Locate a field in a record predicated on locating previous value (awk)

oldfogey · 06-12-2012, 02:58 PM

Hi,

I have a file with several thousand records. I need to extract the value (from a field) within the records that is not always in the same position but does come after a unique, fixed value.

The records are of variable length - min 9 fields, max 12 fields. Somewhere in the record there's a field that contains the string "host". The very next field in the record is the variable length field whose value I need to extract. For example, the records could look like:

date time value value value value host sys12 secure client MAC status
date time value value host server16 client MAC status
date time value value value value host intl33-a client MAC status

So, from the records shown above I need to extract the values sys12, server16 & intl33-a.

I have only the most basic knowledge of awk but I have to believe there's a straightforward way to extract what I need. Can someone be so kind as to assist?

Thanks in advance for any help you can provide!!

schneidz · 06-12-2012, 03:05 PM

Code:

[schneidz@hyper ~]$ grep -o host.*" " oldfogey.txt | awk '{print $2}'
sys12
server16
intl33-a

there probably is a way to do it in one awk maybe usng indexes.

edit: here is the awk solution:

Code:

[schneidz@hyper ~]$ awk '
  { for(i=1;i<=NF;i++){
      if ($i ~ /\<host\>/)
          {print $(i+1)} }
}' oldfogey.txt
sys12
server16
intl33-a

oldfogey · 06-12-2012, 03:16 PM

Awesome! Works perfectly.

Thanks for the quick response.

Regards,

Anthony

David the H. · 06-13-2012, 09:29 AM

Here's a sed solution too:

Code:

sed -rn '/host/ s/.*host ([^ ]+).*/\1/p' infile

BTW, I'd also use a simple string comparison in the awk command, rather than a regex. And you can condense it into a one-liner like this:

Code:

awk '{ for (i=1;i<=NF;i++) { if ($i=="host") { print $(i+1) } } }' infile

grail · 06-13-2012, 11:48 AM

Code:

grep -Eo 'host [^ ]*' file | cut -d' ' -f2

awk -F"host " '{print gensub(/ .*/,"","1",$2)}' file

awk 'match($0,/host ([^ ]*)/,f){print f[1]}' file

awk '{split(RT,a);print a[2]}' RS="host [^ ]*" file

chrism01 · 06-13-2012, 07:28 PM

From scheidz:

Code:

grep -o host.*" " oldfogey.txt | awk '{print $2}'

I like; didn't know/realise about the -o option.

Having done some testing for my own edification, I would use

Code:

grep -o " "host" ".*

otherwise it can match the string 'host' with either/both of trailing/prefix non-space chars

Code:

# extended test file
cat t.t
date time value value value value host sys12 secure client MAC status
date time value value host server16 client MAC status
date time value value value value host intl33-a client MAC status
date time value value value value hostx intl33-a client MAC status
date time value value value value zhost intl33-a client MAC status

# orig code match
grep -o host.*" " t.t
host sys12 secure client MAC
host server16 client MAC
host intl33-a client MAC
hostx intl33-a client MAC
host intl33-a client MAC

# new code match
grep -o " "host" ".* t.t
 host sys12 secure client MAC status
 host server16 client MAC status
 host intl33-a client MAC status