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7z list only filenames
Create folder with batch but only if it doesn't already existSpaces in filenames using forfiles and 7zipRename files to a list with filenames7zip CLI: specify target filename on extractHow to zip list of subfolders from text file and include parent folder name in filename?Cannot create Output directory error: 7zipIssue in Command line argument/String to unzip .7z files when filename contains spacePowershell 7-zip console output trace log modificationNot able to find .7z archives across multiple directories
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;
I'm using 7z version 18.05 and I would like to list only filenames of an archive content.
If I use the command 7z l myArchive.7z i get this output:
7-Zip 18.05 (x64) : Copyright (c) 1999-2018 Igor Pavlov : 2018-04-30
Scanning the drive for archives:
1 file, 146863932 bytes (141 MiB)
Listing archive: myArchive.7z
--
Path = myArchive.7z
Type = 7z
Physical Size = 146863932
Headers Size = 393
Method = LZMA:26
Solid = +
Blocks = 1
Date Time Attr Size Compressed Name
------------------- ----- ------------ ------------ ------------------------
2017-12-06 08:55:47 D...A 0 0 myArchive
2017-12-06 08:55:42 D...A 0 0 myArchivefolder
2017-12-05 19:50:41 ....A 21816530 146863539 myArchivefolderTest.dat
2017-12-06 08:55:42 ....A 21877463 myArchivefolderTest2.dat
2017-12-05 19:51:05 ....A 153953 myArchivefolderTest3.dat
2017-12-05 19:50:41 ....A 4193 myArchivefolderTest4.dat
2017-12-06 08:55:47 ....A 24128956 myArchivelog.txt
2017-12-06 08:55:47 ....A 79980 myArchivereadme.txt
2017-12-05 19:51:05 ....A 3256759999 myArchivefolderzTest.txt
------------------- ----- ------------ ------------ ------------------------
2017-12-06 08:55:47 3324821074 146863539 7 files, 2 folders
I don't know why 7z doesn't have a switch to list only filename. How to get only "Name" column? Any suggest with a dos command?
cmd 7zip
add a comment |
I'm using 7z version 18.05 and I would like to list only filenames of an archive content.
If I use the command 7z l myArchive.7z i get this output:
7-Zip 18.05 (x64) : Copyright (c) 1999-2018 Igor Pavlov : 2018-04-30
Scanning the drive for archives:
1 file, 146863932 bytes (141 MiB)
Listing archive: myArchive.7z
--
Path = myArchive.7z
Type = 7z
Physical Size = 146863932
Headers Size = 393
Method = LZMA:26
Solid = +
Blocks = 1
Date Time Attr Size Compressed Name
------------------- ----- ------------ ------------ ------------------------
2017-12-06 08:55:47 D...A 0 0 myArchive
2017-12-06 08:55:42 D...A 0 0 myArchivefolder
2017-12-05 19:50:41 ....A 21816530 146863539 myArchivefolderTest.dat
2017-12-06 08:55:42 ....A 21877463 myArchivefolderTest2.dat
2017-12-05 19:51:05 ....A 153953 myArchivefolderTest3.dat
2017-12-05 19:50:41 ....A 4193 myArchivefolderTest4.dat
2017-12-06 08:55:47 ....A 24128956 myArchivelog.txt
2017-12-06 08:55:47 ....A 79980 myArchivereadme.txt
2017-12-05 19:51:05 ....A 3256759999 myArchivefolderzTest.txt
------------------- ----- ------------ ------------ ------------------------
2017-12-06 08:55:47 3324821074 146863539 7 files, 2 folders
I don't know why 7z doesn't have a switch to list only filename. How to get only "Name" column? Any suggest with a dos command?
cmd 7zip
add a comment |
I'm using 7z version 18.05 and I would like to list only filenames of an archive content.
If I use the command 7z l myArchive.7z i get this output:
7-Zip 18.05 (x64) : Copyright (c) 1999-2018 Igor Pavlov : 2018-04-30
Scanning the drive for archives:
1 file, 146863932 bytes (141 MiB)
Listing archive: myArchive.7z
--
Path = myArchive.7z
Type = 7z
Physical Size = 146863932
Headers Size = 393
Method = LZMA:26
Solid = +
Blocks = 1
Date Time Attr Size Compressed Name
------------------- ----- ------------ ------------ ------------------------
2017-12-06 08:55:47 D...A 0 0 myArchive
2017-12-06 08:55:42 D...A 0 0 myArchivefolder
2017-12-05 19:50:41 ....A 21816530 146863539 myArchivefolderTest.dat
2017-12-06 08:55:42 ....A 21877463 myArchivefolderTest2.dat
2017-12-05 19:51:05 ....A 153953 myArchivefolderTest3.dat
2017-12-05 19:50:41 ....A 4193 myArchivefolderTest4.dat
2017-12-06 08:55:47 ....A 24128956 myArchivelog.txt
2017-12-06 08:55:47 ....A 79980 myArchivereadme.txt
2017-12-05 19:51:05 ....A 3256759999 myArchivefolderzTest.txt
------------------- ----- ------------ ------------ ------------------------
2017-12-06 08:55:47 3324821074 146863539 7 files, 2 folders
I don't know why 7z doesn't have a switch to list only filename. How to get only "Name" column? Any suggest with a dos command?
cmd 7zip
I'm using 7z version 18.05 and I would like to list only filenames of an archive content.
If I use the command 7z l myArchive.7z i get this output:
7-Zip 18.05 (x64) : Copyright (c) 1999-2018 Igor Pavlov : 2018-04-30
Scanning the drive for archives:
1 file, 146863932 bytes (141 MiB)
Listing archive: myArchive.7z
--
Path = myArchive.7z
Type = 7z
Physical Size = 146863932
Headers Size = 393
Method = LZMA:26
Solid = +
Blocks = 1
Date Time Attr Size Compressed Name
------------------- ----- ------------ ------------ ------------------------
2017-12-06 08:55:47 D...A 0 0 myArchive
2017-12-06 08:55:42 D...A 0 0 myArchivefolder
2017-12-05 19:50:41 ....A 21816530 146863539 myArchivefolderTest.dat
2017-12-06 08:55:42 ....A 21877463 myArchivefolderTest2.dat
2017-12-05 19:51:05 ....A 153953 myArchivefolderTest3.dat
2017-12-05 19:50:41 ....A 4193 myArchivefolderTest4.dat
2017-12-06 08:55:47 ....A 24128956 myArchivelog.txt
2017-12-06 08:55:47 ....A 79980 myArchivereadme.txt
2017-12-05 19:51:05 ....A 3256759999 myArchivefolderzTest.txt
------------------- ----- ------------ ------------ ------------------------
2017-12-06 08:55:47 3324821074 146863539 7 files, 2 folders
I don't know why 7z doesn't have a switch to list only filename. How to get only "Name" column? Any suggest with a dos command?
cmd 7zip
cmd 7zip
asked Mar 26 at 10:58
CyrCyr
1082 silver badges12 bronze badges
1082 silver badges12 bronze badges
add a comment |
add a comment |
1 Answer
1
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votes
If you can install a PowerShell module on your machine, listing the file names is easy enough. This can be done on any modern-day, supported Windows system.
https://www.powershellgallery.com/packages/7Zip4Powershell/1.9.0 describes how to install the module.
Here is a .bat file script showing its usage and output.
C:>TYPE zipfnlist.bat
@ECHO OFF
SET "ZIP_FILENAME=.7zIntf20.zip"
powershell -NoLogo -NoProfile -Command (Get-7Zip -ArchiveFileName "%ZIP_FILENAME%").FileName
C:>CALL zipfnlist.bat
bin
Properties
Ole32.cs
Program.cs
SevenZipFormat.cs
SevenZipInterface.cs
SevenZip.csproj
SevenZip.sln
binDebug
binRelease
binReleaseSevenZip.exe
PropertiesAssemblyInfo.cs
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
If you can install a PowerShell module on your machine, listing the file names is easy enough. This can be done on any modern-day, supported Windows system.
https://www.powershellgallery.com/packages/7Zip4Powershell/1.9.0 describes how to install the module.
Here is a .bat file script showing its usage and output.
C:>TYPE zipfnlist.bat
@ECHO OFF
SET "ZIP_FILENAME=.7zIntf20.zip"
powershell -NoLogo -NoProfile -Command (Get-7Zip -ArchiveFileName "%ZIP_FILENAME%").FileName
C:>CALL zipfnlist.bat
bin
Properties
Ole32.cs
Program.cs
SevenZipFormat.cs
SevenZipInterface.cs
SevenZip.csproj
SevenZip.sln
binDebug
binRelease
binReleaseSevenZip.exe
PropertiesAssemblyInfo.cs
add a comment |
If you can install a PowerShell module on your machine, listing the file names is easy enough. This can be done on any modern-day, supported Windows system.
https://www.powershellgallery.com/packages/7Zip4Powershell/1.9.0 describes how to install the module.
Here is a .bat file script showing its usage and output.
C:>TYPE zipfnlist.bat
@ECHO OFF
SET "ZIP_FILENAME=.7zIntf20.zip"
powershell -NoLogo -NoProfile -Command (Get-7Zip -ArchiveFileName "%ZIP_FILENAME%").FileName
C:>CALL zipfnlist.bat
bin
Properties
Ole32.cs
Program.cs
SevenZipFormat.cs
SevenZipInterface.cs
SevenZip.csproj
SevenZip.sln
binDebug
binRelease
binReleaseSevenZip.exe
PropertiesAssemblyInfo.cs
add a comment |
If you can install a PowerShell module on your machine, listing the file names is easy enough. This can be done on any modern-day, supported Windows system.
https://www.powershellgallery.com/packages/7Zip4Powershell/1.9.0 describes how to install the module.
Here is a .bat file script showing its usage and output.
C:>TYPE zipfnlist.bat
@ECHO OFF
SET "ZIP_FILENAME=.7zIntf20.zip"
powershell -NoLogo -NoProfile -Command (Get-7Zip -ArchiveFileName "%ZIP_FILENAME%").FileName
C:>CALL zipfnlist.bat
bin
Properties
Ole32.cs
Program.cs
SevenZipFormat.cs
SevenZipInterface.cs
SevenZip.csproj
SevenZip.sln
binDebug
binRelease
binReleaseSevenZip.exe
PropertiesAssemblyInfo.cs
If you can install a PowerShell module on your machine, listing the file names is easy enough. This can be done on any modern-day, supported Windows system.
https://www.powershellgallery.com/packages/7Zip4Powershell/1.9.0 describes how to install the module.
Here is a .bat file script showing its usage and output.
C:>TYPE zipfnlist.bat
@ECHO OFF
SET "ZIP_FILENAME=.7zIntf20.zip"
powershell -NoLogo -NoProfile -Command (Get-7Zip -ArchiveFileName "%ZIP_FILENAME%").FileName
C:>CALL zipfnlist.bat
bin
Properties
Ole32.cs
Program.cs
SevenZipFormat.cs
SevenZipInterface.cs
SevenZip.csproj
SevenZip.sln
binDebug
binRelease
binReleaseSevenZip.exe
PropertiesAssemblyInfo.cs
answered Mar 28 at 0:24
litlit
6,6025 gold badges35 silver badges57 bronze badges
6,6025 gold badges35 silver badges57 bronze badges
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add a comment |
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