Shell Programming and Scripting

I am trying to pass the results from a variable gathered from awk, however when I echo the 'PARSE' and 'SUB', the response is blank. This is my command.

awk -F= '/Unit/''{ PARSE=substr($2,1,5) ; SUB=substr($2,1,1) }' inputfile.lst

Is this a kind of valid attempt or am I obligated to declare two lines like:
PARSE=awk -F= '/Unit/''{ print substr($2,1,5) }' inputfile.lst
etc.

Thanks in advance for any replies.

Unfortunately, this doesn't work on all shells - some versions of bash and pdksh execute the read command in a subshell so this will fail.

Under "proper" ksh, sh-posix and zsh, you can do something like (assuming the input file consists of one line)

Code:

awk '{ PARSE=substr($2,1,5) ; SUB=substr($2,1,1); print PARSE , SUB }' inputfile | read a b
echo "PARSE: $a"
echo "SUB:   $b"

If you're using bash, pdksh or some other shell that executes the read in a subshell, you can get around things with a while loop

Code:

awk '{ PARSE=substr($2,1,5) ; SUB=substr($2,1,1); print PARSE , SUB }' inputfile | while read a b
do
  echo "PARSE: $a"
  echo "SUB:   $b"
done

The advantage of the while loop is that you can iterate over multiple lines.

Cheers
ZB

If you do this:
awk 'BEGIN{ print "val1=a val2=b"}' < /dev/null
you will get as output:
val1=a val2=b

That just happens to be the syntax to set val1 and val2 in the shell. So you want to feed that output into the shell for execution. That will set both variables. And it should work with lesser shells.

$ eval `awk 'BEGIN{ print "val1=a val2=b"}' < /dev/null`
$ echo $val1 $val2
a b

zazzybob, I am using Solaris v8, native ksh shell. I had to make some minor customizations to your suggestion, but it works perfect. Now I just have to figure out the looping problem with my script, and it will be a screeeeeamer!! Thanks so much, you rock, and I really appricate it.