Extract a certain file from an archive with 7-Zip from the command line

As a follow-up to surfasb's answer, add a -r flag at the end to recurse:

7z e [archive.zip] -o[outputdir] [fileFilter] -r

Multiple filters support:

7z e [archive.zip] -o[outputdir] [fileFilter_1] [fileFilter_2] -r

Example:

Multiple filters command line:

7z e archive.zip -o outputdir *.xml *.dll -r

PS: I use 7za.exe instead of 7z.exe. This is the actual command I use in my script:

7za.exe x archive.zip -o outputdir *.xml *.pdb *.exe *.ocx *.dll -r

You just add the filename at the end.

7z e [archive.zip] -o [outputdir] [fileFilter]

If you look at the man page for 7z you will find that the following command can be used to extract a file from a 7z archive (though the usage of path is missing from the man page):

7z x <archive> <path to file>

Examples:

7z x backup.7z *.html
7z x backup.7z folderwithin/myfile.html

Alternatively you could use e.

The command line version users guide seems to have more information on the actual usage.

Note that 7z has the following syntax (observe the spaces and quotes surrounding the "-oMy Folder" option to set the output folder name, took me hours to figure out, as I originally did this – the wrong way: * -o "My Folder" *):

7z e "my zip.zip" "-oMy Folder" *.jpg "all of these.*" -r