Command line 7Zip to extract file by specified result path

Question

Is it possible to make 7Zip extract one file by name from archive to specified full path (including new filename)?

I've found only:

7zip e <archive> -o<resultDirectory> <filemask>

But it's not what I need.

Answer 1

You can use the -so option to write the data to stdout and redirect to your target file:

Example:

7z e -so xxxx.7z > yyy.txt

The commandline help of version 9.22 beta:

7-Zip [64] 9.22 beta  Copyright (c) 1999-2011 Igor Pavlov  2011-04-18

Usage: 7z <command> [<switches>...] <archive_name> [<file_names>...]
       [<@listfiles...>]

<Commands>
  a: Add files to archive
  b: Benchmark
  d: Delete files from archive
  e: Extract files from archive (without using directory names)
  l: List contents of archive
  t: Test integrity of archive
  u: Update files to archive
  x: eXtract files with full paths
<Switches>
  -ai[r[-|0]]{@listfile|!wildcard}: Include archives
  -ax[r[-|0]]{@listfile|!wildcard}: eXclude archives
  -bd: Disable percentage indicator
  -i[r[-|0]]{@listfile|!wildcard}: Include filenames
  -m{Parameters}: set compression Method
  -o{Directory}: set Output directory
  -p{Password}: set Password
  -r[-|0]: Recurse subdirectories
  -scs{UTF-8 | WIN | DOS}: set charset for list files
  -sfx[{name}]: Create SFX archive
  -si[{name}]: read data from stdin
  -slt: show technical information for l (List) command
  -so: write data to stdout
  -ssc[-]: set sensitive case mode
  -ssw: compress shared files
  -t{Type}: Set type of archive
  -u[-][p#][q#][r#][x#][y#][z#][!newArchiveName]: Update options
  -v{Size}[b|k|m|g]: Create volumes
  -w[{path}]: assign Work directory. Empty path means a temporary directory
  -x[r[-|0]]]{@listfile|!wildcard}: eXclude filenames
  -y: assume Yes on all queries

More recent alpha versions (9.30) also do not allow to specify a target filename as direct parameter without redirection.