Passing parameters to a Bash function

Question

I am trying to search how to pass parameters in a Bash function, but what comes up is always how to pass parameter from the command line.

I would like to pass param

Answer 1

Thought I'd pipe in with mention of another way to pass named parameters to bash... passing by reference. This is supported as of bash 4.0

#!/bin/bash
function myBackupFunction(){ # directory options destination filename
local directory="$1" options="$2" destination="$3" filename="$4";
  echo "tar cz ${!options} ${!directory} | ssh root@backupserver \"cat > /mnt/${!destination}/${!filename}.tgz\"";
}

declare -A backup=([directory]=".." [options]="..." [destination]="backups" [filename]="backup" );

myBackupFunction backup[directory] backup[options] backup[destination] backup[filename];

An alternative syntax for bash 4.3 is using a nameref

Although the nameref is a lot more convenient in that it seamlessly dereferences, some older supported distros still ship an older version so I won't recommend it quite yet.