Hi,
I am trying to figure out how to open a datafile without knowing the name of
the datafile. For example, using the below code, I can open the datafile
LogFile.dat from the folder where the script is located. Both the script
and LogFile.dat are saved in the same folder therefore the below code works.
$inputFile="LogFile.dat";
# Opening LogFile.dat
open (DATA,$inputFile)||die("Can't open datafile: $!");
However, if I didn't know the name of the datafile how would I open it? I
tried $inputFile="(*).dat"; and $inputFile="*.dat"; but neither works. Does
anyone have any ideas? Thanks,
Allison
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