Regex – SED: multiple patterns on the same line, how to match/parse first one

last-occurrenceparsingregexsed

I have a file, which holds phone number data, and also some useless stuff.
I'm trying to parse the numbers out, and when there is only 1 phone number / line, it's not problem.
But when I have multiple numbers, sed matches the last one (even though everywhere it says it should match only match the first pattern?), and I can't get other numbers out..

My data.txt:

bla bla bla NUM:09011111111 bla bla bla bla NUM:08022222222 bla bla bla

When I parse for the data, my idea was first to remove all the "initial" "bla bla bla" in front of the first phone number (so I search for first occurrence of 'NUM:'), then I remove all the stuff after phone number, and get the number.
After that I want to parse the next occurrence from the leftover string.

So now when I try to sed it, I always get the last number on the line:

>sed 's/.*NUM://' data.txt
08022222222 bla bla bla
> 

Primarily I would like to understand what's wrong with my understanding of SED. Of course more efficient suggestions are welcome!
Doesn't my sed command say, replace all stuff before 'NUM:' with '' (empty)? Why it matches always the last occurrence ?

Thanks!

Best Solution

This might work for you:

echo "bla bla bla NUM:09011111111 bla bla bla bla NUM:08022222222 bla bla bla" |
sed 's/NUM:/\n&/g;s/[^\n]*\n\(NUM:[0-9]*\)[^\n]*/\1 /g;s/.$//'
NUM:09011111111 NUM:08022222222

The problem you have is understanding that the .* is greedy i.e. it matches the longest match not the first match. By placing a unique character (\n sed uses it as a line delimiter so it cannot exist in the line) in front of the string we're interested in (NUM:...) and deleting everything that is not that unique character [^\n]* followed by the unique character \n, we effectively split the string into manageable pieces.