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

Question

I\'m trying to get the contents of a directory using shell script.

My script is:

for entry in `ls $search_dir`; do
    echo $entry
done

Answer 1

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.