[SOLVED] Print only words with Capital Letters (Linux)

Issue

So I am currently reading from this txt file:

Line 961: www-d1.proxy.aol.com - - [01/Aug/1995:00:35:32 -0400] "GET /elv/hot.gif HTTP/1.0" 200 1007
Line 965: www-d1.proxy.aol.com - - [01/Aug/1995:00:35:41 -0400] "GET /elv/PEGASUS/minpeg1.gif HTTP/1.0" 200 1055
Line 966: www-d1.proxy.aol.com - - [01/Aug/1995:00:35:46 -0400] "GET /elv/SCOUT/scout.gif HTTP/1.0" 200 1165
Line 969: www-d1.proxy.aol.com - - [01/Aug/1995:00:35:49 -0400] "GET /elv/DELTA/delta.gif HTTP/1.0" 200 2244
Line 972: www-d1.proxy.aol.com - - [01/Aug/1995:00:35:51 -0400] "GET /elv/ATLAS_CENTAUR/atlas.gif HTTP/1.0" 200 2286
Line 95219: u139.n72.queensu.ca - - [04/Aug/1995:10:40:04 -0400] "GET /elv HTTP/1.0" 302 -

And I am trying to print out only the names in the command line(basically only the ones in capital letters) WITH NO DUPLICATES. For example:

ATLAS_CENTAUR
DELTA
SCOUT
PEGASUS

My codes so far:

grep "/elv" ~/spacestation.txt | awk -F/ '{print $5}' | sort -u

Actual output:

1.0" 302
ATLAS_CENTAUR
DELTA
hot.gif HTTP
SCOUT
PEGASUS

Solution

You need to put regex pattern in your awk script to compare $5:

Solution:

grep "/elv" ~/spacestation.txt | awk -F/ '$5 ~ /^[A-Z_]+/ {print $5}' | sort -u 

1. '~' is for compare $5 with regex pattern matching
2. '^' is first character of word
3. '[A-Z_]' will look for all caps-lock characters including with _
4. '+' is for to continue with matching [A-Z_] if he finds one or more character like this

Answered By - acakojic

Answer Checked By - Marie Seifert (WPSolving Admin)