目录中每个文件的Linux Shell脚本获取文件名并执行程序 [英] Linux Shell Script For Each File in a Directory Grab the filename and execute a program

问题描述

场景:

Linux系统中的文件夹.我想遍历文件夹中的每个.xls文件.

A folder in Linux system. I want to loop through every .xls file in a folder.

此文件夹通常由各种文件夹，各种文件类型(.sh，.pl，.csv等)组成.

This folder typically consists of various folders, various filetypes (.sh, .pl,.csv,...).

我要做的就是遍历根目录中的所有文件，并仅对.xls文件执行程序.

All I want to do is loop through all files in the root and execute a program only on .xls files.

问题是我必须执行的程序是将.xls转换为.csv格式的'xls2csv'.因此，对于每个.xls文件，我必须获取文件名并将其附加到.csv.

The problem is the program I have to execute is 'xls2csv' to convert from .xls to .csv format. So, for each .xls file I have to grab the filename and append it to .csv.

例如，我有一个test.xls文件，而xls2csv的参数为:xls2csv test.xls test.csv

For instance, I have a test.xls file and the arguments fro xls2csv are : xls2csv test.xls test.csv

我说得通吗?

推荐答案

重击:

for f in *.xls ; do xls2csv "$f" "${f%.xls}.csv" ; done

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