7zip

问题 There is a post on this topic already, but it does not have an explicit answer to the fundamental question which I am re-asking here: How do you make 7zip commands from Python? Attempting to use the subprocess module, I implemented the following which runs but does nothing (from what I can tell): import subprocess cmd = ['7z', 'a', '"Test.7z"', '"Test"', '-mx9'] subprocess.Popen(cmd, stderr=subprocess.STDOUT, stdout=subprocess.PIPE) I know that the following 7zip command works, as I have

问题 I'm trying to create a batch script that will zip all the contents in each subdirectory except the latest (or latest few). I'm currently attempting in Windows with 7-Zip but the directory is technically on a Linux server so any suggestions geared towards a Linux command is welcome. The directory structure is like this: Directory-Parent ---Sub-Directory-1 --------File1 --------File2 --------File3 ---Sub-Directory-2 --------File1 --------File2 I would like to run a batch at the Directory-Parent

问题 I am trying to extract files from a password-protected zip, in a USB drive, using PowerShell. I have looked up many ways but the easiest one doesn't seem to work. $7ZipPath = "C:\Program Files\7-Zip\7z.exe" $zipFile = "E:\passwordprotectedtest.zip" $zipFilePassword = "Foo" & $7ZipPath e -oE:\ -y -tzip -p "$zipFilePassword" "$zipFile" I keep getting this error: 7-Zip 9.20 Copyright (c) 1999-2010 Igor Pavlov 2010-11-18 Error Cannot use absolute pathnames for this command I then moved to the

问题 I have Several zip files that Contain multiple filetypes. The ONLY ones I am interested in are the .txt files. I need to extract the .txt files only and place them in a folder of their own, ignoring all other file types in the zips files. All the zip files are in the same folder. Example -foo.zip --1.aaa --2.bbb --3.ccc --4.txt -foo2.zip --5.aaa --6.bbb --7.ccc --8.txt I want to have 4.txt and 8.txt extracted to another folder. I can't for the life of my figure it out and spent ages looking,

问题 I would like to unzip a file with Python using 7zip executable. In Perl this is pretty straightforward: $zip_exe_path = "C:\\Dropbox\\7-zip\\7z.exe"; $logfile_path = "C:\\Temp\\zipped_file.7z"; system ("$zip_exe_path x $log_file_path -y"); I tried this: import subprocess zip_exe_path = "C:\\Dropbox\\7-zip\\7z.exe" logfile_path = "C:\\Temp\\zipped_file.7z" subprocess.call(['zip_exe_path','x','logfile_path','-y']) When I do so I get this error: FileNotFoundError: [WinError 2] The system cannot

问题 Our company develops and distributes a Powerpoint Add-In. From the AddIn I add CustomXml parts (referenced to the presentation and the slide as well) to a Powerpoint presentation via DocumentFormat.OpenXml SDK V 2.10. Everything is OK with the file. Then I copy the file to a SharePoint online. It doesn't matter, if I copy via our WebDAV path, a SharePoint website on a browser or the latest possibility via OneDrive. Subsequently the behavior of that file varies: Powerpoint opens the file

问题 I am trying to zip a folder from an asp page. This is my code: zipFolderName=folderName &"Zipped.zip" command="cd C:\Program Files\7-Zip & " command = command & "7z a -tzip " & zipFolderName & " """ & folderName & """" Response.Write command set objshell = Server.CreateObject("WScript.shell") objShell.exec (command) set objshell=nothing The command that is written in the Response.Write is cd C:\Program Files\7-Zip & 7z a -tzip D:/saveAll/DocumentsZipped.zip "D:/saveAll/Documents" When I run

问题 I want to extract archive files (zip,rar,z7,gz,etc) from PHP. Because the passwords can contain special characters like ¡¿, and PHP exec does not support unicode characters (for some reason). I eneded up using a batch file as explained here php exec() in unicode mode?. But, as lovely as windows 10 is, I can't seem to figure out how to extract an archive that has a single double quote as password... "C:\Program Files\7-Zip\7z.exe" x "C:\Users\me\Desktop\25mb.rar" -o"C:\Users\me\Desktop\" -p

问题 Hi I am quite new to zip format, and I used the Java's util implementation to unzip a file, However it throws a ZipException everytime I try to open the file. I checked to see if the file was corrupted and it is not because I can open it using winRar. So I moved on and tried appache.commons.vfs package to do the same thing and that too resulted in a failure. As a final try I tried the library 7-zip-jbinding and that is able to read the contents of the Zip archive but I am not able to extract

问题 It's zipped and password protected with 7zip, should the mimetype be "application/7zip", or just "data/binary"? Something else? 回答1: According to Wikipedia (7z article), the mime type of a 7zip archive is application/x-7z-compressed , having a password on it will not change the mimetype. If it is a zip (ie. not 7-zip file), the mimetype should be application/zip 来源： https://stackoverflow.com/questions/2203400/whats-the-correct-email-attachment-mime-type-for-a-zipped-and-password-protecte