testing for command line args in bash

Question

I am testing to see if the first argument to my script is --foo

if [ $# > 1 ] then     if [[ "$1" = "--foo" ]]     then         echo "foo is set"         foo = 1     fi fi  if [[ -n "$foo"]] then     #dosomething fi 

Can someone pleaset tell me what is the bash way of testing if --foo is present as one of the arguments, not necessarily the first one?

Answer 1

You should use the external getopt utility if you want to support long options. If you only need to support short options, it's better to use the the Bash builtin getopts.

Here is an example of using getopts (getopt is not too much different):

options=':q:nd:h' while getopts $options option do     case $option in         q  )    queue=$OPTARG;;         n  )    execute=$FALSE; ret=$DRYRUN;; # do dry run         d  )    setdate=$OPTARG; echo "Not yet implemented.";;         h  )    error $EXIT $DRYRUN;;         \? )    if (( (err & ERROPTS) != ERROPTS ))                 then                     error $NOEXIT $ERROPTS "Unknown option."                 fi;;         *  )    error $NOEXIT $ERROARG "Missing option argument.";;     esac done  shift $(($OPTIND - 1)) 

Not that your first test will always show a true result and will create a file called "1" in the current directory. You should use (in order of preference):

if (( $# > 1 )) 

or

if [[ $# -gt 1 ]] 

or

if [ $# -gt 1 ] 

Also, for an assignment, you can't have spaces around the equal sign:

foo=1