Bash script does not allow to handover content to a variable, which is partly the counter of a for loop

Question

I need to replace temporary stored files by variables. I call 6 files from a bitbucket repo and want to place each file content in an own variable.

# declare array for all files
declare -a file_array=("one.txt" "two.txt" "three.txt" "four.json" "five.txt" "six.txt")

for file in "${file_array[@]}"
do 
     echo "$file"
     "file_$file='$(curl -k -s -X GET -H "${var1}" -H "${var2}" "${var3}${var4}")'"

done

This code just gives me an error. It seems like it is always taking the content of the file as a command:

$'file_one.txt=\'<<file content>>'': command not found

Answer 1

You have at least two problems here: you try to put a dot in a variable name, which is invalid, and you try to assign expressions. Plus your quoting which also looks strange.

You could use namerefs (declare -n) and substitute the dots, e.g., with underscores:

# declare array for all files
declare -a file_array=("one.txt" "two.txt" ...)

for file in "${file_array[@]}"; do 
  echo "$file"
  declare -n f="file_${file/./_}"
  f=$(curl -k -s -X GET -H "${var1}" -H "${var2}" "${var3}${var4}")
done
printf '%s\n' "$file_one_txt"