Bash script to uncomment lines with leading spaces on a file with specific pattern

You can try awk and command like this can do the work. If you need to have the specific strings 19 and 38 on specific places you should mention it in question:

awk '/19/ && /38/ {sub(/#/,"")}1' 

This command search for line with 19 and 38 and remove the # symbol. And then print the line (independently if it's match and edited or not)

awk '/19/ && /38/ {sub(/#/,"")}1' input_file
    pd19_ORA svg38
pd19_ORA sil38
#pd29_ORA sil37

If you want to be sure # is the first nonwhite space symbol in the line you can modify the script like this:

 awk '/19/ && /38/ && $1 ~ "^#" {sub(/#/,"")}1' input_file

Here is changed script which search for specific strings, congaing desired numbers:

awk '$1~/pd19_ORA$/ && $2~/[a-z]{3}38$/ && $1 ~ "^#" {sub(/#/,"")}1' 

Assuming you want to uncomment the lines that contain 38 and that follow a line that contain a word-delimited¹ 19 as your attempt suggests you might want to do:

perl -pi -e 's/^\s*\K#// if $flag && /38/; $flag = /\b19\b/' your-file

In your:

sed -n '/\<19\>/,+1p' cmfile|grep '38'|sed -i '/38/s/^#//g' cmfile

The sed -i '/38/s/^#//g' cmfile command doesn't read its input, it only edits cmfile in place, removing a leading # on any line that contains 38 (btw, the g is superfluous as there can only be one substitution).

So the sed -n '/\<19\>/,+1p' cmfile|grep '38' is pointless as nothing is reading its output.

Also note that all of -i, +1, \<\> are non-standard GNU extensions², while the perl command will work on any system³.

Edit

With your updated requirements where you want to uncomment the lines where the first field contains 19 and the second contain 38 (themselves not surrounded by other digits):

perl -api -e 's/^\s*\K#// if $F[0] =~ /(?<!\d)19(?!\d)/ &&
                             $F[1] =~ /(?<!\d)38(?!\d)/' your-file

Here using the -awk mode there the fields are made available in the @F array or, if there may be whitespace on either side of the # which would result in the first field becoming the second field, do the full line matching with a regexp:

perl -pi -e 's/^\s*\K#(?=\s*\S*(?<!\d)19(?!\d)\S*\s+\S*(?<!\d)38(?!\d))//' your-file

Where we have (see perldoc perlrun, perldoc perlop and perldoc perlre for details):

• -p: sed mode, where the -expression is evaluated for each line of input.
• -i: in-place editing
• s/regexp/replacement/: substitute the first match of the regular expression with replacement.
• \s: any whitespace character, \S for non-whitespace.
• <atom>*: the previous atom repeated any number of times including 0, as many as possible
• \K: Keep what's on the left, or IOW resets the start of the match that will end up being substituted.
• (?=...): positive look-ahead, or IOW "provided what follows matches ..."
• (?<!...): negative look-behind, or IOW "provided what precedes does not match ..."
• (?!...): negative look-ahead
• \d: any decimal digit character. Without -C (for Unicode mode), that's limited to 0123456789.

^{¹ In any case, the 19 in pd19_ORA is not word-delimited as both d and _ are word characters; if you wanted to match on 19 that is neither preceded nor followed by a decimal digit, you'd use (?<!\d)19(?!\d) instead of \b19\b (the perl equivalent of ex' \<19\>).}

^{² Well, strictly speaking, \<\> originated in ex/vi, not GNU sed and -i was copied by GNU sed from perl, and ,+1 from ed.}

^{³ Though you'll need perl 5.10.0 (from 2007) or newer for \K (to Keep what's on the left of it). On systems will an older version of perl, you can replace s/^\s*\K#// with s/^(\s*)#/$1/.}