optional exit from loop, shell script

RudraB · 07-17-2008, 02:19 AM

here is an easier version of what i want to do

Code:

#!/bin/bash
echo "Enter file name"
echo "You can enter maximum 3 file"

for (( i=1; i<=3;i++))
until [ "$*" = "" ]
do
 echo "Enter file ${i}"
 read fil_$i
 done
//do the rest of the code

the probleam is, my input needs three input. but what i want to do is to put an optin of optional break, say instead of three, i have only one input ...so, if i put "0", the code should ixit for loop and do the rest of the things.
how can i do that?

colucix · 07-17-2008, 02:56 AM

Check the syntax: to put other statements between for and do is not permitted. Based on your requirements, you have to simply check if the input is equal to 0 and eventually break the loop:

Code:

#!/bin/bash

echo "Enter file name"
echo "You can enter maximum 3 file"

for (( i=1; i<=3; i++ ))
do
  read -p "Enter file ${i}: " tmpfile
  if [ "$tmpfile" = "0" ]
  then
    break
  else
    eval "file_$i=$tmpfile"
  fi
done
echo file_1 is $file_1
echo file_2 is $file_2
echo file_3 id $file_3

As you can see, in this example I used a temporary variable "tmpfile" to read the user's input. Then based on the condition you can either exit the loop using the break statement or assign the value of tmpfile to a variable.

The assigment is done using indirect reference: the variable name contains the value of another variable ($1) so you have to use the eval statement to let the shell expand the variable name and then execute the command. See the Advanced Bash Scripting Guide, here for more details.

RudraB · 07-17-2008, 03:30 AM

thnkx colucix
its just what i was looking for