regex for 24小时unix时间戳

Question

我想为一个24小时的unix时间戳创建一个regex，例如：01/01/2015 00:00:00 **(1420066800)** to 01/01/2015 23:59:59 **(1420153199)**，这是86399秒的差异。unix时间戳格式。在

我使用的是range_regexpython库，但是对于如此大的范围来说，它是有缺陷的。range_to_pattern方法（range_to_pattern(1420066800, 1420153199)）将生成一个正则表达式：1420[0-1][5-6][3-6][1-8]\\d{2} 对于静态边界来创建regex来说，这是很好的，但是对于像：1420159111这样的值，因为从左边开始的7位数字（9）不在第三个范围组（[3-6]）。在

有人能提供一个更好的python3库或者一个解决方法来创建一个86400秒的regex。一天之内？在

Answer 1

# coding=utf8
# the above tag defines encoding for this document and is for Python 2.x compatibility

import re

regex = r"1420([0]([6]([6]([8]([0][0-9])|[9][0-9]{2})|[7-9][0-9]{3})|[7-9][0-9]{4})|[1]([5]([3]([1]([9][0-9]|[0-8][0-9]{1})|[0][0-9]{2})|[0-2][0-9]{3})|[0-4][0-9]{4}))"

test_str = ("01/01/2015 00:00:00 (1420066800) до 01/01/2015 23:59:59 (1420153199)\n\n"
    "1420016799     -no\n"
    "1420066799     -no\n"
    "1420066800     -yes\n"
    "1420066801     -yes\n"
    "1420067820     -yes\n"
    "1420067920     -yes\n"
    "1420073199     -yes\n"
    "1420103199     -yes\n"
    "1420152191     -yes\n"
    "1420153181     -yes\n"
    "1420153199     -yes\n"
    "1420153200     -no\n"
    "1420163199     -no")

matches = re.finditer(regex, test_str)

for matchNum, match in enumerate(matches):
    matchNum = matchNum + 1

    print ("Match {matchNum} was found at {start}-{end}: {match}".format(matchNum = matchNum, start = match.start(), end = match.end(), match = match.group()))

    for groupNum in range(0, len(match.groups())):
        groupNum = groupNum + 1

        print ("Group {groupNum} found at {start}-{end}: {group}".format(groupNum = groupNum, start = match.start(groupNum), end = match.end(groupNum), group = match.group(groupNum)))

# Note: for Python 2.7 compatibility, use ur"" to prefix the regex and u"" to prefix the test string and substitution.

在线：https://regex101.com/r/blnST4/1

Answer 2

根据我上面的评论，你用错了那个库中的函数。在

您应该使用以下方法：

range_to_regex(1420066800, 1420153199)

这将返回正确的正则表达式：

^{pr2}$