Looking for specific user ID's from the passwd file

[Patrick Ware_1]

Hello,

My issue is that I want to look for specific userids that have their first and last characters are alphabetic followed by four numbers.

I've already got the user ID's out of the passwd file:

more passwd | awk -F ":" '{print $1}' > userids.txt

I just need to know how to just pick the users out of the list that are formated:

aannnn

a=alphabet
n=number

like for example:

ab1234

Thanks in advance!

[Steven E. Protter]

Shalom,

awk has a print format function called printf

Take a look at that and you can do pattern matching.

You may also be able to use grep as a pre or post processor for this script.

SEP

[Sandman!]

# grep '^[a-z]\{2\}[0-9]\{4\}$' infile

[James R. Ferguson]

Hi Patric:

# perl -aF":" -nle 'print $F[0] if $F[0]=~/[A-Za-z]+\d{4}$/' /etc/passwd

That is, split the field of the file on the ":" delimiter. If the first (zero-relative) field consists of one or more letters followed by exactly 4-digits, print the userid.

Regards!

...JRF...

[Pete Randall]

Something like:

grep [a-z][a-z][0-9][0-9][0-9][0-9] userids.txt

should work.


Pete

[James R. Ferguson]

Hi (again) Patrick:

Steven > awk has a print format function called printf. Take a look at that and you can do pattern matching

Hardly, 'printf' does formatted printing not pattern matching.

Regards!

...JRF...

[Patrick Ware_1]

Pete's anwser was the closest:

I ended up doing this:

more passwd | awk -F ":" '{print $1}' | grep '[A-Za-z][A-Za-z][0-9][0-9][0-9][0-9]'

[Sandman!]

Yet another way to do it using awk(1):

# awk -F: '$1~"[a-z]{2}[0-9]{4}"{print $1}' /etc/passwd

~hope it helps

[James R. Ferguson]

Hi Patrick:

> ended up doing this:

# more passwd | awk -F ":" '{print $1}' | grep '[A-Za-z][A-Za-z][0-9][0-9][0-9][0-9]'

That's a waste of two processes! Let 'awk' open and read the file instead of having 'cat' or 'more' spawn 'awk'. Then, let 'awk' do the matching that it was designed to do, eliminating the 'grep':

# awk -F":" '/[A-Za-z][A-Za-z][0-9][0-9][0-9][0-9]/ {print $1}' /etc/passwd

Regards!

...JRF...

[Sandman! · Accepted solution]

Ignore last post and use the one below as it anchors the regular expression:

# awk -F: '$1~"^[a-z]{2}[0-9]{4}$"{print $1}' file

~cheers