Process several files and print filename without extension and the value of a specific key

Question

I have ASCII file and I want to output the file name and a specific field from this file beside each other. For example:
The contents of the ASCII file "data1.txt" is as follow:

Name: John
Age: 90
Gender: Male
Date: 20160105

I have a very long list of folders. Inside every folder I have the file data1.txt. The files share the same design like how I showed before but the content is different.

In the parent folder I created I text file with the list of folders and I named it all.txt. Then I wrote the following bash script:

#!/bin/bash
PAT=PATH TO THE PARENT FOLDER
for i in $(cat $PAT/all.txt); do
sed -n '/Name:/p'   $PAT/$i/data1.txt | awk '{ print $2}'

How can I revise the previous code to print the following:

data1: John      

Answer 1

You don't need the all.txt file, sed, or a for loop. find and awk can do it all:

find /path/to/folder -name data1.txt -exec awk -F': ' '/^Name/ {print $2}' +

Using a field separator of : (colon followed by a space character) is the key to getting the output right (no leading space, supports one or more words in the second field)

If you want the filename in the output, change the awk script to something like this:

find ... -exec awk -F': ' '/^Name/ {print FILENAME": "$2}' +

Answer 2

I did the following and the code worked perfectly:

#!/bin/bash
PAT=PATH TO THE PARENT FOLDER
for i in $(cat $PAT/all.txt); do
echo -n "data1" 
sed -n '/Name:/p'   $PAT/$i/data1.txt | awk '{ print $2}'