Bash script using awk doesn't read the entire line, just the first column

There’s a file to be processed, columns are separated by tabs:

$ cat system.log
2       camila  create db
3       andrew  create table
5       greg    update table
6       nataly  update view
7       greg    delete table
9       camila  update table
11      nataly  create view
12      peter   link table
14      andrew  update view
15      greg    update db

I wanted these lines displayed in a form:

Entry No. 7: camila (action: create db)

To do so, I created the following bash script:

#!/bin/bash

filename=$1

while read line; do
        printf $line | awk -F '\t' '{ print "Entry No. ", $1, ": ", $2, " (action: ", $3, ")" }'
done < $filename

However, what I get is:

$ ./log_parser.sh system.log
Entry No.  2 :    (action:   )
Entry No.  3 :    (action:   )
Entry No.  5 :    (action:   )
Entry No.  6 :    (action:   )
Entry No.  7 :    (action:   )
Entry No.  9 :    (action:   )
Entry No.  11 :    (action:   )
Entry No.  12 :    (action:   )
Entry No.  14 :    (action:   )
Entry No.  15 :    (action:   )

Why only the first column gets processed and what to do to have the whole line processed?

>Solution :

You must quote your variables to prevent word splitting. Consider if $line evaluates to the string 2 camila create db. In that case, printf $line is equivalent to printf 2 camila create db which calls printf with 4 arguments. printf correctly parses those arguments and dutifully writes the string 2. If you want to pass a single argument to printf, you could do printf "$line". But that is also incorrect, as the first argument to printf should be a format string, and you don’t want to use input strings as format strings. Instead, you should write printf '%s' "$line". But don’t do that, either. while read; printf | awk is an anti-pattern. Just use awk to read the input.