提取文件名Shell脚本的一部分 [英] Extract part of a filename shell script

问题描述

在bash中，我想提取许多文件名的一部分并将该输出保存到另一个文件中.

In bash I would like to extract part of many filenames and save that output to another file.

文件的格式为coffee_ {我想输入一些数字} .freqdist.

The files are formatted as coffee_{SOME NUMBERS I WANT}.freqdist.

#!/bin/sh
for f in $(find . -name 'coffee*.freqdist)

该代码将找到所有coffee_ {我想要的数字} .freqdist文件.现在，如何制作仅包含{我想要的一些数字}的数组并将其写入文件?

That code will find all the coffee_{SOME NUMBERS I WANT}.freqdist file. Now, how do I make an array containing just {SOME NUMBERS I WANT} and write that to file?

我知道写文件一会以以下结尾.

I know that to write to file one would end the line with the following.

  > log.txt

尽管缺少如何过滤文件名列表的中间部分，但我仍然缺少.

I'm missing the middle part though of how to filter the list of filenames.

推荐答案

您可以在bash中以本机方式进行操作，如下所示:

You can do it natively in bash as follows:

filename=coffee_1234.freqdist
tmp=${filename#*_}
num=${tmp%.*}
echo "$num"

这是一个纯bash解决方案.不涉及任何外部命令(如sed)，因此速度更快.

This is a pure bash solution. No external commands (like sed) are involved, so this is faster.

使用以下命令将这些数字附加到文件中:

Append these numbers to a file using:

echo "$num" >> file

(开始循环之前，您需要删除/清除文件.)

(You will need to delete/clear the file before you start your loop.)

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