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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