Behold, Ow Mun Heng <Ow.Mun.Heng@xxxxxxx> hath decreed:
> Guys,
>
> Writing a batch_create_new_user script.
> input is a file with this format
>
> Username:Full Name:Default Group:Default Shell:Home Directory:Password
>
> and the using those fields ($1 $2 $3 etc) and passing then to the useradd program
>
> I'm having some trouble getting bash to get the positional fields into variables which I
> can then plug into useradd
>
> eg:
> pseudo code
> for each line in input file
> do
> username=$1
> name=$2
> useradd -c "$2" $1
> done
>
> I ended up using awk, which works but seems complicated
>
> eval `awk -F ':' \
> '{printf ("useradd -c \"%s\" ",$2)}; \
> length($3)>0 {printf ("-g %s ",$3)}; \
> length($4)>0 {printf ("-s %s ",$4)}; \
> length($5)>0 {printf ("-d %s ",$5)}; \
> {printf ("%s\n;",$1)};' $1`
>
> Can it be done using bash itself? (without calling awk)
> Actually I was hoping I can get awk to get the fields for the variables
You could try this:
read -d: USERNAME
read -d: NAME
read -d: GROUP
etc..
Keep in mind that when using -d, read will not treat a newline
as a delimiter, so you have to use more trickery if you want
optional fields.
> > psuedo code
> username = awk -F: {'print $1}'
> name = awk -F: {'print $2}'
>
> Any takers??
>
> BTW, what's the best way to check the number of fields in the input file?
> I was thinkning of counting how many ":" there are in each line. But can't
> figure out how to actually _do_ it.
>
>
>
> --
> Ow Mun Heng
> Fedora GNU/Linux Core 2 on D600 1.4Ghz CPU kernel
> 2.6.7-2.jul1-interactive
> Neuromancer 17:13:44 up 5:56, 10 users, load average: 1.65, 1.32, 0.97
>
>
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"Every man is a mob, a chain gang of idiots."
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