Command to zip a directory using a specific directory as the root

I'm writing a PHP script that downloads a series of generated files (using wget) into a directory, and then zips then up, using the zip c开发者_如何学Command.

The downloads work perfectly, and the zipping mostly works. I run the command:

zip -r /var/www/oraviewer/rgn_download/download/fcst_20100318_0319.zip /var/www/oraviewer/rgn_download/download/fcst_20100318_0319

which yields a zip file with all the downloaded files, but it contains the full /var/www/oraviewer/rgn_download/download/ directories, before reaching the fcst_20100318_0319/ directory.

I'm probably just missing a flag, or something small, from the zip command, but how do I get it to use fcst_20100318_0319/ as the root directory?

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I don't think zip has a flag to do that. I think the only way is something like:

cd /var/www/oraviewer/rgn_download/download/ && \
   zip -r fcst_20100318_0319.zip fcst_20100318_0319

(The backslash is just for clarity, you can remove it and put everything on one line.)

Since PHP is executing the command in a subshell, it won't change your current directory.

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I have also get it worked by using this command

exec('cd '.$_SERVER['DOCUMENT_ROOT'].' && zip -r com.zip "./"');

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cd /home/public_html/site/upload/ && zip -r sub_upload.zip sub_upload/

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Use the -j or --junk-paths option in your zip command.

From the zip man page:

-j

--junk-paths

Store just the name of a saved file (junk the path), and do not store directory names. By default, zip will store the full path (relative to the current directory).