I have a log file that has lines that look like this:

blah blah blah Photo for (702049679 - blah blah blah

Now I know I could get all the lines like that from the file by doing:

grep "Photo for" logFile

But how can I take those lines and get a comma seperated list of each number after the parenthesis in a single output line (these are going to be pasted into an SQL query)?

The numbers in question will be the first occurrence of a string of numeric characters 9 or more digits long. Ideally it could be matched using that criteria, or the criteria of the first number occurring after the "Photo for (" text.

  • Could you be more specific about the input data? blah blah blah is pretty nondescript and vague...
    – jasonwryan
    Commented Jan 31, 2014 at 0:31
  • @jasonwryan Updated.
    – Muhd
    Commented Jan 31, 2014 at 0:34

3 Answers 3


An alternative that uses only Perl and matches the criteria you specified:

perl -ne '
    /Photo for/ && /([0-9]{9,})/ && push @numbers,$1;
    END{ $" = ","; print "@numbers" }
' logFile

This will print out a comma-separated list of the first occurrence of a contiguous digit string made up of 9 or more digits on each line matching Photo for.


A regex this complicated is better handled with Perl, e.g.

grep "Photo for" logFile | perl -pe 's/.*Photo for ((\d+).*/\1/' | tr '\n' ','

If Perl is out of the question:

grep "Photo for" logFile | awk '{sub(/.*Photo for \(/,"",$0);sub(/[ ].*/,"");print $0}' | tr '\n' ','
  • First one works... second one gives awk: fatal: Unmatched ( or \(: /.*Photo for (/. Thanks though, problem solved. EDIT: Second just needs that parenthesis escaped with \.
    – Muhd
    Commented Jan 31, 2014 at 0:42
  • I fixed the second answer (I added "Photo for" to the regex at the last minute when you edited the post)
    – samiam
    Commented Jan 31, 2014 at 0:59

1) need gnu grep

grep -iPo "(?<=Photo for \()[0-9]+" logFile|tr "\n" ","

2) Using awk

awk -F \( '/Photo for/{split($2,a,OFS);printf a[1] ","}' logFile

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