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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;








1















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?










share|improve this question




























    1















    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?










    share|improve this question
























      1












      1








      1








      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?










      share|improve this question














      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






      share|improve this question













      share|improve this question











      share|improve this question




      share|improve this question










      asked Mar 26 at 10:58









      CyrCyr

      1082 silver badges12 bronze badges




      1082 silver badges12 bronze badges






















          1 Answer
          1






          active

          oldest

          votes


















          -1














          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





          share|improve this answer






















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            1 Answer
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            active

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            active

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            votes






            active

            oldest

            votes









            -1














            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





            share|improve this answer



























              -1














              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





              share|improve this answer

























                -1












                -1








                -1







                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





                share|improve this answer













                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






                share|improve this answer












                share|improve this answer



                share|improve this answer










                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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