Find a specific file then pipe to stdout/awk

Question

I'm looking for a way to traverse directories recursively to find a specific file, then stop the search and pipe the filename path to an awk function or something similar. I asked a question earlier that was similar, but after testing on machines other than mine it turns out the locate command isn't going to work since not everyone uses it on their system.

Code that I used with locate:

dir="/path/to/destination/";
mkdir "$dir";
locate -l 1 target_file.txt | \
   awk -v dir="$dir" '{printf "cp \"%s\" \"%s\"\n", $1, dir}' | \
   sh

Answer 1

The find(1) command will do it. To only get one line, use head(1).

dir="/path/to/destination/";
mkdir "$dir";
find /path/location -name target_file.txt |
    head -n 1 |
    awk -v dir="$dir" '{printf "cp \"%s\" \"%s\"\n", $1, dir}' |
    sh

Answer 2

If you know only one file exists then you can use

find ./ -name "target_file.txt" -exec cp -r {} $dir \;

And if you are not sure, use head to limit 1 and use xargs

find ./ -name "target_file.txt" | head -1 | xargs -I {} cp -r {} $dir/