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以此格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);
}

然后我应该得到：

这不是数字：a

Answer 1

您可以从例外中提取它：

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

Answer 2

您可以尝试这样的事情

    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

应该是直截了当的：

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 4

您可以尝试/捕获NumberFormatException，在这种情况下可能会发生这种情况，并通过包含自定义消息的抛出将异常传播到“ main”。

try { 
    combinationLength = Integer.parseInt(strCombinationLength);
} catch (NumberFormatException e) {
    throw new NumberFormatException("The parameter \"combinationLength \" must be a number");
}

主要

try {
    //some code
} catch (NumberFormatException e) {
    System.out.println(e.getMessage());
}

如何在NumberFormatException时使用有问题的输入字符串打印自定义消息

4 个答案: