Hai!
Hello,
My requirement is to open file with 666 permissions.[If fine doesn't exists it should get created ].Iam doing as below,is this ok.
=========================== sysopen(LOG,"$main::TRACELOGFILE",O_CREATE,0666) or die "Can't open trace file $main::TRACELOGFILE"; ============================
ami doing any thing wrong ,if yes correct me.If file is created once it is working fine.For the first time it is giving below error =========================== Can't open trace file /var/opt/XXXXX/log/XXXXX08013.log ==========================
You should include the $! variable in your error message so you know *why* it failed. The third argument to sysopen must include *one* of either O_RDONLY or O_WRONLY or O_RDWR (read only OR write only OR read and write.)
Read the section "Open A la C" in perlopentut for more details on sysopen.
perldoc perlopentut
John -- use Perl; program fulfillment
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