How to process string in using awk in shell script - bash

Question

I am very new to shell scripting and have to do so many tasks around it. I am trying to learn as fast a possible but some times shell scripting makes a task look very easy and at other times it just toys with me. And I am facing similar situation now.
I have a command which gives me an output like this.
File Dependents
----------------------------------------------------------------------------
<File> is a requisite of <Dependents>
Path: /usr/lib/obj
Java 1.0.0.0 analysis 0.0.0.2
runtime 1.2.0.0
client 1.2.0.0
framework 6.1.9.100
sguide 1.9.10.0
sysmgt 6.1.9.100
dsm 6.1.9.200
Path: /etc/obj
Java 1.0.0.0 analysis 1.2.0.2
runtime 2.0.0.0
client3 6.1.9.0
sysmgt 6.1.9.0
dsm2 6.1.9.0
Now I want to get the list of dependencies into an array for further processing. This is what I am able to do so far:
<command> | cut -f1 | grep '[a-z]' | grep -v File | grep -v : | awk '{ print $1}'
output is:
Java<<< I want this to be analysis
runtime
client
framework
sguide
sysmgt
dsm
Java<<< want this to be analysis
runtime
client3
sysmgt
dsm2
I have to capture these two lists in two separate arrays.
Can someone please help me in achieving this output in an elegant way. I don't want to butcher this code with my brute force method involving lot of conditions and comparisions.

Answer 1

awk to the rescue!
$ arr1=$(command ... | awk -v c=1 '!NF{f=0} f && s==c{print $1} /Java/{f=1; s++; if(s==c) print $(NF-1)}')
$ arr2=$(command ... | awk -v c=2 '!NF{f=0} f && s==c{print $1} /Java/{f=1; s++; if(s==c) print $(NF-1)}')
$ echo $arr1
analysis runtime client framework sguide sysmgt dsm
$ echo $arr2
analysis runtime client3 sysmgt dsm2
perhaps better if you run the command once and split the results into two arrays.
Explanation
awk -v c=1 set awk variable c to 1 (describes group instance number)
'!NF{f=0} if there are no fields (empty line) reset f
f && s==c{print $1} if f is set and counter equals to c print the first field
/Java/{f=1; s++; when pattern matched to Java, set f and increment counter and
...if(s==c) print $(NF-1)}' if counter matches c print the penultimate field.

Answer 2

You can fix your solution by removing the substring with Java first:
command | sed 's/Java [^ ]*//' | cut -f1 | grep '[a-z]' | grep -v File | grep -v : | awk '{ print $1}'
When you use awk, you can better use the full strength of awk. Just say you want the print the second last field of any line with a number:
command | awk '/[0-9]/ { print $(NF-1) }'
This is better than trying to use sed (do you have tabs or spaces?)
command | sed -n '/[0-9].[0-9]/ s/^.* \([^ ]*\) .*/\1/p'
A funny solution is using rev to revert your text. That way cut can find the second field.
command | grep '[0-9].[0-9]' | rev | cut -d " " -f2 | rev
For people who only read the last line, I will repeat the awk solution:
command | awk '/[0-9]/ { print $(NF-1) }'