Improving this validate function

Shell Programming and Scripting
1

Hi guys, I use this function which was provided to me by someone at this site. It works perfectly for validating a users input option against allowed options..

example:

validateInput "1" "1 3 4 5" would return 0 (success)

function validateInput {
 input=$1
 allowedInput=$2
 
 for allowedOption in ${allowedInput}
 do
  if [ ${input} -eq ${allowedOption} ]
  then
   return 0
  fi
 done
 
 return 1
}

but now we allow users to enter more than 1 number and so this validate needs to be improved

validateInput "16" "1 3 4 5" this should return fail (1) as 6 is not in the allowedInput string.. currently the function return success because "1" is in the allowedInput string.. but I somehow need it to verify all the numbers in the first parameter

any ideas guys?

2

Well, now you need to validate the number of arguments, and validate each one separately.

validateInputs(){

 inputs=$1
 allowedInput=$2
 min_args=$3
 max_args=$4
 arg_ct=0
 inval_args=""

 for input in $inputs qzzq
 do
  if [ $input = qzzq ]
  then
   break
  fi

  for allowedOption in ${allowedInput}
  do
   if [ ${input} -eq ${allowedOption} ]
   then
    inval_args="$inval_args $input"
   fi
  done
  (( $arg_ct += 1 ))
 done

 if [ "$inval_args" != "" ]
 then
  echo "Invalid:$inval_args" >/dev/tty
  return 1
 fi

 if (( $arg_ct > $max_args || $arg_ct < $min_args ))
 then
  return 2
 fi

 return 0
}
3 
$
$
$ cat -n f38
     1  #!/usr/bin/bash
     2  x=$1
     3  y=$2
     4  i=0
     5  while [ $i -lt ${#x} ]; do
     6    d=${x:$i:1}
     7    if ! [[ " $y " =~ " $d " ]]
     8    then
     9      echo "$d does not exist in $y"
    10      return 1
    11    fi
    12    i=`expr $i + 1`
    13  done
    14  echo "$x exists in $y"
    15  return 0
$
$
$ . f38 "1" "1 2 3 4"
1 exists in 1 2 3 4
$
$
$ . f38 "123" "1 2 3 4"
123 exists in 1 2 3 4
$
$
$ . f38 "0123" "1 2 3 4"
0 does not exist in 1 2 3 4
$
$
$ . f38 "1236" "1 2 3 4"
6 does not exist in 1 2 3 4
$
$
$ . f38 "1237" "1 2 3 77 4"
7 does not exist in 1 2 3 77 4
$
$

tyler_durden

4

And another one:

validateInput() {
  input=$1 allowedInput=$2
  while [ -n "$input" ]; do
  c=${input%${input#?}} input=${input#$c}
    case $allowedInput in
      ( *$c* )          ;;
      (   *  ) return 1 ;;
    esac
  done
  }
5

You want to be able to validate any word in a space delimited list, not every byte, right?

You might change your input validation to a case string, somewhat more varsatile with | (or), [abc] (list of char) and such:

#!/usr/bin/ksh

valInp(){

for i in $1 qzzq
do
 bad_args=""

 case "$i"  in
 (qzzq)
   ;;
 ($2)
   ;;
 (*)
   bad_args="$bad_args $i"
   ;;
 esac
done

if [ "$bad_args" != "" ]
then
 return 1
fi
}

Using a grep -E regex-pattern would be even more powerful.

#!/usr/bin/ksh

valInp(){

bad_args=""

echo $1 | tr ' ' '\12' | grep -Ev "$2" | while read bad
do
 bad_args="$bad_args $bad"
done

if [ "$bad_args" != "" ]
then
 return 1
fi
}