Unhandled exception type ParseException

这个错误通常是因为代码中抛出了 ParseException 异常，但该异常没有被处理。要解决这个问题，可以使用 try-catch 块来捕获 ParseException 异常，并在 catch 块中处理它。例如：

try {
    // 可能会抛出 ParseException 异常的代码
} catch (ParseException e) {
    // 处理 ParseException 异常的代码
}

在 catch 块中，你可以根据具体情况选择如何处理该异常，例如输出错误信息，或者将异常转换为其他类型的异常并抛出。

相关问题

public static void main(String[] args) { String[][] t = new String[13][6]; t[0] = new String[]{"悉尼","13512345678","2023-06-12","2023-06-20","飞机","2"}; t[1] = new String[]{"上海","1351234567c","2023-06-12","2023-06-20","飞机","2"}; t[2] = new String[]{"上海","1351234567","2023-06-12","2023-06-20","飞机","2"}; t[3] = new String[]{"上海","23512345678","2023-06-12","2023-06-20","飞机","2"}; t[4] = new String[]{"上海","13512345678","2023-06-2","2023-06-20","飞机","2"}; t[5] = new String[]{"上海","13512345678","2023-06-12","2023-06-11","飞机","2"}; t[6] = new String[]{"上海","13512345678","2023-06-12","2023-06-20","飞机","2"}; t[7] = new String[]{"上海","13512345678","2023-06-12","2023-06-20"," ","2"}; t[8] = new String[]{"上海","13512345678","2023-06-12","2023-06-20","自驾","-1"}; t[9] = new String[]{"上海","13512345678","2023-06-12","2023-06-20","火车","2.5"}; t[10] = new String[]{"上海","13512345678","2023-06-12","2023-06-20","自驾","2"}; t[11] = new String[]{"上海","13512345678","2023-06-12","2023-06-20","火车","2"}; t[12] = new String[]{"上海","13512345678","2023-06-12","2023-06-20","飞机","0"}; String[] a={"出差地无效","联系电话无效","联系电话无效","联系电话无效","日期无效", "日期无效","日期无效","出行方式无效","同行人数无效","同行人数无效","提交申请","提交申请","提交申请"}; String expected,actual,result; System.out.println("编号"+"\t"+"出差地"+"\t"+"联系电话"+"\t"+"开始日期"+"\t"+"结束日期"+"\t"+"出行方式"+"\t"+"同行人数"+"\t"+"预期输出"+"\t"+"实际输出"+"\t"+"测试结果"); for(int i=0;i<=12;i++){ expected=a[i]; actual=application(t[i][0],t[i][1],t[i][2],t[i][3],t[i][4],t[i][5]); if(expected.equals(actual)) result="Passed"; else result="Failed"; System.out.println("T["+(i+1)+"]"+"\t"+t[i][0]+"\t"+t[i][1]+"\t"+t[i][2]+"\t"+t[i][3]+"\t"+t[i][4]+"\t"+t[i][5]+"\t"+expected+"\t"+"\t"+actual+"\t"+result); } } }如何处理Unhandled exception type ParseException

在代码中出现了解析异常ParseException，需要在代码中使用try-catch语句来捕获并处理ParseException异常，以防止程序终止运行。可以在调用application方法的地方添加try-catch语句来处理异常。例如：

for(int i=0;i<=12;i++){
    expected=a[i];
    String result = "";
    try {
        actual=application(t[i][0],t[i][1],t[i][2],t[i][3],t[i][4],t[i][5]);
        if(expected.equals(actual))
            result="Passed";
        else
            result="Failed";
    } catch (ParseException e) {
        result = "ParseException";
    }
    System.out.println("T["+(i+1)+"]"+"\t"+t[i][0]+"\t"+t[i][1]+"\t"+t[i][2]+"\t"+t[i][3]+"\t"+t[i][4]+"\t"+t[i][5]+"\t"+expected+"\t"+"\t"+actual+"\t"+result);
}

在try语句块中执行application方法，如果出现ParseException异常，则会跳转到catch语句块中执行相关操作，例如将result设置为"ParseException"。这样可以避免程序出现Unhandled exception type ParseException错误。

Unhandled Exception: type 'String' is not a subtype of type 'int'

这个错误通常发生在你试图将一个字符串类型的值赋给一个要求整数类型的变量或者函数参数。可以检查一下你的代码，看看是否有这样的情况出现。如果有，请确认该变量或参数的类型是否正确，或者使用 `int.parse()` 或 `double.parse()` 方法将字符串转换为数字类型。如果问题仍然存在，可以提供更多的上下文和代码细节，以便我能够更好地帮助你解决问题。