I am trying to find a way to use awk to start printing only once it finds the search parameter and print only a couple of columns following it.

To try to explain a little better I am looking over a log file that prints a lot of information as a single line. I want to take and extrapolate only certain aspects of that line. IE there's a big file with this line somewhere in it:

blahblah blah blah there is a lot of information here for name: John Doe and it just keeps on going and doesn't stop.

I want it to search for name and only print out:

name: John Doe

I know I can use the column segment, but I don't actually know where in that line name will be in different files.

  • How will you be able to tell then, which word is the name? There might be multiple names, too, or no names on a line?
    – Alex
    Aug 25, 2017 at 18:24
  • Why not use grep -o for this? grep -o 'name: ([A-Z][a-z]* )*' /path/to/input
    – DopeGhoti
    Aug 25, 2017 at 18:29
  • There is always going to be a name listed somewhere within the file and inside of the long line of just dumped information from a database. It will also only contain one name as it is just the name for the individual database table. Ultimately, there is a lot of good information in these files that is just poorly formatted and I am trying to create a quick bash script I can run on those logs so that it is in a more reader friendly format. Aug 25, 2017 at 18:30
  • @DopeGhoti The point is, if he knows which name he's looking for, then he doesn't even have to search: he can just display it. His question is, how can he discern that single name from a lot of good information. I mean, will it always be prefixed by name:?
    – Alex
    Aug 25, 2017 at 18:31
  • @Alex Yes, it's always going to be prefixed by name. I don't know what the name will be when I start looking though. Aug 25, 2017 at 18:34

3 Answers 3


The simplest answer would NOT involve awk. It would use grep, to output any matches of the word "name", and up to two words that follow:

grep -o 'name [A-Za-z]* [A-Za-z]*' filename

The -o switch includes only the parts that match, not the entire line; and the [A-Za-z]* tells that you are asking for a single word.

Let me know if you prefer to do this in awk-- but that would be much more complicated.

  • I'm glad-- I wasn't entirely sure whether this was what you wanted, from your post...
    – Alex
    Aug 25, 2017 at 18:46
  • awk would not really be that much more complicated: awk /name [A-Z][a-z]* [A-Z][a-z]*/ { print substr( $0, RSTART, RLENGTH) }' inputfile.
    – DopeGhoti
    Aug 25, 2017 at 19:22
  • @DopeGhoti Hmm, I wasn't aware of the / string / functionality. In fact, it doesn't seem to work on my machine... As your quote is written, it's a syntax error; if I add an opening single quote before /name, it yields no results; if I put the opening quote before { print, I get the error awk: non-terminated regular expression and... at source line 1
    – Alex
    Aug 25, 2017 at 19:49
  • I omitted what to search in.. the syntax is awk /regex/ ~ $0 { code to run if present } to search the entire line.
    – DopeGhoti
    Aug 25, 2017 at 19:55
  • 1
    awk RSTART,RLENGTH works only for the explicit match(...) function as in DopeGhoti's separate answer but not for a regexp as a pattern or the ~ operator. Aug 26, 2017 at 12:28

This can also be done with awk:

$ awk 'match( $0, /name [A-Z][a-z]* [A-Z][a-z]*/ ) { print substr( $0, RSTART, RLENGTH ) }' /path/to/inputfile

Perl solution:

perl -lne '/(name:\s+\w+\s+\w+)/ and print $1'

  • \s matches whitespace characters (whitespaces per see, tabulations, etc.)
  • \w matches letters and numbers
  • -lne means don't print anything (n), except when print is used, output a newline in the end (l) and execute (e) the following

So the whole line will match name: followed by one or more whitespace character and one or more letter, again one or more whitespace character and one or more letter, and will then print that.

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