I need to be able to parse out a file name. For example Review_Quarterly_20101130_20101209.zip, based on the file name I need to have visual cron extract the file to the Quarterly folder under the 20101130 folder. Another file would be Review_Monthly_20101130_20101209.zip and that would need to go to the 20101130\Monthly folder.

I was looking at using a variable which I know I can grab the file name but then I can't figure out how to parse it.

Thanks!

I think you need to use the "regex" variable. But I am no expert on regex. There are a lot of forums for that.

{REGEX(Replace|a b c d|\s+|, )}

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