Linux Shell Script for Each File in a Directory Grab the Filename and Execute a Program

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

bash:

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

How to loop over files in directory and change path and add suffix to filename

A couple of notes first: when you use Data/data1.txt as an argument, should it really be /Data/data1.txt (with a leading slash)? Also, should the outer loop scan only for .txt files, or all files in /Data? Here's an answer, assuming /Data/data1.txt and .txt files only:

#!/bin/bash
for filename in /Data/*.txt; do
    for ((i=0; i<=3; i++)); do
        ./MyProgram.exe "$filename" "Logs/$(basename "$filename" .txt)_Log$i.txt"
    done
done

Notes:

• /Data/*.txt expands to the paths of the text files in /Data (including the /Data/ part)
• $( ... ) runs a shell command and inserts its output at that point in the command line
• basename somepath .txt outputs the base part of somepath, with .txt removed from the end (e.g. /Data/file.txt -> file)

If you needed to run MyProgram with Data/file.txt instead of /Data/file.txt, use "${filename#/}" to remove the leading slash. On the other hand, if it's really Data not /Data you want to scan, just use for filename in Data/*.txt.

Execute command on all files in a directory

The following bash code will pass $file to command where $file will represent every file in /dir

for file in /dir/*
do
  cmd [option] "$file" >> results.out
done

Example

el@defiant ~/foo $ touch foo.txt bar.txt baz.txt
el@defiant ~/foo $ for i in *.txt; do echo "hello $i"; done
hello bar.txt
hello baz.txt
hello foo.txt

Bash script to iterate files in directory and pattern match filenames

The easiest way is probably just to iterate each group separately. This side-steps the parsing issue entirely.

DIRECTORY=.

for i in $DIRECTORY/YYYYMMDD_*_bulk_import.csv; do
    # Process $i
done

for i in $DIRECTORY/YYYYMMDD_*_genstats_import.csv; do
    # Process $i
done

for i in $DIRECTORY/YYYYMMDD_*allstats.csv; do
    # Process $i
done

Set DIRECTORY to whatever directory you want to search. The default . will search the current working directory.

Shell Script to list files in a given directory and if they are files or directories

Your line:

for file in $dir; do

will expand $dir just to a single directory string. What you need to do is expand that to a list of files in the directory. You could do this using the following:

for file in "${dir}/"* ; do

This will expand the "${dir}/"* section into a name-only list of the current directory. As Biffen points out, this should guarantee that the file list wont end up with split partial file names in file if any of them contain whitespace.

If you want to recurse into the directories in dir then using find might be a better approach. Simply use:

for file in $( find ${dir} ); do

Note that while simple, this will not handle files or directories with spaces in them. Because of this, I would be tempted to drop the loop and generate the output in one go. This might be slightly different than what you want, but is likely to be easier to read and a lot more efficient, especially with large numbers of files. For example, To list all the directories:

find ${dir} -maxdepth 1 -type d

and to list the files:

find ${dir} -maxdepth 1 -type f

if you want to iterate into directories below, then remove the -maxdepth 1

How to get the list of files in a directory in a shell script?

search_dir=/the/path/to/base/dir/
for entry in "$search_dir"/*
do
  echo "$entry"
done

For files in directory, only echo filename (no path)

If you want a native bash solution

for file in /home/user/*; do
  echo "${file##*/}"
done

The above uses Parameter Expansion which is native to the shell and does not require a call to an external binary such as basename

However, might I suggest just using find

find /home/user -type f -printf "%f\n"

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