Linux Grep or Awk to find strings and store into array - awk

Question

I would like to print the string in the following pattern. And I would like to store it in a array. Please help me, I need O/p as follows
test11
orcl
My commands/Tries
egrep -i ":Y|:N" /etc/oratab | cut -d":" -f1 | grep -v "\#" | grep -v "\*" | tr -d '\n' | sed 's/ /\n/g' | awk '{print $1}'
Above commands O/p:
test11orcl
Contents of Oratab will be as follows,
[oracle#rhel6112 scripts]$ cat/etc/oratab
#
# This file is used by ORACLE utilities. It is created by root.sh
# and updated by the Database Configuration Assistant when creating
# a database.
# A colon, ':', is used as the field terminator. A new line terminates
# the entry. Lines beginning with a pound sign, '#', are comments.
#
# Entries are of the form:
# $ORACLE_SID:$ORACLE_HOME:<N|Y>:
#
# Multiple entries with the same $ORACLE_SID are not allowed.
#
#
test11:/u01/app/oracle/product/11.2.0/dbhome_1:N
orcl:/u01/app/oracle/product/10.2.0/db_1:N
End of Cat Output
From the above file am trying to extract the STRING before the :/

Answer 1

As a start, try this:
$ cat input.txt
test11:/u01/app/oracle/product/11.2.0/dbhome_1:N
orcl:/u01/app/oracle/product/10.2.0/db_1:N
$ awk -F: '{print $1}' input.txt
test11
orcl
update
Using bash:
#!/bin/bash
ARRAY=()
while read -r line
do
[[ "$line" = \#* ]] && continue
data=$(awk -F: '{print $1}' <<< $line)
ARRAY+=($data)
done < input.txt
for i in "${ARRAY[#]}"
do
echo "$i"
done
In action:
$ ./db.sh
test11
orcl

Answer 2

You could use sed also,
sed -r 's/^([^:]*):.*$/\1/g' file
Example:
$ cat cc
test11:/u01/app/oracle/product/11.2.0/dbhome_1:N
orcl:/u01/app/oracle/product/10.2.0/db_1:N
$ sed -r 's/^([^:]*):.*$/\1/g' cc
test11
orcl
OR
$ sed -nr 's/^(.*):\/.*$/\1/p' file
test11
orcl