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Top Forums Shell Programming and Scripting Problem with double quote and string variable Post 302176335 by yunccll on Tuesday 18th of March 2008 05:31:01 AM
Old 03-18-2008
if you want to do this job with script, you can try code as follow:
Code:
#!/bin/bash

INFILE=output.txt

#replace the double quote with space
LIST=$(sed -e 's/\"/ /g' $INFILE)

zip archive.zip $LIST

exit 0

 

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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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