On Thu, Nov 27, 2008 at 2:37 PM, Stephen Harris <lists@xxxxxxxxxx> wrote:
> 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
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>
Stephen, you are right. i learned a bit more about "for" cycles and
the blank space was being stripped.
Rainer, some backquotes as you said were missing.
So the code ends like this, and it works :))
#while read linia
#do
# user=`echo $linia | awk -F : '{ print $1 }'`
# nombre=`echo $linia | awk -F : '{ print $5 }'`
# echo "el nombre es $nombre"
# echo "zmprov ma $user@$domain displayName $nombre">>$file
#
# x=$[x+1]
#done < /etc/passwd
Stephen, your single line awk command is wonderful.
the while cycle takes about a minute (it is a very very long passwd
file) the awk took less than 10 seconds
.........I have so much to learn....
--
------------------------------------------------------------
Erick Perez
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