在 Perl 中提取日期掩码的正则表达式是什么？

发布于 2024-12-20 13:28:55 字数 389 浏览 0 评论 0原文

我有一个 perl 字符串，其中包含目录规范。如果字符串包含构成日期掩码的任何单个子字符串或子字符串组合，我想提取该子字符串。例如，目录规范可能是：

/mydir/data/YYYYMMDD

我希望能够提取“YYYYMMDD”字符串。但是，路径的该部分可以是以下字符串的任何单个或组合：

YY
YYYY
MM
DD

因此目录规范字符串可以读取：

   /mydir/data/DD/data2

并且我希望作为正则表达式比较的结果返回“DD”。当字符串必须包含一个或多个日期掩码字符串并且该字符串必须位于两个“/”字符之间或存在于字符串末尾时，如何捕获该字符串？

原文

I have a string in perl that contains a directory specification. If the string contains any individual or combination of substrings that comprise a date mask, I want to extract that substring. For example, the directory spec may be:

/mydir/data/YYYYMMDD

I want to be able to extract the "YYYYMMDD" string. However that portion of the path could be any individual or combination of the following strings:

YY
YYYY
MM
DD

So the directory spec string could read:

   /mydir/data/DD/data2

and I want the "DD" returned as a result of the regex comparison. How do I capture the string when it must contain one or more of those date mask strings and that string must be between two "/" characters or exist at the end of the string?

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一刻暧昧 2024-12-27 13:28:55

我假设 YYYY 和 YY 不应同时出现在同一模式中，因为否则就没有意义。

use Data::Munge qw(list2re);
use List::MoreUtils qw(uniq);
use Algorithm::Combinatorics qw(variations);
use Perl6::Take qw(gather take);

list2re
uniq
gather {
    for my $n ([qw(YYYY MM DD)], [qw(YY MM DD)]) {
        for my $k (1..scalar @$n) {
            take map { join q(), @$_ } variations($n, $k)
        }
    }
}

I'm making the assumption that YYYY and YY shall not both appear in the same pattern, because otherwise it does not make sense.

use Data::Munge qw(list2re);
use List::MoreUtils qw(uniq);
use Algorithm::Combinatorics qw(variations);
use Perl6::Take qw(gather take);

list2re
uniq
gather {
    for my $n ([qw(YYYY MM DD)], [qw(YY MM DD)]) {
        for my $k (1..scalar @$n) {
            take map { join q(), @$_ } variations($n, $k)
        }
    }
}

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别念他 2024-12-27 13:28:55

我假设只有一个“日期”组件，或者如果没有，您需要第一个：

#!/usr/bin/perl
use warnings;
use strict;

my @paths = qw(
    /mydir/data/YYYYMMDD
    /mydir/data/YY/data2
    /mydir/data/YYMM/data2
    /mydir/data/DD/data2
);

foreach my $path (@paths) {
    my($date) = grep /^(([YMD])\2)+$/, split '/', $path;
    print "$path: $date\n";
}

I assume that there is only one "date" component, or if not, that you want the 1st one:

#!/usr/bin/perl
use warnings;
use strict;

my @paths = qw(
    /mydir/data/YYYYMMDD
    /mydir/data/YY/data2
    /mydir/data/YYMM/data2
    /mydir/data/DD/data2
);

foreach my $path (@paths) {
    my($date) = grep /^(([YMD])\2)+$/, split '/', $path;
    print "$path: $date\n";
}

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魄砕の薆 2024-12-27 13:28:55

假设掩码字段始终按 Y - M - D 的顺序排列，这将满足您的需要：

my ($mask) = $path =~ m{ / ( (?:YY){0,2} (?:MM)? (?:DD)? ) (?:/|$) }x;

Assuming the mask fields are always in the order Y - M - D, this will do what you need:

my ($mask) = $path =~ m{ / ( (?:YY){0,2} (?:MM)? (?:DD)? ) (?:/|$) }x;

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咽泪装欢 2024-12-27 13:28:55

~~我会使用~~

my ($date) = m{/([0-9]{2,8})(?:/|$)}
并检查是否
not(length($date) % 2) # $date has even length

~~，也许还有一些检查是否有效的组合。~~

更新： 好的，要只获取掩码，而不是数字，您可以将其更改为“

my ($date) = m{/([YMD]{2,8})(?:/|$)};
my $check = $date;
$check =~ s/YYYY/y/;
$check =~ s/MM//;
$check =~ s/DD//;
print "Matches $date\n" if grep $_ eq $check, (q{}, 'y', 'YY');

这应该排除所有无效组合，例如 YYDDYY 或 YYYYMMYY 等”。

~~I'd use~~

my ($date) = m{/([0-9]{2,8})(?:/|$)}
and check whether
not(length($date) % 2) # $date has even length

~~and maybe some checks for valid combinations.~~

Update: OK, to just get the mask, not the numbers, you can change this to

my ($date) = m{/([YMD]{2,8})(?:/|$)};
my $check = $date;
$check =~ s/YYYY/y/;
$check =~ s/MM//;
$check =~ s/DD//;
print "Matches $date\n" if grep $_ eq $check, (q{}, 'y', 'YY');

This should exclude all invalid combinations like YYDDYY or YYYYMMYY and so on.

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~没有更多了~