Parsing only first regex match in a line with several matches

Question

Is it possible to have a regex that parses only a1bcdea1 from this line a1bcdea1ABCa1DEFa1 ?

This grep command does not work:

Answer 1

How about using a ^ start anchor and restricting character set used:

grep -o '^[A-Za-z]1[A-Za-z]*1'

See this Bash demo or Regex Pattern at regex101

If you expect more digits or other characters in between, go with this

grep -oP '^[A-Za-z]1.*?[A-Za-z]1'

The lazy matching requires perl compatible mode. For not at line start, go with this

grep -oP '^.*?\K[A-Za-z]1.*?[A-Za-z]1'

\K resets beginning of the reported match and is a PCRE feature as well.

Answer 2

Here is a gnu awk solution using split function:

awk '(n = split($0, a, /[a-zA-Z]1/, b)) > 1 {print b[1] a[2] b[2]}' file

a1bcdea1

This awk command splits each line on regex /[a-zA-Z]1/ and stores split tokens in array a and delimiters in array b.