We may use either of sed or awk to completely solve the problem.
With sed:
$ sed 's/^.\./0&/' file.txt When & occurs in the replacement part of the substitution command (s), it will be expanded to the part of the input line that matches the pattern part of the command.
The regular expression ^.\. means "match all lines that starts with (^) an arbitrary character (.) followed by a literal dot (\.)".
If the line is 1.02.2017 23:40:00, the pattern will match, and 1. would be replaced by 01. at the start of the line.
With awk:
Building on the partial awk code in the question...
This will, as stated, print the second character of each line of input:
$ awk '{ print substr($0, 2, 1) }' file.txt We can use the fact that substr($0, 2, 1) returns the second character and use that as the condition:
$ awk 'substr($0, 2, 1) == "." { ... }' file.txt What goes into { ... } is code that prepends $0, which is the contents of the current line, with a zero if the preceding condition is true:
$ awk 'substr($0, 2, 1) == "." { $0 = "0" $0 }' file.txt Then we just need to make sure that all lines are printed:
$ awk 'substr($0, 2, 1) == "." { $0 = "0" $0 } { print }' file.txt The condition substr($0, 2, 1) == "." may of course be changed into a regular expression too (we use exactly the same expression as we used in the sed solution):
$ awk '/^.\./ { $0 = "0" $0 } { print }' file.txt Some people who thinks "shorter is always better" would write that as
$ awk '/^.\./ { $0 = "0" $0 } 1' file.txt (and probably also remove most spaces: awk '/^.\./{$0="0"$0}1' file.txt)