Question

我编写的方法检查给定的String输入是否可以解析为正整数。

有没有更简洁的方法来写这个，所以我不重复拒绝该值的代码？

try {
  int num = Integer.parseInt(value);

  if (num <= 0) {
    errors.rejectValue(FIELD_FILE, INVALID_MESSAGE_KEY, new Object[]{lineNumber, fieldName}, "Line {0}: {1} must be a positive integer");
  }
} catch (NumberFormatException e) {
    errors.rejectValue(FIELD_FILE, INVALID_MESSAGE_KEY, new Object[]{lineNumber, fieldName}, "Line {0}: {1} must be a positive integer");
}

Answer 1

一个简单的方法：

int num = 0;
try {
   num = Integer.parseInt(value);
 } catch (NumberFormatException e) {
    num = -1;
 }
 if (num <= 0) {
   errors.rejectValue(FIELD_FILE, INVALID_MESSAGE_KEY, new Object[]{lineNumber, fieldName}, "Line {0}: {1} must be a positive integer");
 }

Answer 2

我相信这是你能得到的清洁工。即使它传递给try，如果它不在你期望的结果范围内，也会强制它进入catch。

try {
  int num = Integer.parseInt(value);

  if (num <= 0) {
    throw new NumberFormatException();
  }
} catch (NumberFormatException e) {
    errors.rejectValue(FIELD_FILE, INVALID_MESSAGE_KEY, new Object[]{lineNumber, fieldName}, "Line {0}: {1} must be a positive integer");
}

验证给定的字符串是否可以转换为正整数

2 个答案: