$dir = "./";

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Please explain what the "./" represents here? I assume it means present working directory? If so how would I rewrite this to take an input? 
$dir = "./";
opendir( DIRS, $dir );

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I tried: 
Print "Enter directory";
$dir=<STDIN>;      
chomp $dir;
opendir (DIRS, $dir);

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In reply to Changing to input by Anonymous Monk

Title:
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