Printing all matches in a line using regular expression in awk - linux

Question

Say i have a line:
Terminal="123" Pwd="567"
I want to select only number portion using awk
awk 'match($1, /[0-9]+/){print substr($1, RSTART, RLENGTH)};match($2, /[0-9]+/){print
substr($2, RSTART, RLENGTH)}' file
This gives the desired result.
123 567.
However there must be other better way to select both numbers without writing two match statements.
Thanks.

Answer 1

does grep work for you?
kent$ echo 'Terminal="123" Pwd="567"'|grep -o '[0-9]\+'
123
567
quick and dirty with awk:
awk -F'[^0-9]*' '{$1=$1}7'
test:
kent$ awk -F'[^0-9]*' '{$1=$1}7'<<< 'Terminal="123" Pwd="567"'
123 567
or:
kent$ awk '{gsub(/[^0-9 ]/,"")}7'<<< 'Terminal="123" Pwd="567"'
123 567

Answer 2

Here is a nice little solution with awk:
awk '{gsub("[^0-9]+"," "); print}'
Just converts all consecutive non-digit characters into one space, so it leaves one space before the digit sequence 123.

Answer 3

Here is another way to do it with awk. We set the field separator to "
$ echo 'Terminal="123" Pwd="567"' | awk -F\" '{print $2, $4}'
123 567

Answer 4

I ran into a similar problem but my patterns were more complex so I couldn't brush off my problems with gsub or such. I wrote a recursive function and a wrapper to it. It finds multiple matches in one variable and prints them out separated with a space:
awk '
function rec_wrap(str)
{
matches=""
return rec_func(str)
}
function rec_func(str2)
{
where=match(str2, /RE/)
if(where!=0) {
matches=(matches substr(str2, RSTART, RLENGTH) " ")
rec_func(substr(str2, RSTART+RLENGTH, length(str2)))
}
return matches
}
{print rec_wrap($1)}
' file.txt
The wrapper rec_wrap is needed to empty the variable matches. Function match writes the position and length of the leftmost match to variables RSTART and RLENGTH and the match is extracted with substr and appended to variable matches. Then the function rec_func calls itself with the rest of the string str2 as parameter until match fails to find anymore matches.