Shell Programming and Scripting

Hello,

I want to convert a string to interger.
but its not working..
can anyone help me

thanks in advance

esham

There is no "int" function in bash/ksh, and even if there was, that syntax is incorrect in the way you're invoking the "function" and in the way you're trying to assign to the variable. The correct method would be, assuming an "int" function existed, to do something like

foo=`int $i`

Anyway, back to the matter at hand, something like this should get you started in ensuring that the input is an integer

Code:

#!/bin/ksh

for i in "$@"
do
  echo "$i" | grep '^[0-9]*$' >/dev/null 2>&1
  if [ "$?" -ne "0" ]; then
     echo "$i Not an integer"
  elif [ "$i" -gt "255" ]; then
     echo "$i Invalid"
  else
     echo "$i okay...."
  fi
done
exit 0

Modify that to suit your needs.
A sample run....

Code:

$ ./testint.ksh 1 2 3 10.1 abcd
1 okay....
2 okay....
3 okay....
10.1 Not an integer
abcd Not an integer

One more thing, if you're going to add exit codes, it doesn't make sense to return an exit code of zero after some invalid action is performed. Always return a non-zero exit code if your script exits as the result of some error condition.

Cheers
ZB

You can type your variable as an integer using the command typeset -i. Example: typeset -i FOO. Now the shell will expect FOO to be set with an integer. UNIX will throw an error if you attempt to set FOO to any non-integer. Capture $? and evaluate it for non-zero status. Also, be sure to discard any output error messages to /dev/null during the assignment. If $? is non-zero, FOO was not set with an integer.

typeset -i FOO

read FOO 2>&1 /dev/null
ERR=$?

if [ $ERR -ne 0 ]
then
echo "Integer input expected"
else
echo "Thanks for the integer"
fi

awk '{for(i=1;i<=NF;i++) print int($i)}'

then strings will be convert into zero.