advanced awk


 
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# 1  
Old 03-31-2010
advanced awk

Hi all

Input
Code:
group1	user1
	user2
	user3
group2	user4
	user5
	user1
group3	user6
	user7
	user8

Desired output
Code:
group1
group2

So far I've tried, but did not work
Code:
BEGIN
{
  i=0;                                  # declaring variable
}

{
  if (NF==2){currgrp=$1;}               # check whether awk actually processing 2 column record, if yes save the group name
  else
     {
      if ($1~/uname/)                   # else if awk is processing 1 column record and is found user name, save the group name to array and increment i
       {
         netgroups[i]=currgrp;
         i++;
       }
     }
}

END                                     # at the end of processing whole file print the array (all saved groups)
{
  for(j=0;j<i;j++)
  {
    print netgroups[i];
  }
}

skript was called
Code:
awk -v uname="user1" -f awkscript.awk group_file

what is wrong with script above ?
# 2  
Old 03-31-2010
It was a bit complicated, I think for the task?

Code:
$ cat Test1
awk '
 $2 { G = $1 }
 $NF == U && ! _[G]++ { print G }
' U="$1" file1

$ cat file1
group1	user1
	user2
	user3
group2	user4
	user5
	user1
group3	user6
	user7
	user8

$ ./Test1 user1
group1
group2

# 3  
Old 03-31-2010
This one should work:
Code:
awk -v uname="user1" 'END {
  while (++j <= i) print netgroups[j]
  }
NF == 2 { currgrp = $1 }
$0 ~ uname { netgroups[++i] = currgrp }
' group_file

Use gawk, nawk or /usr/xpg4/bin/awk on Solaris.

---------- Post updated at 07:47 PM ---------- Previous update was at 07:39 PM ----------

Quote:
Originally Posted by scottn
It was a bit complicated, I think for the task?
[...]
Sure, no END block is needed Smilie


Code:
awk  'NF == 2 { g = $1 }
$NF ~ u && $0 = g x
' u="user1" group_file

One of the previous solutions will fail if the username is 0 Smilie

... and my code assumes unique group names.

Last edited by radoulov; 03-31-2010 at 02:52 PM..
# 4  
Old 03-31-2010
Thank you guys,
To radulov: your solution is so similar to mine, don't you know why mine is not working ? Seems more similar to me Smilie
# 5  
Old 03-31-2010
Code:
$ u=user1
$ sed -n '/^group/h; /'"$u"'/{g;s/[[:blank:]].*$//;p;}' file
group1
group2

I would recommend against using this type of solution unless you can guarantee that the user name will not contain any sed regular expression special characters ("/", ".", "*", "\" .. come immediately to mind).

You have been warned Smilie

Alister

Last edited by alister; 03-31-2010 at 04:21 PM..
# 6  
Old 03-31-2010
Quote:
Originally Posted by wakatana
Thank you guys,
To radulov: your solution is so similar to mine, don't you know why mine is not working ? Seems more similar to me Smilie
Comments inline.

Code:
BEGIN {                          # do not put BEGIN on a line by itself   
  i = 0                          # no need to initialize the counter,
                                 # awk does this automatically
  }                                 
{
  if (NF == 2) {
    currgrp = $1
    if ($2 ~ uname) {
      netgroups[i] = currgrp     # you shoud check if the username matches 
      i ++                       # when NF == 2 too
      }
    }        
  else
     {
      if ($1 ~ uname) {           # /uname/ matches litteral "uname"
         netgroups[i] = currgrp
         i ++
       }
     }
}

END {                             # do not put END on a line by itself
  for(j = 0; j < i;j ++)             
  {
    print netgroups[j]            # j not i!
  }
}

# 7  
Old 03-31-2010
Quote:
Originally Posted by radoulov
Code:
BEGIN {                          # do not put BEGIN on a line by itself   
...snip...
END {                             # do not put END on a line by itself

There's nothing wrong with those two lines. BEGIN and END are syntactically no different from any other pattern. So long as the corresponding action begins on the same line (the opening curly brace is sufficient), it's fine.

Example:
Code:
$ cat beginend.awk 
BEGIN {
    print "begin"
}

END {
    print "end"
}

$ echo | awk -f beginend.awk 
begin
end

Regards,
Alister
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