Search a line(string) between other two diferent lines

Question

I'm searching for one liner command (in best way) that can find if specific string/line is presented between other two lines. I search for it and I only found commad to get content between two lines, but how can i check if someting is present or not..

.....
 1 a 2 b 3
 4
   5
.....
 1 c 2 d 3
 4
   5
.....
 1 e 2 f 3
   5
.....

I found this:

sed -n '/^ 1 .* 2 .* 3$/,/^ 5$/p'

the result with this command cut the unwanted lines good for start (cut lines "...."), but still not know how to check if "4" is presented between :

 1 a 2 b 3
 4
   5
 1 c 2 d 3
 4
   5
 1 e 2 f 3
   5

The output should look like this:

 "4" is missing after "1 e 2 f 3"

OR only (is even better):

"1 e 2 f 3"

Answer 1

sed is not the right tool for the job. However, we can probably still sort of do it in sed:

sed -n '/^ 1 .* 2 .* 3$/,/^ 5$/ { /^ 1 .* 2 .* 3$/ { h }; /^ 4$/ { x; s/.*//; x;}; /^ 5$/ { x; p; x} }' filename | grep -v -e '^$'

Here's only the part that I added into your range block:

/^ 1 .* 2 .* 3$/ { h }; /^ 4$/ { x; s/.*//; x;}; /^ 5$/ { x; p; x}

Read this as:

if line matches regex /^ 1 .* 2 .* 3$/, then h (store line into hold buffer)
if line matches regex /^ 4$/, then x (exchange buffer, i.e. make operations apply to hold buffer rather than standard buffer), then replace everything in hold buffer*, then x again (to switch back to standard buffer)
if line matches ^ 5$/, then switch to hold buffer, p (print contents of hold buffer), and switch back to standard buffer

* Unfortunately, s/.*// doesn't delete lines in the hold buffer. Deleting lines in the hold buffer seems difficult, so we get rid of them by piping into grep -v -e '^$' instead.

Update

This version prints the filename after the match (using the F command), and manages without piping into grep. Thank you, Paulo!

sed -n '/^ 1 .* 2 .* 3$/,/^ 5$/ { /^ 1 .* 2 .* 3$/ { h }; /^ 4$/ { x; s/.*//; x;}; /^ 5$/ { x; /^$/ !{ p; F }; x} }' data

Answer 2

Thanx @sneep for mentioning perl:

perl -lane 'if($n=/^ 1 .* 2 .* 3$/../^ *5$/) {$s=$_ if $n==1; $s="" if /^ *4$/; print "$ARGV: $s" if $s && $n=~/E/}' /otherdir/*

I hope I understand the problem correctly: please add to the OP to specify any further details.

Explanation:

• perl Practical Extraction and Reporting Language.
• -lane switches commonly helpful for one-liners.
• ' start of the actual program instructions
• if(/^ 1 .* 2 .* 3$/../^ *5$/) { execute the actions between curly braces only for pieces of text that start with a line matching /^ 1 .* 2 .* 3$/ and end with a line that contains a single 5 digit, possibly preceded by any number of spaces.
• $n=/^ 1 .* 2 .* 3$/../^ *5$/ keep track of the line number within the text portion.
• $s=$_ if $n==1; save the first line of the text portion in the $s variable.
• $s="" if /^ *4$/; look for the desired string: a single 4 digit that comes after any number of spaces. If found, delete the previously saved line for the $s variable.
• print "$ARGV: $s" if $s && $n=~/E/ print the filename, a colon, a space and the line that started the current text portion if the variable $s contains any text and the $n counter shows that the last line of the text portion has been reached.
• } end of the actions to be performed on the text portion.
• ' end of the program instructions.
• /otherdir/* process all files in the /otherdir/ path