string 在 Perl 中将字符串拆分为数组

Split a string into array in Perl

提问by Rahul Reddy

my $line = "file1.gz file2.gz file3.gz";
my @abc = split('', $line);
print "@abc\n";

Expected output:

预期输出：

file1.gz
file2.gz
file3.gz

I want the output to be file1.gzin $abc[0], file2.gzin $abc[1], and file3.gzin $abc[2]. How do I split $line?

我希望输出为file1.gzin $abc[0]、file2.gzin$abc[1]和 file3.gzin $abc[2]。我如何拆分$line？

回答by raina77ow

Splitting a string by whitespace is very simple:

按空格拆分字符串非常简单：

print $_, "\n" for split ' ', 'file1.gz file1.gz file3.gz';

This is a special form of splitactually (as this function usually takes patterns instead of strings):

这是split实际上的一种特殊形式（因为此函数通常采用模式而不是字符串）：

As another special case, splitemulates the default behavior of the command line tool awkwhen the PATTERNis either omitted or a literal string composed of a single space character (such as ' 'or "\x20"). In this case, any leading whitespace in EXPRis removed before splitting occurs, and the PATTERNis instead treated as if it were /\s+/; in particular, this means that any contiguous whitespace (not just a single space character) is used as a separator.

作为另一种特殊情况，当被省略或由单个空格字符（例如或）组成的文字字符串时，split模拟命令行工具的默认行为。在这种情况下，任何前导空格 in在拆分发生之前被删除，而是被视为; 特别是，这意味着任何连续的空格（不仅仅是单个空格字符）都用作分隔符。awkPATTERN' '"\x20"EXPRPATTERN/\s+/

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Here's an answer for the original question (with a simple string without any whitespace):

这是原始问题的答案（带有一个没有任何空格的简单字符串）：

Perhaps you want to split on .gzextension:

也许您想拆分.gz扩展名：

my $line = "file1.gzfile1.gzfile3.gz";
my @abc = split /(?<=\.gz)/, $line;
print $_, "\n" for @abc;

Here I used (?<=...)construct, which is look-behind assertion, basically making split at each point in the line preceded by .gzsubstring.

在这里，我使用了后视断言的(?<=...)构造，基本上在子字符串前面的行中的每个点进行拆分。.gz

If you work with the fixed set of extensions, you can extend the pattern to include them all:

如果您使用固定的扩展集，您可以扩展模式以包含它们：

my $line = "file1.gzfile2.txtfile2.gzfile3.xls";
my @exts = ('txt', 'xls', 'gz');
my $patt = join '|', map { '(?<=\.' . $_ . ')' } @exts;
my @abc = split /$patt/, $line;
print $_, "\n" for @abc;

回答by e2-e4

Having $lineas it is now, you can simply split the string based on at least one whitespace separator

有$line像现在，你可以基于至少一个空格分隔简单地分割字符串

my @answer = split(' ', $line); # creates an @answer array

then

然后

print("@answer\n");               # print array on one line

or

或者

print("$_\n") for (@answer);      # print each element on one line

I prefer using ()for split, printand for.

我更喜欢使用()for split,print和for。

回答by void

I found this one to be very simple!

我发现这个非常简单！

my $line = "file1.gz file2.gz file3.gz";

my @abc =  ($line =~ /(\w+[.]\w+)/g);

print $abc[0],"\n";
print $abc[1],"\n";
print $abc[2],"\n";

output:

输出：

file1.gz 
file2.gz 
file3.gz

Here take a look at this tutorial to find more on Perl regular expressionand scroll down to More matchingsection.

在这里查看本教程以查找有关Perl 正则表达式的更多信息并向下滚动到更多匹配部分。

回答by Thanos

You already have multiple answers to your question, but I would like to add another minor one here that might help to add something.

您的问题已经有多个答案，但我想在这里添加另一个可能有助于添加一些内容的次要答案。

To view data structures in Perl you can use Data::Dumper. To print a string you can use say, which adds a newline character "\n"after every call instead of adding it explicitly.

要在 Perl 中查看数据结构，您可以使用Data::Dumper. 要打印字符串，您可以使用say，它"\n"在每次调用后添加一个换行符，而不是显式添加它。

I usually use \swhich matches a whitespace character. If you add +it matches one or more whitespace characters. You can read more about it here perlre.

我通常使用\swhich 匹配空白字符。如果添加+它匹配一个或多个空白字符。您可以在此处阅读更多相关信息perlre。

#!/usr/bin/perl

use strict;
use warnings;

use Data::Dumper;

use feature 'say';

my $line = "file1.gz file2.gz file3.gz";
my @abc  = split /\s+/, $line;

print Dumper \@abc;
say for @abc;

回答by user3682640

Just use /\s+/ against '' as a splitter. In this case all "extra" blanks were removed. Usually this particular behaviour is required. So, in you case it will be:

只需使用 /\s+/ 对 '' 作为分隔符。在这种情况下，所有“额外”的空白都被删除了。通常需要这种特殊行为。所以，在你的情况下，它将是：

my $line = "file1.gz file1.gz file3.gz";
my @abc = split(/\s+/, $line);