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February 9th, 2010, 06:04 PM #1cron job unzipping protocol
Not sure if this is the right place to post this but it is related to datafeeds.
If one is using a cron job to unzip a datafeed that has been delivered to a server, how do you do it in such a way that the unzipped file(s) go to a specified file folder? I used the following years ago but it put the unzipped file into the home directory, not a public directory. I'd like to do it in such a way that the file is public for easy accessibility.
unzip -o /home/domainusername/public_html/domainfolder/filename.zip
February 9th, 2010, 06:39 PM #2
I think -d is destination, this is what I use:
unzip -o /data/incoming.zip -d /data/unzipped_path/
February 9th, 2010, 07:57 PM #3
If you type "unzip -?", you'll see the usage instructions. It looks like "-d" is right.
I'd say the post technically should go in the Programmer's Corner subforum, but since most people trying to unzip files are probably doing it for datafeeds, this subforum is fine as well.
February 10th, 2010, 10:41 AM #4
Worked like a charm! Thank you both!!
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