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Posted by rlee0001 on 02/12/06 20:11
Echo Echo,
At the beginning output $file. For some reason the file is reading
itself as input. If you are using:
php myprog.php infile.csv
....on the command line then I think you should be trying to get
"infile.csv" (or whatever) using $argv[1] not $argv[0]. Otherwise you
are passing the files own source to itself and the line
print "$data[$c]<br> \n" ;
Is causing the source to print.
-Robert
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