AWK reading path from text file

You could build the same logic with the print/no-print flag inside the while(getline) loop. Something like this:

awk '{ while((getline a < $0) > 0) { 
    if (a ~ /BEGIN/) p=1;
    if (p) print a;
    if (a ~ /END BEGIN/) p=0;
} }' path_resume.txt > output.txt

The only thing here is that you need to explicitly use if statements, instead of implicit conditionals like you can on the top level of the AWK script.

Or in Bash, slurp the list of filenames to an array and pass them all to AWK in one go:

readarray -t filenames < path_resume.txt
awk '/BEGIN/ {p=1}; p; /END BEGIN/ {p=0}' "${filenames[@]}" > output.txt

(I put the print action before the END BEGIN check to also print that end delimiter.)

Something like the following, untested, will do what you want using any awk:

awk '
    NR == FNR { ARGV[ARGC++]=$0; next }
    $1 == "BEGIN" { f=1 }
    f
    $0 == "END BEGIN" { f=0 }
' path_resume.txt > output.txt

or this might be very slightly faster but see http://awk.freeshell.org/AllAboutGetline so you don't fall into the trap of using getline unnecessarily or incorrectly:

awk '
    BEGIN {
        file = ARGV[--ARGC]
        while ( (getline line < file) > 0 ) {
            ARGV[ARGC++] = line
        }
    }
    $1 == "BEGIN" { f=1 }
    f
    $0 == "END BEGIN" { f=0 }
' path_resume.txt > output.txt

Using xargs (and without using double awk to read each line from the file) you can pass all the paths you have in your text file:

xargs awk  '/BEGIN/{flag=1}/END BEGIN/{print;flag=0}flag'  < path_resume.txt > out.txt

I'm assuming the your file path_resume.txt does not contain spaces in any path

In awk I used {print;flag=0} to also print the line END BEGIN.

If you had paths with spaces in their filenames then you can use this:

tr '\n' '\0' < path_resume.txt | xargs -0  awk  '/BEGIN/{flag=1}/END BEGIN/{print;print "";flag=0}flag' > out.txt

Assuming all pathnames in the input file are properly quoted, concatenate all the listed files together and use sed to extract the parts between lines starting with BEGIN and end with END BEGIN:

xargs cat <path_resume.txt |
sed -n '/^BEGIN/,/^END BEGIN/p' >output.txt

This uses xargs to call cat with the sisted files in path_resume.txt as operands. This creates a single stream of all data concatenated from all the files in the order listed in the input file.

The stream is then filtered by sed which prints all lines in the blocks we're interested in.

One could reformulate the sed expression so that it instead deletes all lines that occur outside of the blocks we're interested in (allowing us to drop the -n option):

xargs cat <path_resume.txt |
sed '/^BEGIN/,/^END BEGIN/!d' >output.txt

You could obviously also call sed directly from xargs:

xargs sed '/^BEGIN/,/^END BEGIN/!d' <path_resume.txt >output.txt

Or, using awk,

xargs awk '/^BEGIN/,/^END BEGIN/' <path_resume.txt >output.txt

How about this approach? It's kinda lazy, but it will do the job. You read the lines of path_resume.txt and use every line as a file for awk to work with

for i in `cat path_resume.txt`; do awk '/BEGIN/{flag=1}/END BEGIN/{flag=0}flag' $i >> output; done