拆分字符串转换成在Perl数组 [英] Split a string into array in Perl

问题描述

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

期望的输出：

file1.gz
file2.gz
file3.gz

我想要的输出为 file1.gz 在 $ ABC [0] 和 $ ABC [1] 应该有 file2.gz ; $ ABC [2] 应该有 file3.gz 。我该如何分割？

I want the output to be file1.gz in $abc[0] and $abc[1] should have file2.gz; $abc[2] should have file3.gz. How do I split?

推荐答案

将字符串分割用空格是很简单的：

Splitting a string by whitespace is very simple:

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

这是拆分实际上是一种特殊形式（如该功能通常需要模式，而不是字符串）：

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

作为另一种特殊情况，拆分模拟的默认行为
  命令行工具 AWK 在 PATTERN 是省略或文字
  单个空格字符组成的字符串（如或\\ X20）。在这种情况下，在 EXPR 任何前导空格是
  除去之前发生分裂，而 PATTERN 作为代替处理
  如果它是 / \\ s + / ;特别是，这意味着任何连续
  空白（不只是单个空格字符）用作隔膜。

As another special case, split emulates the default behavior of the command line tool awk when the PATTERN is either omitted or a literal string composed of a single space character (such as ' ' or "\x20"). In this case, any leading whitespace in EXPR is removed before splitting occurs, and the PATTERN is 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.

下面是原问题的答案（用一个简单的字符串，而没有任何空格）：

---

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

也许你想拆就。广州扩展：

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

下面我用（小于？= ...）结构，这是的查找背后断言，基本上在按。广州子pceded行$ p $每个点使分裂。

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

如果您与固定集扩展的工作，你可以扩展模式，包括他们所有的：

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

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;

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