目录中每个文件的Linux Shell脚本获取文件名并执行程序

Question

Scenario :

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

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

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

Edit :

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.

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

Did I make sense?

Answer 1

庆典：

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

Answer 2

for i in *.xls ; do 
  [[ -f "$i" ]] || continue
  xls2csv "$i" "${i%.xls}.csv"
done

The first line in the do checks if the "matching" file really exists, because in case nothing matches in your for , the do will be executed with "*.xls" as $i . This could be horrible for your xls2csv .

Answer 3

Look at the find command.

What you are looking for is something like

find . -name "*.xls" -type f -exec program 

Post edit

find . -name "*.xls" -type f -exec xls2csv '{}' '{}'.csv;

will execute xls2csv file.xls file.xls.csv

Closer to what you want.

Answer 4

find . -type f -name "*.xls" -printf "xls2csv %p %p.csv\n" | bash

bash 4 (recursive)

shopt -s globstar
for xls in /path/**/*.xls
do
  xls2csv "$xls" "${xls%.xls}.csv"
done