java的DateTimeFormatter比SimpleDateFormat更严格吗？以毫秒为单位分析日期

Question

我有一个简单的测试用例：

public static void main(String[] args) {
  String pattern = "yyyy-MM-dd HH:mm:ss.SSS";
  String date = "2017-01-15 15:15:15.5";

  SimpleDateFormat simpleDateFormat = new SimpleDateFormat(pattern);
  DateTimeFormatter dateTimeFormatter = DateTimeFormatter.ofPattern(pattern);
  FastDateFormat fastDateFormat = FastDateFormat.getInstance(pattern);
  try {
    System.out.println("SFD: " + simpleDateFormat.parse(date));
  }catch (Exception e) {
    System.err.println("SDF failed");
  }
  try {
    System.out.println("DTF: " + dateTimeFormatter.parse(date));
  }catch (Exception e) {
    System.err.println("DTF failed");
  }
  try {
    System.out.println("FDF: " + fastDateFormat.parse(date));
  }catch (Exception e) {
    System.err.println("FDF failed");
  }
}

输出如下：

SFD: Thu Jan 15 15:15:15 CET 1970
DTF failed
FDF: Thu Jan 15 15:15:15 CET 1970

根据结果，Java的8 DateTimeFormatter比SimpleDateFormat更严格。我的问题是，为什么在这种情况下，最好的方法是用.S作为毫秒或.SSS接受两个日期，比如用try/catch多次解析

Answer 1

您基本上是在问为什么SimpleDateFormat在模式显示为“.SSS”时接受一位毫秒值

原因在javadoc中SimpleDateFormat：

Pattern letters are usually repeated, as their number determines the exact presentation:

...

Number: For formatting, the number of pattern letters is the minimum number of digits, and shorter numbers are zero-padded to this amount. For parsing, the number of pattern letters is ignored unless it's needed to separate two adjacent fields.

显然，您可以通过调用setLenient(false)来改变这一点。然而，javadocs并没有具体说明宽大的实际效果

相比之下，javadoc for DateTimeFormatter只是简单地说“模式字母的计数决定了格式”，而不区分格式和解析

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为什么它们不同？您需要询问DateTimeFormatterAPI的设计人员，但我认为设计者认为SimpleDateFormat（默认）宽松解析模式的^{特别的和未指定的性质是有害的

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如何简单地让DateTimeFormatter接受具有一位或三位毫秒值的日期

一种方法是使用DateTimeFormatterBuilder创建格式化程序。这允许您为格式中的任何字段指定最小和最大宽度

Answer 2

SimpleDateFormat不是每个默认值都严格，因为属性lenient是每个默认值true。但是您可以将属性lenient设置为false以使其严格

SimpleDateFormat simpleDateFormat = new SimpleDateFormat(pattern);
simpleDateFormat.setLenient(false);
DateTimeFormatter dateTimeFormatter = DateTimeFormatter.ofPattern(pattern);
try {
    System.out.println("SFD: " + simpleDateFormat.parse(date));
} catch (Exception e) {
    System.err.println("SDF failed");
}
try {
    System.out.println("DTF: " + dateTimeFormatter.parse(date));
} catch (Exception e) {
    System.err.println("DTF failed");
}

结果将是

SDF failed
DTF failed

见DateFormat.parse(String, ParsePosition)

By default, parsing is lenient: If the input is not in the form used by this object's format method but can still be parsed as a date, then the parse succeeds. Clients may insist on strict adherence to the format by calling setLenient(false).