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Full Discussion: My variable cannot be passed through into my path
Top Forums Shell Programming and Scripting My variable cannot be passed through into my path Post 302348055 by cyberfrog on Thursday 27th of August 2009 08:15:17 AM
Old 08-27-2009
don't get any I'm affraid, i only realise it hasn't copied the directory when looking into the target one and its empty. to confirm the source and target are correct I have echoed these pathnames and they appear correct in the terminal and when I do the same command cp -r source target on the command line it works? by the way I'm on the linux too
 

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LEARN ABOUT DEBIAN

zipmerge

ZIPMERGE(1)						      General Commands Manual						       ZIPMERGE(1)

NAME

       zipmerge - merge zip archives

SYNOPSIS

       zipmerge [-DhIiSsV] target-zip source-zip Op source-zip ...

DESCRIPTION

       zipmerge  merges  the  source zip archives source-zip into the target zip archive target-zip.  By default, files in the source zip archives
       overwrite existing files of the same name in the target zip archive.

       Supported options:

	      -D   Ignore directory components in file name comparisons.

	      -h   Display a short help message and exit.

	      -I   Ignore case in file name comparisons

	      -i   Ask before overwriting files.  See also -s.

	      -S   Do not overwrite files that have the same size and CRC32 in both the source and target archives.

	      -s   When -i is given, do not before overwriting files that have the same size and CRC32.

	      -V   Display version information and exit.

EXIT STATUS

       zipmerge exits 0 on success and 1 if an error occurred.

SEE ALSO

       zipcmp(1), ziptorrent(1), libzip(3)

AUTHORS

       Dieter Baron <dillo@giga.or.at> and Thomas Klausner <tk@giga.or.at>

NiH								   June 4, 2008 						       ZIPMERGE(1)
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