unzip .zip file and list the files included in the .zip archive


 
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# 1  
Old 04-08-2008
unzip .zip file and list the files included in the .zip archive

Hello,
I am trying to return the name of the resulting file from a .zip archive file using unix unzip command.

unzip c07212007.cef7081.zip
Archive: c07212007.cef7081.zip
SecureZIP for z/OS by PKWARE
inflating: CEP/CEM7080/PPVBILL/PASS/G0063V00

I used the following command to unzip in quite mode (-q), without making directories(-j) and overwrite existing files(-o)

unzip - qjo c07212007.cef7081.zip

The above cmd unzippes the file "G0063V00" in the current working directory.

In addition to unzipping the file in the current working dir, I would like to store the filename resulting filename "G0063V00" in a variable for further processing. How do I get the filename that was just unzipped ? I tried using -l option but I get a lot more information than needed.

unzip -qjol c07212007.cef7081.zip

Length Date Time Name
-------- ---- ---- ----
265 07-22-07 03:32 CEP/CEM7080/PPVBILL/PASS/G0063V00
-------- -------
265 1 file

So I tried the following:

basename "`unzip -qojl c07212007.cef7081.zip`"

G0063V00
-------- -------
265 1 file


As you see, I still get more information than I need. I just want to return the filename which in this case is "G0063V00".

Thanks for you help in advance.
# 2  
Old 04-08-2008
you could sort by last modified, grab first entry into variable then echo out the variable.
# 3  
Old 04-08-2008
There are other processes concurrently creating files in the same directory.
# 4  
Old 04-08-2008
Code:
basename "`unzip -qojl c07212007.cef7081.zip | awk 'NR==3{print $4}'`"

maybe?
# 5  
Old 04-08-2008
Thanks ! That did the trick.

If you dont mind, can you briefly explain
awk 'NR==3{print $4}'

Thanks for you help !
# 6  
Old 04-08-2008
Fourth field of third line. See the awk manual page or any introductory Unix book for details.
 
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