Linux Shell Script For Each File in a Directory Grab the filename and execute a program

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

Answer 1

bash:

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

Answer 2

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 3

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

Answer 4

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.