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XST1 01-30-2005 10:19 AM
shell scripting: variable problem
 
I'm writing code that takes the arguments presented to the script and prints them back out again. heres my code:

Code:
#!/usr/bin/bash

count=0
number=$#

while [ $count -lt $number ]
do
        count='expr $count + 1'
       
        token='$'$count
       
        echo $token
       
        shift
done
for some reason count never increses by 1 but instead contains expr $count + 1 and token always prints $expr $count + 1. Because of this, my while loop only executes once then crashes. Anyone have any ideas?

Duudson 01-30-2005 10:27 AM
Code:
count=`expr $count + 1`

XST1 01-30-2005 10:35 AM
that fixed it, but i noticed another thing.... when i do echo $token, it will print out $1 instead of actually printing out the value in $1. Any way to fix that?

homey 01-30-2005 11:04 AM
Try this...
Code:
#!/bin/bash
#Example: ./test 5
count=0
number=$1
while [ $count -lt $number ]
do
  count=`expr $count + 1`
  token=$count
  echo $token
  shift
done

Duudson 01-30-2005 11:26 AM
Code:
#!/bin/bash

count=0
number=$#

while [ $count -lt $number ]
do
        count=`expr $count + 1`
        echo $1
        shift
done

Duudson 01-30-2005 11:32 AM
or

Code:
#!/bin/bash

count=0
number=$#

while [ $count -lt $number ]
do
        count=`expr $count + 1`

        token=${!count}

        echo $token

done

XST1 01-30-2005 01:06 PM
when we say number=$# what does $# imply? What does it mean?

Duudson 01-30-2005 01:45 PM
Quote:
Originally posted by XST1
when we say number=$# what does $# imply? What does it mean?
how many arguments were given


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