How to pass function parameter to do loop?

Shell Programming and Scripting
1

Hi All,

I have created one function for KSH and was running well with one parameter input since I just had to use $1 to get the parameter.
Now I want to do loop for each parameter(actually filenames) . I have try to use do loop, but $i does not resolve to parameter instead it resolves to 1, 2, 3 etc...

example,

func()
echo There are ${#} parameters..
i=1
while (( i <= ${#} ))
do
if  [ ! -f $("$i")];
then 
    echo "Error $i does not exist!"
else
echo "Error $i does exist, please proceed".
fi
     (( i+=1 ))
done
}

Please advice.

2 
#!/bin/ksh

function process_files
{
   echo There are ${#} parameters..
   k=1
   while [ $k -le ${#} ]
   do
      eval i=\${$k}
      if [ ! -f "$i" ]
      then
         echo "Error $i does not exist!"
      else
         echo "Error $i does exist, please proceed".
      fi
      (( k = k + 1 ))
   done
}
1 Like
3

Hello rdrtx1,

I think in shell unlike awk there is no need for function function_name , you could put directly as function_name {....} , so it could be written as follows.

#!/bin/ksh
process_files ()
{
   echo There are ${#} parameters..
   for i in $*
   do
      if [ ! -f "$i" ]
      then
         echo "Error $i does not exist!"
      else
         echo "Error $i does exist, please proceed".
      fi
   done
}

Thanks,
R. Singh

4

Include parentheses in the other function definition format:

process_files ()
5

Worked like charm. thanks!:b:

---------- Post updated at 02:57 PM ---------- Previous update was at 02:41 PM ----------

It seems you edited and replaced with your previous answer. It worked for me. Thanks anyways!

6

Note: the natural loop through the arguments is the for loop.

func(){
# the for loop defaults to 'in "$@"'
  for arg
  do
    if [ -f "$arg" ]
    then
      echo "Error '$arg' does exist, please proceed."
    else
      echo "Error '$arg' does not exist!"
    fi
  done
}
# pass all arguments of this shell to func()
func "$@"
func file1 file2 "filename with spaces"

Note that "$@" is preferrable to $* because it does not split on words.