pass command with stdout to a bash function

Question

ok.. this is the deal. i have a bash script like the following. the script is only to show you what i mean .. it may look strang... but is exactly what i need.:

#/bin/bash runcommand () { message="$1" shift echo "$message" $@ > /tmp/logfile if [ $? -gt 0 ]; then cat /tmp/logfile fi } runcommandwrapper () { wrapperoptions="$1" shift $* } rm -f /tmp/test ; rm -f /tmp/logfile runcommand "echo into file" echo "SUCCESS" > /tmp/test echo "-----------------" echo "test file:" echo "-----------------" cat /tmp/test echo "-----------------" echo echo "-----------------" echo "logfile file:" echo "-----------------" cat /tmp/logfile echo "-----------------" echo echo echo rm -f /tmp/test ; rm -f /tmp/logfile runcommand "echo into file" 'echo "SUCCESS" > /tmp/test' echo "-----------------" echo "test file:" echo "-----------------" cat /tmp/test echo "-----------------" echo echo "-----------------" echo "logfile file:" echo "-----------------" cat /tmp/logfile echo "-----------------" echo 

this works

runcommand "running command mount" mount 

this does not work

runcommand "running command fdisk" fdisk > /tmp/fdiskoutput 

in this case the text in quotation marks is not treated as a whole argument within the wrapper script. try it, you'll see what I mean. --> SOLVED

So Running the above script returns:

----------------- test file: ----------------- echo into file ----------------- ----------------- logfile file: ----------------- SUCCESS ----------------- echo into file ----------------- test file: ----------------- cat: /tmp/test: No such file or directory ----------------- ----------------- logfile file: ----------------- "SUCCESS" > /tmp/test ----------------- 

but the expected outcome is:

----------------- test file: ----------------- SUCCESS ----------------- ----------------- logfile file: ----------------- ----------------- echo into file ----------------- test file: ----------------- SUCCESS ----------------- ----------------- logfile file: ----------------- ----------------- 

how can i pass commands with redirection or pipe lining as a command to another function in bash ?

and help and tips would be very much appreciated! I have no idea how to get this working, or if this is possible at all?

Answer 1

From the bash manpage:

 * Expands to the positional parameters, starting from one. When the expansion occurs within double quotes, it expands to a single word with the value of each parameter separated by the first character of the IFS special variable. That is, "$*" is equivalent to "$1c$2c...", where c is the first character of the value of the IFS variable. <snip> @ Expands to the positional parameters, starting from one. When the expansion occurs within double quotes, each parameter expands to a separate word. That is, "$@" is equivalent to "$1" "$2" ... <snip> 

You therefore need to use "$@" (including the double quotes) instead of $* since you want to retain the original parameter quoting.