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

Solution 1:

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

Solution 2:

This is a way to do it where the syntax is simpler for me to understand:

yourfilenames=`ls ./*.txt`
for eachfile in $yourfilenames
do
   echo $eachfile
done

./ is the current working directory but could be replaced with any path
*.txt returns anything.txt
You can check what will be listed easily by typing the ls command straight into the terminal.

Basically, you create a variable yourfilenames containing everything the list command returns as a separate element, and then you loop through it. The loop creates a temporary variable eachfile that contains a single element of the variable it's looping through, in this case a filename. This isn't necessarily better than the other answers, but I find it intuitive because I'm already familiar with the ls command and the for loop syntax.

Solution 3:

The other answers on here are great and answer your question, but this is the top google result for "bash get list of files in directory", (which I was looking for to save a list of files) so I thought I would post an answer to that problem:

ls $search_path > filename.txt

If you want only a certain type (e.g. any .txt files):

ls $search_path | grep *.txt > filename.txt

Note that $search_path is optional; ls > filename.txt will do the current directory.

Solution 4:

for entry in "$search_dir"/* "$work_dir"/*
do
  if [ -f "$entry" ];then
    echo "$entry"
  fi
done