在cmd中解压缩文件时出错(在Windows中使用7z)

Question

我试图通过命令行使用7 Zip.如下所示,使用该命令7z l列出目标zip文件中的3个文件.

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

但是,当我尝试实际解压缩文件时,我得到一个"没有文件来处理错误".我以前从未试过从cmd解压缩.我是否必须尝试挖掘zip文件来提取这3个文件？

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

Answer 1

选项-o不是o-.像这样运行命令:

7z e recording_20130731180507.zip -o"C:\users\User1\documents\folder1\test"