Linux extract rar part

How do you extract a particular directory from a multi-part RAR using unrar CLI

I can’t find any way to do this after searching for hours and trying various things. I’m using rarlab’s free unrar CLI on OS X. I want to extract a single folder and its contents, preserving paths, from a multi-part RAR archive. Seems like a simple enough task but… The help listing is of no help. Someone must have figured this out by now.

I didn’t include anything I tried because nothing worked. I don’t want people to have to wade through posts to get to the code listing that works. I think in a case like this, someone either knows how to do this or they don’t pretty much. But if you insist, the only thing I tried that might be relevant is: unrar l -ap/dir1 filename.part*.rar and variations of this. I got the ‘-ap’ switch from the unrar help listing: ap Set path inside archive . Kind of vague at best but the only thing that seems like it might work.

So much to say; so little space. First of all, I don’t know what you mean by “I don’t want people to have to wade through posts to get to the code listing that works.” Secondly, members of Super User like to see what you’ve tried (1) so we know that you’re invested in the question, and you aren’t just asking us to “do your homework for you,” (2) because it might clarify your problem, and (3) so we can avoid suggesting solutions that have already failed for you. Next … OK, I’m running out of room here, and this is looking more and more like an answer, so I’m going to post the rest as an answer.

Yeah, well many times on these forums, I find a long thread with lots of code blocks that I have to wade through before I get to the right one that has the answer. There’s what the OP tried, then someone else’s speculative code, then yet another one’s speculative code. It’s not obvious which one is the final solution so you have to read many posts before you find the right one. I hoped to avoid that. (part 1)

As far as your comment (1) about being ‘invested’, as you put it, or asking you to do my homework, I find that a bit offhanded. I think «after searching for hours and trying various things» clearly indicates my efforts and I shouldn’t have to be questioned further about that. As for (2), maybe… I’m not sure about (3) because I think someone either knows this or they don’t. Certainly people have encountered this problem and have learned to deal with it and can just give an answer without speculating. At least that’s what I was hoping for. (part 2)

It may shock you to learn this, but some people lie. If you hang around Super User long enough to earn some reputation, people may be more inclined to take you at your word.

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How can I extract a multi-part rar file in linux?

Question: I want to validate multipart compressed files like Zip because when any part missing for compressed files then it raises an error, but I want to validate it before extraction and different software creates a different naming structure. Here is a simple Windows command: Here is the batch file version ( instead of ): This will echo the filename and the «Volume Index» field from the output of the (list) command for that file like the example below: To rename all of the files in order, you can use the following batch file.

How can I extract a multi-part rar file in linux?

I need to extract a multi part rar archive in the form of archive_1.exe archive_2.rar archive_3.rar. How can I do this in linux?

I have tried unrar and 7z but unrar x archive_1.exe is extracting files with the reverse path format (eg. \file\dir) and 7z is unable to extract beyond the first file.

Any suggestions are welcome!

cd /path/to/your/download unrar x filename.of.part1.rar 

How to fix Win-RAR corrupt files, Hi guys! Welcome to another video. Today I have provided a very easy solution to a famous error «The archive is in unknown format or damaged» of Win Rar Arch

Extract Multi-part Files to One! Multi-part Compress

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How to extract multiple zip,rar or 7zip files

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How to get RAR files volume index using 7-zip commands (or any alternative solutions)

I have some files (207 RAR files downloaded), and they don’t have a proper name to refer each part/index, and if I want to extract them I should name them manually by checking volume index (from Properties section), and as you notice that can be too time-consuming

I thought that I might do it using a batch or a python script, so the first thing I needed to do is to get the volume index from CMD «7z i FileName.rar», but I couldn’t find the volume index number (or maybe HEX) from the output Am I using the wrong command? Or maybe it is printed but I can’t find it?

I attach one of RAR files output (Volume index is 18 [It is 19th part]) https://www.pastiebin.com/5cee55060c642

I would be thankful if you can help me.

You should be able to accomplish this with the l (list) command.

Here is a simple Windows command:

for %a in (*.rar) do @ECHO %a & "C:\Program Files\7-Zip\7z.exe" l %a | FIND "Index" & ECHO. 

Here is the batch file version ( %% instead of % ):

@ECHO off for %%a in (*.rar) do ECHO %%a & "C:\Program Files\7-Zip\7z.exe" l %%a | FIND "Index" & ECHO. 

This will echo the filename and the «Volume Index» field from the output of the l (list) command for that file like the example below:

myFile.part1.rar Volume Index = 0 myFile.part2.rar Volume Index = 1 myFile.part3.rar Volume Index = 2 

To rename all of the files in order, you can use the following batch file.

Warning This will affect all files in the directory it’s run in and the original file names will be lost.

@ECHO off SET PATH=%PATH%;C:\Program Files\7-Zip\ for %%a in (*) do CALL:Index "%%~a" GOTO:EOF :Index ECHO %~1 SETLOCAL EnableDelayedExpansion for /f "tokens=3 delims== " %%a in ('7z.exe l "%~1" ^| FIND "Index"') do CALL:Rename "%~1" %%a ENDLOCAL GOTO:EOF :Rename SET /A Index=%2+1 REN "%~1" file%Index%.rar 

How can I extract a multi-part rar file in linux?, Improve this question. I need to extract a multi part rar archive in the form of archive_1.exe archive_2.rar archive_3.rar. How can I do this in linux? I …

How to validate multi part compressed (i.e zip) files have all parts or not in C#?

I want to validate multipart compressed files like Zip because when any part missing for compressed files then it raises an error, but I want to validate it before extraction and different software creates a different naming structure.

I also refer one DotNetZip related questions.

The below screenshot is from 7z software.

And the second screenshot is from DotNetZip from C#.

One more thing is that I also want to test that it’s also corrupted or not like 7z software. Please refer below screenshot for my requirements.

Please help me with these issues.

I am not sure if you will be able to see the exact error as shown in your snapshot. But I have a code which may help you to find if the multipart file is readble.

I have used nuget Package CombinationStream.

The ZipArchive constructor throws ArgumentException or InvalidDataException if the stream is not readable.

public static bool IsZipValid() < try < string basePath = @"C:\multi-part-zip\"; Listfiles = new List < basePath + "somefile.zip.001", basePath + "somefile.zip.002", basePath + "somefile.zip.003", basePath + "somefile.zip.004", basePath + "somefile.zip.005", basePath + "somefile.zip.006", basePath + "somefile.zip.007", basePath + "somefile.zip.008" >; using (var zipFile = new ZipArchive(new CombinationStream(files.Select(x => new FileStream(x, FileMode.Open) as Stream).ToList()), ZipArchiveMode.Read)) < // Do whatever you want >> catch(InvalidDataException ex) < return false; >return true; > 

I am not sure if this is what you are looking for or you need more details in the error. But hope this helps you to come to solution of your issue.

From your comments I understood that the issue you have is to identify the files (get the list of parts belonging together). You can get a list of files like

List files = System.IO.Directory.EnumerateFiles(@"D:\Zip\ForExtract\multipart\", "500mbInputData.*", SearchOption.TopDirectoryOnly).OrderBy(x => x).ToList(); 

List files = System.IO.Directory.EnumerateFiles(@"D:\Zip\ForExtract\multipart\", "500mbInputData.zip.*", SearchOption.TopDirectoryOnly).OrderBy(x => x).ToList(); 

and then use the file list in your CombinationStream . The rest of the code would look like Manoj Choudhari wrote. You could also put the path and the file name with wild card into a parameter, so I’d suggest to add the following parameters to the function:

public static bool IsZipValid(string basePath, string fileNameWithWildcard) < try < Listfiles = System.IO.Directory.EnumerateFiles( basePath, fileNameWithWildcard, SearchOption.TopDirectoryOnly).OrderBy(x => x).ToList(); using (var zipFile = // . rest is as Manoj wrote 

if (IsZipValid(@"D:\Zip\ForExtract\multipart\", "500mbInputData.*")) < // . >

if (IsZipValid(@"D:\Zip\ForExtract\multipart\", "500mbInputData.zip.*")) < // . >

To find out which kind of files you have in the basepath, you could write a helper function like

List getZipFormat(string path) < bool filesFound(string basePath, string pattern) =>System.IO.Directory.EnumerateFiles( basePath, pattern, SearchOption.TopDirectoryOnly).Any(); var isTar = filesFound(path, "*.tar. "); var isZip = filesFound(path, "*.z??"); var is7Zip = filesFound(path, "*.7z. "); var result = new List(); if (isTar) result.Add("TAR"); if (isZip) result.Add("ZIP"); if (is7Zip) result.Add("7ZIP"); return result; > 

Modify it to your needs — it will return a list of strings containing «TAR», «ZIP» or «7ZIP» (or more than one of them), depending on the patterns matching against the files in the base directory.

Usage (example for multi-zipformat check):

 var isValid = true; var basePath = @"D:\Zip\ForExtract\multipart\"; foreach(var fmt in getZipFormat(basePath)) switch (fmt)

Note: As per my experiments with this, it could happen that the files remain open although your program has ended — meaning your files will still be locked the next time you run your code. So, I’d strongly suggest to explicitly close them, like

 var fStreams = files.Select(x => new FileStream(x, FileMode.Open) as System.IO.Stream).ToList(); using (var cStream = new CombinationStream(fStreams)) using (var zipFile = new ZipArchive(cStream, ZipArchiveMode.Read)) < // Do whatever you want. // . but ensure you close the files fStreams.Select(s =>< s.Close(); return s; >); >; 

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How to extract files from a split 7zip archive?

I downloaded a series of packages that all compress one thing. The files end in .0 , .1 , .2 , etc.

The User Lists of these packages I downloaded say to use the file.bat to successfully extract, or to use 7zip. I do not understand how to do that with 7zip.

Install 7zip from the Ubuntu Software Center.

Command-line solution:

In terminal, cd to the directory that contains the volumes and run 7z command on the first file from the sequence:

cd ~/Downloads 7z x asdf.iso.0 

7zip should notice that you have a multi-volume archive and unpack everything. If you want the directory structure flattened, use 7z e instead of 7z x .

GUI Solution

• Right-click the first package of those you downloaded (eg file.zip.001 ).
• Select «Extract Here».

Using 7zip to extract iso file is extremely simple even it has multiple parts. Let’s assume you have an iso archive with three parts named by a.iso.001 , a.iso.002 , a.iso.003 .

The standard command line is in this way:

Passing the -t flag with the «split» indication will specifically tell 7z that the iso has multi-parts, the preceding «iso» or «udf» will depend on your iso container format.

Install 7zip, if its not installed, using this command:

sudo apt install p7zip-full 

Now move to directory which contains split files and run this:

How to get RAR files volume index using 7-zip, This will echo the filename and the «Volume Index» field from the output of the l (list) command for that file like the example below: …

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