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php-general | 2005051
Date: 05/12/05
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[snip]
Thank you for your response Jay, but that is not working. My program
will not run at all with the following:
var $exportFile = "Export." . date("mdy") . ".txt";
I seem to be able to use the date function is I am not starting the
declaration with "var", but then my program is not working correctly.
[/snip]
You may have to assemble it beforehand sort of ...
$exportFileName = "Export." . date("mdy") . ".txt";
var $exportFile = $exportFileName;
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