在Perl中删除换行符的最巧妙方法 [英] Neatest way to remove linebreaks in Perl

问题描述

我正在维护一个脚本，该脚本可以从各种来源获取其输入，并且可以每行对其进行处理.根据实际使用的来源，换行符可能是Unix风格，Windows风格，甚至对于某些聚合输入来说，还可能是mix(！).

I'm maintaining a script that can get its input from various sources, and works on it per line. Depending on the actual source used, linebreaks might be Unix-style, Windows-style or even, for some aggregated input, mixed(!).

从文件中读取时，会出现以下内容:

When reading from a file it goes something like this:

@lines = <IN>;
process(\@lines);

...

sub process {
    @lines = shift;
    foreach my $line (@{$lines}) {
        chomp $line;
        #Handle line by line
    }
}

所以，我要做的是用消除Unix风格或Windows风格的换行符的方式替换chomp. 我想出了太多的方法来解决这个问题，这是Perl的常见弊端之一:)

So, what I need to do is replace the chomp with something that removes either Unix-style or Windows-style linebreaks. I'm coming up with way too many ways of solving this, one of the usual drawbacks of Perl :)

您对切断通用换行符的最简洁方法有何看法?什么是最有效的?

What's your opinion on the neatest way to chomp off generic linebreaks? What would be the most efficient?

一个小小的澄清-方法'process'从某处获取行列表，不必从文件中读取.每行可能都有

A small clarification - the method 'process' gets a list of lines from somewhere, not nessecarily read from a file. Each line might have

• 没有尾随的换行符
• Unix风格的换行符
• Windows风格的换行符
• 仅返回回车(当原始数据具有Windows样式的换行符并以$/='\ n'读取时)
• 线条不同样式的集合集

推荐答案

经过 perlre 文档，到目前为止，我将提出我的最佳建议，该建议似乎效果很好. Perl 5.10添加了\ R字符类作为通用换行符:

After digging a bit through the perlre docs a bit, I'll present my best suggestion so far that seems to work pretty good. Perl 5.10 added the \R character class as a generalized linebreak:

$line =~ s/\R//g;

与以下相同:

(?>\x0D\x0A?|[\x0A-\x0C\x85\x{2028}\x{2029}])

我将让这个问题待一会儿，只是为了看看是否还有更多漂亮的方法值得建议.

I'll keep this question open a while yet, just to see if there's more nifty ways waiting to be suggested.

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