Iterate through list return by find command in shell script

Question

I am trying to find .csv file in directory and assign its name to a variable in a shell script.

I have written following code.

x="$(find extracted_data/*.csv)"
echo "$x"
IFS = $'\n'
for i in "${x[@]}"; do
        echo "${i}"
done

However, after running this code I get following output:

extracted_data/another_file.csv 
extracted_data/sample.csv 
sheet.sh: 4: IFS: not found 
sheet.sh: 5: Bad substitution

What is wrong done in code as it is giving those two errors?

I have followed following two links for my code loop through file names returned by find and for array

Please consider adding a tag to indicate which shell you are using, it will help identify errors that are more subtle due to shell-specific syntax. — AdminBee
– AdminBee, Commented Jan 10, 2023 at 14:08

AdminBee · Accepted Answer · 2023-01-10 14:02:12Z

To start off, I will reiterate the recommendation to not loop over find's output.

Now, to address your question: There is a rather simple syntax error in your script. The variable assignment

IFS = $'\n'

must not have spaces around the = sign. Shell variable assignments must be of the form

IFS=$'\n'

You may want to check out shellcheck, also available as standalone tool in many Linux distributions, to guard against such errors.

That said, what you want to accomplish can be done in a simpler and less error-prone way by simply stating

for x in extracted_data/*.csv
do
  echo "$x"
done

This wouldn't even break if the filenames contained newlines (which unfortunately is a valid character for filenames).

1 Answer 1