Awk/grep/sed get comma separated list of numbers from lines of text

Question

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.

Answer 1

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.

Answer 2

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' ','

Answer 3

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