Printing the actual field delimiter value not the regular expression

Question

Given the following input:

check1;check2
check1;;check2
check1,check2

and the awk command:

awk -F';+|,' '{print $1 FS $2}'

FS should contain the selected delimiter?

How can you print the delimiter which is selected i.e. either of ;, ;; or , not the regular expression that the describes the delimiters.

If the input is check1;check2 then the output should be check1;check2.

Answer 1

I don't think awk stores the matched delimiter anywhere. If you use GNU awk, you can do it yourself:

gawk '{match($0, /([^;,]*)(;+|,)(.*)/, a); print a[1], a[2], a[3]}'

Answer 2

If you're using GNU Awk (gawk) you can use the 4th argument of split():

gawk '{split($0, a, /;+|,/, seps); print a[1] seps[1] a[2]}' file

Output:

check1;check2
check1;;check2
check1,check2

Using it within a loop is also easy to handle:

gawk '{nf = split($0, a, /;+|,/, seps); for (i = 1; i <= nf; ++i) printf "%s%s", a[i], seps[i]; print ""}' file

22011,25029;;3331,25275
6740,16516;;27292,1217
13480,31488;;7947,18804
328,30623;;12470,6883

If you only need the fields you would only have to touch a. Separators would be separated in seps and the indices of those are aligned with a.

Answer 3

GNU awk has this feature for records not fields so you could also do something like this:

$ awk '{printf "%s%s",$0,RT}' RS=';+|,|\n' file
check1;check2
check1;;check2
check1,check2

Where RT is the value match by RS for the given record which you can see by:

$ awk '{printf "%s",RT}' RS=';+|,|\n' file
;
;;
,