Bash function not returning expected output

Question

I'm trying to write a function for my Bash script to keep it DRY but for some reason the output of the code is not the same as when it's not inside a function.

What am I missing ?

Working:

#Get file name from file path
fileName="$(basename "$file")";
#Remove " ' and white space from name
fileName=${fileName//[\"\'\ ]/};
convert "$file" -resize $RESOLUTION\> "$OUTPUT_PATH"$fileName;

Not working:

function cleanUpName() {
  #Get file name from file path
  fileName="$(basename "$1")";
  #Remove " ' and white space from name
  echo ${fileName//[\"\'\ ]/};
}

convert "$file" -resize $RESOLUTION\> "$OUTPUT_PATH"$( cleanUpName $file);

you are not quoting the second $file reference. In fact, you are missing a lot of quotes, e.g. in the function's echo statement and around the $(…) subshell in the last line. This makes all these strings subject to multiple shell expansions. — Robin479
– Robin479, Commented Oct 1, 2019 at 15:37
Can you provide sample input that fail, expected output, and actual output ? — dash-o
– dash-o, Commented Oct 1, 2019 at 15:39

glenn jackman · Accepted Answer · 2019-10-01 16:33:18Z

1

As @Robin479 suggested in the comments, I was missing quotes for my file variable working code is as follow:

function cleanUpName() {
  #Get file name from file path
  fileName="$(basename "$1")";
  #Remove " ' and white space from name
  echo "${fileName//[\"\'\ ]/}"
}
convert "$file" -resize $RESOLUTION\> "$OUTPUT_PATH$( cleanUpName "${file}")"

edited Oct 1, 2019 at 16:33

glenn jackman

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answered Oct 1, 2019 at 15:40

Philou

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1 Comment

Robin479 Over a year ago

You should quote the echo argument and the subshell in the last line, i.e. echo "${fileName//[\"\'\ ]/}" and "${OUTPUT_PATH}$( cleanUpName "${file}")"

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