正则表达式语法中的问题

您的日期正则表达式：

^(((0[1-9]|[12]'d|3[01])'/(0[13578]|1[02])'/((19|[2-9]'d)'d{2}))|((0[1-9]|[12]'d|30)'/(0[13456789]|1[012])'/((19|[2-9]'d)'d{2}))|((0[1-9]|1'd|2[0-8])'/02'/((19|[2-9]'d)'d{2}))|(29'/02'/((1[6-9]|[2-9]'d)(0[48]|[2468][048]|[13579][26])|((16|[2468][048]|[3579][26])00))))$

非常有趣。我决定对此进行分析，看看它到底匹配什么。事实证明，这个正则表达式匹配所有格式为DD/MM/YYYY的有效日期，从1900年到9999年。有趣的是，它还正确匹配了从1597年到9999年的所有有效闰日。此正则表达式了解每个月的有效天数。它知道5月有31天，6月只有30天。它还知道二月有28天，闰年有29天。在这里，它被分解，以便凡人可以阅读：

$re_date = '%
    # Match all valid DD/MM/YYYY dates from 1900 to 9999 and
    #   all leap days from year 1597 to 9999.
    ^                           # Anchor to start of string.
    ( # $1:
      (  # $2: Date format alternative 1: (months having 31 days)
        ( 0[1-9]|[12]'d|3[01])  # $3: Day: 01-09,10-19,20-29,30,31
        '/
        (0[13578]|1[02])        # $4: Month: 01,03,05,07,08,10,12
        '/
        ((19|[2-9]'d)'d{2})     # $5,$6: Year: 1900-9999
      )                         # End $2:
    | (  # $7: Date format alternative 2: (months having 30 days)
        (0[1-9]|[12]'d|30)      # $8: Day: 01-09,10-19,20-29,30
        '/
        (0[13456789]|1[012])    # $9: Month: 01,03-09,10-12
        '/
        ((19|[2-9]'d)'d{2})     # $10,$11: Year: 1900-9999
      )                         # End $7:
    | (  # $12: Date format alternative 3: (month having 28 days)
        (0[1-9]|1'd|2[0-8])     # $13: Day 01-09,10-19,20-28
        '/
        02                      # Month: 02
        '/
        ((19|[2-9]'d)'d{2})     # $14,$15: Year: 1900-9999
      )                         # End $12:
    | (  # $16: Date format alternative 3: (leap days)
        29                      # Day: 29
        '/
        02                      # Month: 02
        '/ # Match all valid leap day dates from year 1597 to 9999.
        (                       # $17: Year alt 1 (divisible by 4 but not 100)
          (1[6-9]|[2-9]'d)      # $18: Century part: 16-19,20-99
          ( 0[48]               # $19: Year part: Either 04-08
          | [2468][048]         # or 20,24,28,40,44,48,60,64,68,80,84,88
          | [13579][26]         # or 12,16,32,36,52,56,72,76,92,96,
          )                     # End $19:
        | (                     # or $20: Year alternative 2 (divisible by 400)
            ( 16                # $21: Century part: Either 16
            | [2468][048]       # or 20,24,28,40,44,48,60,64,68,80,84,88
            | [3579][26]        # or 32,36,52,56,72,76,92,96
            )                   # End $21:
            00                  # Year part: 00
          )                     # End $20:
        )                       # End $17:
      )                         # End $16:
    )                           # End $1:
    $                           # Anchor to end of string.
    %x';

为了解决眼前的问题，这里有一个更精确的正则表达式：

$count = preg_match_all('%
    # Match version/date sub-string
    'b          # Anchor to word boundary.
    (           # $1: Version number.
      [Vv]      # Version identifier (allow V or v).
      [ ]+      # One or more spaces.
      [0-9]+    # Major version number is one or more digits.
      (?:       # Group minor version numbers.
        '.      # Minor versions separated by dot.
        [0-9]+  # Minor version is one or more digits.
      )*        # Zero or more minor versions.
    )           # End $1: Version number.
    [ ]+        # One or more spaces.
    (           # $2: Date.
      [0-9]{4}  # Year is four digits.
      /         # / Separator.
      [0-9]{2}  # Month is two digits.
      /         # / Separator.
      [0-9]{2}  # Day is two digits.
    )           # End $2: Date.
    %x', $text, $matches);
if ($count > 0) {
    $versions = $matches[1];
    $dates    = $matches[2];
    printf("Found %d matches:'n", $count);
    for ($i = 0; $i < $count; ++$i) {
        printf("  Match%3d:  Version: %-15s  Date: %s'n",
            $i + 1, $versions[$i], $dates[$i]);
    }
} else {
    echo("No matches found.'n");
}

注意：当处理像这样的非平凡正则表达式时，最好使用'x'自由间距模式来编写它们。这允许添加大量的注释和缩进，使其更易于阅读。