I am attempting to use 7 Zip through the command line. As you can see below, using the command 7z l
lists the 3 files in the target zip file.
C:\Users\User1\Downloads>7z l recording_20130731180507.zip
--
Path = recording_20130731180507.zip
Type = zip
Physical Size = 311686
Date Time Attr Size Compressed Name
------------------- ----- ------------ ------------ ------------------------
2013-07-31 18:05:06 ..... 655 655 SD_DISK\20130731\18\2013073
1_180505_A4BC_00408CC2B40B\recording.xml
2013-07-31 18:05:06 ..... 309752 309752 SD_DISK\20130731\18\2013073
1_180505_A4BC_00408CC2B40B\20130731_18\20130731_180505_59EB_00408CC2B40B.mkv
2013-07-31 18:05:06 ..... 279 279 SD_DISK\20130731\18\2013073
1_180505_A4BC_00408CC2B40B\20130731_18\20130731_180505_59EB_00408CC2B40B.xml
------------------- ----- ------------ ------------ ------------------------
310686 310686 3 files, 0 folders
However, when I attempt to actually unzip the file, I get a "no files to process error". I've never tried unzipping from cmd before. Do I have to try to dig into the zip file to extract those 3 files?
C:\Users\User1\Downloads>7z e recording_20130731180507.zip o-C:\users\User1\do
cuments\folder1\test
No files to process
Files: 0
Size: 0
Compressed: 311686
The option is -o
, not o-
. Run the command like this:
7z e recording_20130731180507.zip -o"C:\users\User1\documents\folder1\test"