nfe) {
    System.out.println("This is not a number: " + input_string_which_causing_the_error);
}

那我应该得到：

这不是一个数字：a

Answer 1

它应该是直接的：

String some_string = "12";//retrieval logic
try {
    int num = Integer.parseInt(some_string);
} catch (NumberFormatException nfe) {
    System.out.println("This is not a number: " + some_string);
}

Answer 2

你可以试试这样的

    String input="abc"; // declare input such that visible to both try and catch
    try {
        int a = Integer.parseInt(input);
    } catch (NumberFormatException e) {
        System.out.println("This is not a number: " + input);
    }

Answer 3

您可以从异常中提取它：

} catch (NumberFormatException e) {
    System.err.println(e.getMessage().replaceFirst(".*For input string: ", "This is not a number"));
}