比较时间字符串在Java中的格式

Question

我有两个时间字符串，它们可以采用任何格式（12小时或24小时，带有或不带有时区）。如果它们的格式在Java中不同，并且数据不匹配，该如何比较？

PS]我准备了一个正则表达式列表，并使用这些表达式匹配字符串以获取格式，然后使用字符串的equals（）方法检查数据差异。这种方法的问题是（20：01：02,20 01 01）返回格式差异，而预期结果应该是数据差异。请帮助，我在这里呆了很长时间。

正则表达式的映射-

private static final Map<String, String> TIME_FORMAT_REGEXPS = new HashMap<String, String>() {{
    put("^(1[0-2]|0?[1-9]):([0-5]?[0-9])(●?[AP]M)?$", "1");
    put("^(2[0-3]|[01]?[0-9]):([0-5]?[0-9])$", "2");
    put("^(1[0-2]|0?[1-9]):([0-5]?[0-9]):([0-5]?[0-9])(●?[AP]M)?$", "3");
    put("^(2[0-3]|[01]?[0-9]):([0-5]?[0-9]):([0-5]?[0-9])$", "4");
    put("^(2[0-3]|[01][0-9]):?([0-5][0-9])$", "5");
    put("^(?<hour>2[0-3]|[01][0-9]):?(?<minute>[0-5][0-9])$", "6");
    put("^(2[0-3]|[01][0-9]):?([0-5][0-9]):?([0-5][0-9])$", "7");
    put("^(?<hour>2[0-3]|[01][0-9]):?(?<minute>[0-5][0-9]):?(?<second>[0-5][0-9])$", "8");
    put("^(Z|[+-](?:2[0-3]|[01][0-9])(?::?(?:[0-5][0-9]))?)$", "9");
    put("^(2[0-3]|[01][0-9]):?([0-5][0-9]):?([0-5][0-9])(Z|[+-](?:2[0-3]|[01][0-9])(?::?(?:[0-5][0-9]))?)$", "10");
    put("^(?<hour>2[0-3]|[01][0-9]):?(?<minute>[0-5][0-9]):?(?<second>[0-5][0-9])(?<timezone>Z|[+-]"
                    + "(?:2[0-3]|[01][0-9])(?::?(?:[0-5][0-9]))?)$",
            "11");

}};

检查字符串格式的功能-

private String determineTimeFormat(String dateString) {
    for (String regexp : TIME_FORMAT_REGEXPS.keySet()) {
        if (dateString.toLowerCase().matches(regexp)) {
            return TIME_FORMAT_REGEXPS.get(regexp);
        }
    }
    return "100"; // Unknown format.
}

Answer 1

您可以尝试这样的事情：

// create a formatter
DateTimeFormatter formatter = new DateTimeFormatterBuilder()
        .appendPattern("[dd/MM/yyyy HH:mm:ss]")  // Add as many optional patterns as required
        .appendPattern("[dd-MM-yyyy HH:mm:ss]")
        .appendPattern("[dd-MM-yyyy[ [HH[:mm][:ss]]]]")   // nested optional tokens specified with square brackets []
        .appendOptional(DateTimeFormatter.ISO_DATE_TIME)  // can use standard Java DateTimeFormatters as well
        .parseDefaulting(ChronoField.HOUR_OF_DAY, 0)      // supply default values for missing fields
        .parseDefaulting(ChronoField.MINUTE_OF_HOUR, 0)
        .parseDefaulting(ChronoField.SECOND_OF_MINUTE, 0)
        .toFormatter();

// each of these will parse successfully
LocalDateTime time1 = LocalDateTime.parse("2020-02-01T01:02:00+00:00", formatter);
LocalDateTime time2 = LocalDateTime.parse("01-02-2020 01:02:00", formatter);
LocalDateTime time3 = LocalDateTime.parse("01-02-2020 01:02", formatter);
LocalDateTime time4 = LocalDateTime.parse("01-02-2020", formatter);


// use .equals() method to compare times
assert time1.equals(time2);  // true
assert time2.equals(time3);  // true
assert time3.equals(time4);  // true

此相关答案的帮助https://stackoverflow.com/a/39685202/7174786

有关其他DateTimeFormatter的详细信息，请参见JavaDocs：https://docs.oracle.com/javase/8/docs/api/java/time/format/DateTimeFormatter.html