vi /remote/scm/rsync/builds/v_dialer/FailedInformal/3.0.0.0.118/build.log
will opens a buildlog,i assigned this absolute path to $path variable
$path1="$path1/build.log";
open(F,"$path1)or die "can't open the file $!";
i want to know what was the reason why open()is notable to open the build.log by same path from command prompt its opning...where i am doing wrong....seeking solutions from wise monks.

In reply to By using open() unable to open the file by perladdict

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