error in running the shell script

bhandu · 07-10-2007, 12:33 AM

I have written a shell script which replaces the password of a user in File2 if that user is also present in File1.Following are the contents of two files

File 1

palanspu:3aLdJ8mtCEXJI,A/SS:24563:/usr/bin/ksh
rahul:7j6Zm/48.6rs,A/1S:24669:/usr/bin/ksh

File 2

rahul:ydMaodGCHIzxU,A/jN:25100:/usr/bin/ksh
anukul:YX6UQQl6OpSaU,A/US:25112:/usr/bin/ksh

The script is as follows

#!/bin/bash

#creates array of passwords from File1
pass1=`cut -d ":" -f2 first`
cnt=0

if [ -s tmp-sec ] ; then
rm tmp-sec
else
touch tmp-sec
fi

for usr1 in `cut -d ":" -f1 first` ; do
sed -e "/^$usr1:/ s/:.*$/:${pass1[$cnt]}/" sec >> tmp-sec
cnt=$(($cnt+1))
done

After executing the shell script the output is taken in the file tmp-sec. But I get the error as follows and also the contents of tmp-sec file is not as expected

$ sh temp.sh
sed: Function /^palanpsu:/ s/:.*$/:3aLdJ8mtCEXJI,A/SS cannot
be parsed.
$ cat tmp-sec
rahul:
anukul:YX6UQQl6OpSaU,A/US:25112:/usr/bin/ksh
$

Can anyone help?

jschiwal · 07-10-2007, 01:16 AM

I didn't look at your entire script. I did notice this line:
pass1=`cut -d ":" -f2 first`

If you want an array use:
pass1=(`cut -d ":" -f2 first`)

Awk might be a better tool for the job. Remember that the /etc/passwd file also contains system users as well, so you need to examine the value of the uid as well.

Linux distro's will also use the shadow suite, so that the encrypted password won't be in the /etc/passwd file.