CentOS - Nov 2008 - OT: bash script for passwd and shadow manipulation

Hi,
i need to extract some information from the /etc/passwd file to be
used as a command input in a mail software. My /etc/passwd looks like:
k.thomas:x:1918:100:Kimaura Thomas:/home/users/k.thomas:/bin/usersh
My main issue here is that the fifth field contains spaces and spanish
chars with accent. I currently do not posess the skill to understand
how to use cut to extract all the field.
So far, cut returns the name up to the space, so in this case it will
return Kimaura and not Kimaura Thomas.
Can you please point me to internet examples of tools (sed, awk, grep
or cut) that will help me accomplish this, or maybe provide the code?

And here is the code (not the best...im still learninng..)
#!/bin/bash
# Passwd to Zimbra import
# This script modifies the displayName field in the Zimbra mailserver
# This script will parse /etc/passwd and extracts field 1 and field 5
# field 1 is the username.
# field 5 is the name in long format of the username
#
# domain="oj.gob.pa"
# file="zimbranames.file"
# x=0
# echo ''>$file
#
# for linia in `cat /etc/passwd`
# do
#    user=`echo $linia|cut -f1 -d":"`
#    nombre=`echo $linia|cut -f5 -d":"`
#    echo "zmprov ma $user@$domain displayName $nombre">>$file
#    x=$[x+1]
# done
# echo "$x accounts exported to \"$PWD/$file\""
# sleep 5

Thanks in advance.

-- 
------------------------------------------------------------
Erick Perez
------------------------------------------------------------

Erick Perez wrote:
> Hi,
> i need to extract some information from the /etc/passwd file to be
> used as a command input in a mail software. My /etc/passwd looks like:
> k.thomas:x:1918:100:Kimaura Thomas:/home/users/k.thomas:/bin/usersh
> My main issue here is that the fifth field contains spaces and spanish
> chars with accent. I currently do not posess the skill to understand
> how to use cut to extract all the field.
> So far, cut returns the name up to the space, so in this case it will
> return Kimaura and not Kimaura Thomas.
> Can you please point me to internet examples of tools (sed, awk, grep
> or cut) that will help me accomplish this, or maybe provide the code?
>
> And here is the code (not the best...im still learninng..)
> #!/bin/bash
> # Passwd to Zimbra import
> # This script modifies the displayName field in the Zimbra mailserver
> # This script will parse /etc/passwd and extracts field 1 and field 5
> # field 1 is the username.
> # field 5 is the name in long format of the username
> #
> # domain="oj.gob.pa"
> # file="zimbranames.file"
> # x=0
> # echo ''>$file
> #
> # for linia in `cat /etc/passwd`
> # do
> #    user=`echo $linia|cut -f1 -d":"`
> #    nombre=`echo $linia|cut -f5 -d":"`
> #    echo "zmprov ma $user@$domain displayName
$nombre">>$file
> #    x=$[x+1]
> # done
> # echo "$x accounts exported to \"$PWD/$file\""
> # sleep 5
>
> Thanks in advance.
>   
You should be able to do this easily with awk.  Look at the -F option 
for defining fields.  The follow command line will pluck field 1 and 
field 5 from your example.
$ echo "k.thomas:x:1918:100:Kimaura 
Thomas:/home/users/k.thomas:/bin/usersh" | awk -F : '{ print $1",
"$5 }'
Or, in your script
.
.
.
user=echo $linea | awk -F :  '{ print $1 }'
nombre=echo $linea | awk -F :  '{ print $5 }'
.
.
.
That's about as inelegant as it comes but it should be easy to 
understand.  There is a fantastic (g)awk manual at 
http://www.gnu.org/software/gawk/manual/ presented in a number of formats.

On Thu, Nov 27, 2008 at 01:56:11PM -0500, Erick Perez
wrote:
> So far, cut returns the name up to the space, so in this case it will
> return Kimaura and not Kimaura Thomas.
No, it doesn't.
> # for linia in `cat /etc/passwd`
This is your mistake.  Think about it
  for linia in `cat /etc/passwd`
  do
    echo Line just read: $linia
  done

That shows what is going wrong; the "in" is splitting at the white
space.

What you should be doing
  cat /etc/passwd | while read linia
instead.

Even better would be
  while read linia
  do
   ....
  done < /etc/passwd

Or, to rewrite the whole program in a one line awk script:

awk -F: '{printf("zmprov ma %s@'$domain' displayName
%s\n",$1,$5)}' /etc/passwd > $file

-- 

rgds
Stephen