let's say I have this file format

and another file
i only want to get the lines between "test" and "*", so how do I do that and then assign to a variable with a space in between?

I'm not entirely sure I understand the question but one way, fairly clunky but hey that's me all over

If the input is all single column without separators;
start=grep -ni '^test' | awk '{FS=":"};{print $1}'
end=grep -ni '^*' | awk '{FS=":"};{print $1}'

start will then equal the line before you want to begin and end the line after

Using expr or any of the other tools you can then add one to start and subract one from end and have the range you need to work with.

If you're using perl etc then in a while loop look for test and then process until the variable equals *

If you grab the values between the separators into a variable then echoing that variable without quoting it will "magically" condense the whitespace into one space per run of whitespace.

That doesn't work too well if the input is supposed to contain the asterisk, though, because that will be expanded as a wildcard.

Instead, you can use tr to change newlines to spaces:

variable1=`sed -n '/^test$/,/^\*\*$/p file1 | tr '\012' ' '`
variable2=`sed -n '/^test$/,/^\*$/p' file2 | tr '\012' ' '`

Here is one way of doing what you want to do - but requires ksh93

#!/bin/ksh93
#
#  showme
#

TMP=file.$$

cat <<'EOT' >$TMP
first
test
dfgsdgs
sdgsg
3
4
sd
6
sdgsdg
8
9
**
last
EOT

# read required lines into var
var=$(
exec 3< $TMP
3<#'test'
3<# ((CUR + 5))
3<##'\*\**'
exec 3<&-
)

rm $TMP

# remove newlines
tmp=$(print $var)

print "var: $tmp"

exit 0

$ ./showme
dfgsdgs sdgsg 3 4 sd 6 sdgsdg 8 9
$

sed -n '/test/,/*/p' file

Actually plain * is a syntax error, you need to backslash it.