How to print file content only if the first line matches a certain pattern?

With POSIX tools chest:

{ head -n 1 | grep pattern && cat; } <file

If you're writing a shell script, you could so something like

for file in ./*; do head -n 1 "$file" | grep -q 'PATTERN' && cat "$file"; done

Or, in Perl:

perl -Tlne '$f = /PATTERN/ if $. == 1; print if $f; $. = 0 if eof' ./*

 awk '/pattern/{print FILENAME}; {nextfile}' ./*.txt

would print the name of the non-hidden txt files in the current directory whose first line matches the extended regular expression pattern with those awk inplementations that support nextfile.

If instead of printing the file name, you want to print the whole file content, you can do:

 awk 'FNR == 1 && ! /pattern/ {nextfile}; {print}' ./*.txt

That's efficient in that it runs only one command, but awk being not the most efficient command to dump the content of a file, with large files, you could possibly obtain better performances by doing something like:

 awk '/pattern/{printf "%s\0", FILENAME}; {nextfile}' ./*.txt |
   xargs -r0 cat

That is, only use awk to print the list of files that match (0-delimited) and rely on cat to dump their content.

Oldschool, just translate your sentence into standard commands:

for file in *; do
    if head -n 1 "${file}" | grep -q 'PATTERN'; then
        cat "${file}"
    fi
done

For learning bash that is a good start. If you just need a quick solution, try the sed-, awk- or perl-answers. Both nice, but they are own languages you need (and probably want) to learn.

It's a pretty simple example, so if you want to learn more, you could also try the same in ruby, php, js (e.g. in nodejs) or any other language that allows file access. Even C/C++ or Java should be easy to manage with a small task.