- 1 year ago
Extract 7-zip files into their containing directory from a batch file?
To extract 7-zip files into their containing directory from a batch file, you can use the following approach:
batch@echo off
rem Set the path to the 7-Zip executable
set "7zipPath=C:\Path\To\7z.exe"
rem Set the path to the directory containing the 7-Zip files
set "sourceDir=C:\Path\To\7zFiles"
rem Extract 7-Zip files into their containing directory
for /r "%sourceDir%" %%F in (*.7z) do (
echo Extracting "%%F"...
"%7zipPath%" x -o"%%~dpF" "%%F"
)
echo Extraction complete.
In the above script, make sure to set the 7zipPath
variable to the actual path of the 7-Zip executable (7z.exe
). Set the sourceDir
variable to the path of the directory containing the 7-Zip files you want to extract.
The script uses a loop to iterate through each 7-Zip file (*.7z
) in the specified directory and its subdirectories. It extracts each file into its containing directory using the 7-Zip command-line tool (7z.exe
) with the -o
option specifying the output directory.
After running the script, the 7-Zip files will be extracted into their respective containing directories.
Note: Make sure that the 7-Zip executable (7z.exe
) is installed and accessible from the specified 7zipPath
. Adjust the paths in the script according to your specific setup.