Scripting help

[Terrence]

My scripting skills are woeful I'm afraid. I generate a file that contains only a list of user id's. Whenever I generate this file, I need to remove the same 10 user id's that always appear and leave the remaining id's. I've been doing this manually in vi but now I need to automate the process.

Simplest and/or most elegant and/or easiest to understand solution gets the full 10!!.

[RAC_1 · Accepted solution]

grep -Ev "user1|user2|upto_10_users" "input_file" > new_file

If they appear after certain line in file, then

head -"line" inout_file OR tail -"line" input_file

Anil

[Jean-Luc Oudart]

grep -v user1 | grep -v user2 | grep -v user3 | ...
>

Regards,
Jean-Luc

[H.Merijn Brand (procura]

# your_script_that_generates_the_list | grep -v -w -e 203 -e merijn -e skipme -e baduser -e 488 > zzz

# your_script_that_generates_the_list | perl -pe's/\b(203|merijn|skipme|baduser|488)\b//g' >zzz

Enjoy, Have FUN! H.Merijn [ who doen't know if -w is supported in all grep's on your machine ]

[Simon Hargrave]

As above, but I'd also put anchors in place, eg

grep -vE "^uid1$|^uid2$|^uid3$" file > newfile

The ^ and $ anchors ensure that for example removing "fred" doesn't also remove "frederick".

If you want to edit the file in-situ, you could do: -

ed file </^uid1$
d
/^uid2$
d
/^uid23$
d
w
q
EOF

This will let you do it without a temporary "newfile".

[Jeff_Traigle]

Several ways depending what IDs you're trying to eliminate...

If they are the reserved users with UID < 100, you can simply use awk as follows:

awk '{if ($3 < 100) print}' /etc/passwd > newfile

If there are specific normal user's you're talking about, you can do similar to above specifying the UIDs of those users combined with the and operator:

awk '{if ($3 != 250 && $3 != 300) print}' /etc/passwd > newfile

or by username:

awk '{if ($1 != "joebob" && $1 != "nancysue") print}' /etc/passwd > newfile

[Jeff_Traigle]

Oops... first one should have been >, not <

[Jeff_Traigle]

Argh... actually >= with 100 as the right-hand operand... maybe I should just go home now. :)

[Rick Garland]

Can use the "uniq" funtion as well.

cat /etc/passwd | awk -F: '{print $1, $2}' | uniq

The uniq function will get rid of repeats

[Terrence]

Many thanks to all who played!

10 points to RAC for his insanely fast and simple answer. (Doh! grep! I know grep! Why didn't I think of that!)

10 points to Simon for his excellent addition to the simple answer that prevented a problem that I've had in the past where too much is removed. I'll be writing this solution down for other scripts I have as well.

10 points to Rick who unknowingly gave me a solution to a completely different scripting problem I was mulling.

[Muthukumar_5]

We can do with awk script as,

awk '{ for ( i=1; i<=NF; i++) if ( $i != "user1" && $i != "user2" && $i != "user3" ) print $0 }'

IT will give the way to get user lists except your users.

HTH.

[H.Merijn Brand (procura]

Terence, not fishing for points here, but that -w option is exactly the same as the proposed anchors, only much cleaner: it means "match words only"

And in the perl statement, I did exactly the same: \b is word bound

Enjoy, Have FUN! H.Merijn

[Andrew Merritt_2]

You also have the option to put the list of user ids in a file, rather than putting them in the command, which may be less prone to mistyping if you edit the script, e.g. if the list ever changes.

So, if the list of user ids (one per line) is /tmp/uids:

grep -Fwv -f /tmp/uids file

should do what you want.

(As you're dealing with fixed strings, -F is probably faster than -E).

Andrew