On Thu, October 18, 2007 2:47 pm, Chas. Owens wrote: > On 10/18/07, Paul <[EMAIL PROTECTED]> wrote: > snip >> >>>> The output is: >> >>>> text >> >>>> 0 > snip >> The function is like this: >> >> my$variable = (system "/usr/sfw/bin/openssl rsautl -decrypt -inkey >> private.pem -in cryptedfile") > snip > > There is no way that $variable will have the data you want. The > system function only returns the exit code and exec status of the > subshell it spawns. The reason you are seeing the data you want in > the output is because the subshell is writing to STDOUT. You need to > be using the qx// (or backtick) operators to capture the subshell's > output: > > my $var = qx(/usr/sfw/bin/openssl rsautl -decrypt -inkeyprivate.pem > -in cryptedfile); > print "I got [$var]\n"; > Thank you, the "qx" worked. Forgot all about that thing.
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