Question

我试着弄清楚这一段时间，所以我有一个方法调用一个count（）方法，假设要抛出异常

count（）方法

public int count() throws ParseException {
  return something that may throw the ParseException
}

然后从这里打电话

ParseQuery<ParseObject> query = ParseQuery.getQuery(className);
query.fromLocalDatastore();

int result = 0;

try {
   result = query.count();
} catch (ParseException e) {
   result = 0;
}

return result;

现在我一直在尝试不同的场景，但无论IDE仍然没有编译并给我以下错误

Error:(254, 11) error: exception ParseException is never thrown in body of corresponding try statement

Error:(253, 33) error: unreported exception ParseException; must be caught or declared to be thrown

行result = query.count（）;

我不知道我做错了什么，谢谢你的帮助

Answer 1

你不能捕获你的try块永远不会抛出的异常，因为建议错误

try {
   result = query.count(); // this statement not throwing ParseException
} catch (ParseException e) { // you are trying to catch ParseException that never gonna throw. 
   result = 0;
}

就像

    try {
           .... code // throws ExceptionA
        }
    catch (ExceptionB e) { // and calling ExceptionB

        }

Android未处理的异常错误

1 个答案: