print FS (field separator) in awk

wakatana · 11-04-2009, 05:10 PM

Hi gurus is possible to print string that matches the FS in awk ?

Input:

Code:

bbbb454ccc999vvv
aaaa444ggg777bbb
mmmm789jjj456kkk

Desired Output:

Code:

bbbb454vvv
aaaa444bbb
mmmm789kkk

Maybe this can be done another way but I am interesting how to print FS.

I tried that:

Code:

awk 'BEGIN {FS="\([0-9]*\)"} {print $1 "\1" $3}' input
bbbbvvv
aaaabbb
mmmmkkk

tuxdev · 11-04-2009, 05:20 PM

So what you want to do is cut out those middle 6 chars?

Code:

#!/bin/bash
while read -r LINE ; do
   echo "${LINE:0:7}${LINE:13}"
done

ghostdog74 · 11-04-2009, 05:24 PM

what do you mean match the FS. FS is the field separator. by default its white space including tabs. From your output, you just want to get rid of 7th to 12th position , in that case use substr(). Please start reading the gawk documentation (my sig)

wakatana · 11-04-2009, 06:00 PM

I know what is FS for, but is also special variable like $1 so when I can print $1 why cannot I print FS ? I thing problem Is that I have to acces FS throught array because FS has more values (another for each column) but i dont know how??? maybe from some temporary variables I dont know.

Quote:

Originally Posted by ghostdog74

From your output, you just want to get rid of 7th to 12th position

Yes this example looks like this but I just simplified It cause of better understanding.

Lets say I have file /etcg/roups which contains DIFFERENT count of users

Code:

group1:*:23:jani,feri,jozi,pali,name1,name2,name3...
users2:*:32:user,user2,user3...

And also I have file which contains logins which should be deleted from groups.

1. I am writing script which use awk with delimiter ":[^:]*:[^:]*:"
2. so in $1 i will have name of group and in $2 the comma separated users.
3. then I will use some gsub() or something to $2 string for deleting users
4. and finaly I will need to "compose" variables to previous format (print $1 "FS" "$2")

Thank you

PS perl and other utils cannot be used just pure AWK or SED or some GNU utils

ghostdog74 · 11-04-2009, 06:05 PM

It doesn't work that way since you are giving regular expression to FS. how do you expect print to know which FS you want ?

do it another way, since you are using /etc/groups and its a structured file format, just use ":" as delimiter and get field1 and field4.

Code:

# awk -F":" '{print $1,$4}' file
group1 jani,feri,jozi,pali,name1,name2,name3...
users2 user,user2,user3..

wakatana · 11-05-2009, 08:17 AM

Thanks

this should work, Also print FS wolud be nice if AWK supports it

I want replace all "mar" string with ","

Still have one little bug

Code:

$cat testovaci
users:*:20:marek,marecek,mar,martin,mar,jomar,martinka,martinko,mar,jomar,mar,mar
$awk -F':' '{gsub("(^mar,)|(,mar$)|(,mar,)",",",$4); print $1":"$2":"$3":"$4}' testovaci
users:*:20:marek,marecek,martin,jomar,martinka,martinko,jomar,mar

Why I still have ",mar" at the end of output ?

Thanks a lot