AWK在模式上将文件拆分为较小的文件 [英] AWK split files into smaller files on pattern

问题描述

我有一个类似于以下文件:

I have a file similar to:

ISA.00. <rest of line>
information lines
information lines
...
...
...
ISA.00. <rest of line>
information lines
information lines
...
...
ISA.00. <rest of line>

我正在使用:

awk "/ISA.00./{file=(FILENAME)(++i)}{print > file}" %LOCATION%\%CURRFILE%

要遍历文件文件夹，请将它们拆分为仅包含一个"ISA"数据集的文件.我想保留原始源文件名，并在末尾添加1，例如subfile1，subfile2，subfile3.我都做到了.

To loop through a folder of files, splitting them into files containing one "ISA" data set only. I want to keep the original source file name and add 1 to the end like subfile1, subfile2, subfile3. I have accomplished both.

我将如何定义一个输出目录，子文件"将进入该目录，将原始源文件保留在原始位置，而子文件目录"中仅包含子文件".

How would I define an output directory, where the "subfiles" would go, leaving the original source file in the original location, and only having the "subfiles" in the "subfile directory".

我陷入了种种死胡同，并寻求紧急帮助.

I have run into a variety of dead ends and am asking for help as a last resort.

提前谢谢！

推荐答案

在文件名之前添加目录名称:

Add the directory name before the filename:

print > "subfiles/"file

如果在Windows上，则应使用反斜杠\分隔目录，如下所示:

If you are on windows, you should use the backslash \ to separate the directory, like this:

print > "subfiles\\"file

(反斜杠应转义)

这假定目录已经存在.另外，在使用awk时，请使用单引号，例如awk 'commands'文件

This assumes that the directory already exists. Also, use single quotes when with awk, like awk 'commands' file

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