Question

In this example I want it to know if $@ contains two words/symbols “load” and “/”

for one word/symbol this works

    case "$@" in */*)
    ;;
    echo "going to do stuff"
*)
    echo "will do something else"
    ;;
esac

or

string='My string';

if [[ "$string" == *My* ]]
then
  echo "It's there!";
fi

But if two words/symbols appear at random places I cant figure out how to do it.

Update:

The input will the module command. In this case I want to know if it is the module load with or without / that indicate version. the command will look like this

1) module load appname/1.1.1 or
2) module load appname
3) module (not load) (list, avail etc)

It is number 1 I am interested in for now.
3 will in some cases be variation of 1.
2 will be run as is but will include a message to the user

Answer 1

Something like this is possible:

if  [[ ${@} =~ .*/.* && ${@} =~ ((^)|([ ]))load(($)|([ ])) ]]
then
  echo both
fi

-or-

if LOAD=0 && SLASH=0 && \
   for ARG in ${@};
   do
     if [ "${ARG#*/}" != "${ARG}" ]; then SLASH=1; fi
     if [ "${ARG}" = "load" ]; then LOAD=1; fi
   done && [ "${LOAD}${SLASH}" = "11" ];
then
  echo both
fi

-or-

function loadslash()
{
  LOAD=0 && SLASH=0
  for ARG in ${@};
  do
    if [ "${ARG#*/}" != "${ARG}" ]; then SLASH=1; fi
    if [ "${ARG}" = "load" ]; then LOAD=1; fi
  done
  test "${LOAD}${SLASH}" = "11"
}

if loadslash ${@}
then
  echo both
fi