Clarification needed in this reg exp/awk code
 

In my bash script, the $ERLDIR contains following:

/lhome/mselvam/vpn-5-5/otp/

and I have the code as:

bt_cv_erts_vsn="`ls ${ERLDIR} ${ERLDIR}/lib | \
awk -F- '/^erts-/ { if ($2 > v) v=$2; }
END { print v; }'`"

case "$bt_cv_erts_vsn" in
"")
{ echo "configure: error: "Could not figure out version Erlang library: erts"" 1>&2; exit 1; }
;;
*)
;;
esac

and I am getting the eror as:

checking erts version... configure: error: Could not figure out version Erlang library: erts

My understanding is, the above code, searches for the pattern erts- at the beginning of each line of the output of the ls command of 2 directories, and
prints the version v onto the bt_cv_erts_vsn

before doing so it is replacing the v with the second field of the matching line if the second field of the matching line was > v.

Is my understanding right? or am i wrong?

also my doubt is what is the second field it will look?

it can match erts (and not erts-) in one of the lines below (ls output), but there is no second field in it

config.log
config.status
configure
configure.in
configureout
erts
install
lib
make
makeinstallerr
makeinstallout
makeinstallwarn

if i echo v or bt_cv_erts_vsn, both of them are empty.
what will be the original value of v, in my program v is not initialised to anything

meena

I'm not sure I understand what you're after.

Post an example of some files that are in $ERLDIR and $ERLDIR/lib and what you expect to extract from it.

searching for erts in the ls output
 

The erts is searched on the ls output right? in my case, ls output contains erts and which is a directory

does the awk actions, will be to extract from within files? i dont think so/but i am not sure..

this code is written by someone else, but when i am making a build, i get this error.

erts is erlang library, which has been installed already by build procedures, but i do not know where version is available?


By the way when I give ls -l the second column happens to be an integer, is this the version of the file? what does this second column of ls -l means?

BUT IN MY program, it is not searching from ls -l output; instead it is meant to search on ls output which will not have a second field against erts?

meena

The second column of ls -l is the link count, which shows how many directory entries are referring to the file/directory. ls on its own, when not outputting to stdout, will show each filename on a single line; your awk command is splitting that line on a hyphen character, so the second field will be (presumably) the version number of the library named erts-n.n.n.so or whatever.