Print a variable column using awk

Question

I'm trying to filter some mail logs and I need to get some specific information from certain columns that are not always the same. The information can be on different columns each line. I always use awk to print only the columns I want, like:

cat file.log | awk '{print $1" "$2" "$3}' >> output.txt

but in this case, I don't know which column contains the string I'm looking for.

Example file.log where I'm looking for columns that contain the string "5":

A B C 222 586 999 724 644  
A B C 510 333 987 678 633  
A B C 348 488 920 566 240

Result I want:

A B C 586  
A B C 510  
A B C 566

Any help is appreciated

if at all you can consider perl then you can do $perl -pe "s/^(\D+).*\b(5\d*).*/$1$2/" your_file.txt — Onyambu
– Onyambu, Commented Aug 30, 2018 at 5:08

Mévatlavé Kraspek · Accepted Answer · 2018-08-30 03:57:15Z

2

Using a regex ~ /5/ and iterating on each columns on each rows/lines (KISS) :

$ awk '
    {
        for (i=1; i<=NF; i++) {
            if (i==1 || i==2 || i==3 || $i ~ /5/) {
                printf "%s ", $i
            }
         }
         print ""
    }
' file

Output:

A B C 586 
A B C 510 
A B C 566

edited Aug 30, 2018 at 3:57

answered Aug 29, 2018 at 20:46

Mévatlavé Kraspek

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Comments

glenn jackman · Accepted Answer · 2018-08-29 22:06:20Z

1

a tricky perl one-liner

perl -anE 'say "@{[ @F[0..2], grep {/5/} @F[3..$#F] ]}"' file

where

-n iterates over the input file by line
-a splits each line into words, and stores into the @F array
@F[0..2] is the first 3 words; @F[3..$#F] is the list of the 4th up to the last word
grep {/5/} filters the list, returning only the words containing a "5"
"@{[ ... ]}" is a syntax trick to stringify a list as space-separated.

Can also use this, just as tricky but a little less "syntax-y"

perl -anE 'push @F, grep {/5/} splice @F,3; say "@F"'

answered Aug 29, 2018 at 22:06

glenn jackman

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2 Comments

Onyambu Over a year ago

I guess perl -pe "s/^(\D+).*\b(5\d*).*/$1$2/" will do the job

Onyambu Over a year ago

perl -ape "s/\d.*//s;$_=$_. join(' ', grep{/5/} @F).\"\n\"" file

James Brown · Accepted Answer · 2018-08-30 04:54:14Z

1

$ awk '{
    for(i=4;i<=NF;i++)    # iterate fields starting from the 4th
        if($i~/5/)        # if there is a 5 in the field
            $3=$3 OFS $i  # append field value to the 3rd field
    print $1,$2,$3        # output 1st thru 3rd
}' file

Output:

A B C 586
A B C 510
A B C 566

edited Aug 30, 2018 at 4:54

answered Aug 29, 2018 at 21:23

James Brown

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1 Comment

karakfa Over a year ago

perhaps OFS instead of " "?

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