Perl 正则表达式从命令行对文件进行操作 [英] Perl regex to act on a file from the command line

问题描述

在一个文件中，说 xyz.txt 我想替换任何数字的模式，后跟一个点示例:1.,2.,10.,11.等等.有一个空格.怎么在命令行写一个perl命令作用于文件做上面的事情，应该用什么正则表达式?

In a file, say xyz.txt i want to replace the pattern of any number followed by a dot example:1.,2.,10.,11. etc.. with a whitespace. How to compose a perl command on the command line to act on the file to do the above, what should be the regex to be used ?

请帮忙谢谢.

推荐答案

这必须是 Perl oneliner 吗?

This HAS to be a Perl oneliner?

perl -i -pe  's/\d+\./ /g' <fileName>

Perl 命令行选项:-i 用于指定对输入文件执行的操作.如果你不给它一个文件扩展名，原始文件就会丢失并被 Perl 修改过的输出替换.例如，如果我有这个:

The Perl command line options: -i is used to specify what happens to the input file. If you don't give it a file extension, the original file is lost and is replaced by the Perl munged output. For example, if I had this:

perl -i.bak -pe  's/\d+\./ /g' <fileName>

原始文件将以 .bak 后缀存储， 本身将包含您的输出.

The original file would be stored with a .bak suffix and <fileName> itself would contain your output.

-p 意味着将您的 Perl 程序封装在一个打印循环中，看起来有点如下:

The -p means to enclose your Perl program in a print loop that looks SOMEWHAT like this:

while ($_ = <>) {
    <Your Perl one liner>
    print "$_";
}

这是对正在发生的事情的稍微简化的解释.您可以通过从命令行执行 perldoc perlrun 来查看实际的 perl 循环.主要思想是它允许您像 sed 或 awk 一样对文件的每一行进行操作.

This is a somewhat simplified explanation what's going on. You can see the actual perl loop by doing a perldoc perlrun from the command line. The main idea is that it allows you to act on each line of a file just like sed or awk.

-e 只包含您的 Perl 命令.

The -e simply contains your Perl command.

您也可以进行文件重定向:

You can also do file redirection too:

perl -pe  's/\d+\./ /g' < xyz.txt > xyz.txt.out

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