Bash script - Modify output of command and print into file

Question

Im trying to get text output of specified command, modify it somehow (e.g. add prefix before output) and print into file (.txt or .log)

LOG_FILE=...
LOG_ERROR_FILE=..
command_name >> ${LOG_FILE} 2>> ${LOG_ERROR_FILE}

I would like to do it in one line to modify what command will return and print it into files. The same situation for error output and regular output.

Im beginner in bash scripts, so please be understading.

Answer 1

Create a function to execute commands and capture sterr an stdout to variables.

function execCommand(){
  local command="$@"
  {
    IFS=$'\n' read -r -d '' STDERR;
    IFS=$'\n' read -r -d '' STDOUT;
  } < <((printf '\0%s\0' "$($command)" 1>&2) 2>&1)
}

function testCommand(){
    grep foo bar
    echo "return code $?"
}

execCommand testCommand
echo err: $STDERR
echo out: $STDOUT

execCommand "touch /etc/foo"
echo err: $STDERR
echo out: $STDOUT

execCommand "date"
echo err: $STDERR
echo out: $STDOUT

output

err: grep: bar: No such file or directory
out: return code 2
err: touch: cannot touch '/etc/foo': Permission denied
out:
err:
out: Mon Jan 31 16:29:51 CET 2022

Now you can modify $STDERR & $STDOUT

execCommand testCommand &&  { echo "$STDERR" > err.log; echo "$STDOUT" > out.log; }

Explanation: Look at the answer from madmurphy