EDIT Given: Using batch script.
You may wish to use a third party zip tool (highly recommend 7-ZIP command line version called 7z.exe) to accomplish this.
with 7z, the syntax is as follows:
7z <command> [<switch>...] <base_archive_name> [<arguments>...]
To extract the command would be:
7z e file.zip -y
the -y switch assume "Yes" answer to any questions that may come up during extraction such as overwrite requests.
So your command will read
CD "C:\Location\Of\ZipFiles"
FOR /F "USEBACKQ tokens=*" %%F IN (`DIR /b *.zip`) DO (7z e "%%F" -y)
If you want to output them into a different location, you can use the -o switch and specify the directory:
7z e "%%F" -y -oC:\Some\Other\Folder\
EDIT:
To perform the extract with full paths and specifying all ZIP archives only, use this:
7z x -tzip "C:\Location\of\zips\*"
Or even nutzier... all ZIP files on C: drive:
7z x -r -tzip "C:\*"
EDIT2:
Making it compatible with your output file, this:
dir /s /b *.zip > allzips.txt
for /F %%x in (allzips.txt) do (7z x -tzip "%%x")
Best Answer
This is my own take on the problem:
for %%F in (*.zip) do ( "C:\Program Files\7-Zip\7z.exe" x -y -o"%%F_tmp" "%%F" * & pushd %%F_tmp & "C:\Program Files\7-Zip\7z.exe" a -y -r -t7z ..\"%%~nF".7z * & popd & rmdir /s /q "%%F_tmp" )Save this to a
zip to 7z.batfile, place it into the directory with all the zip files you want to convert and double-click it there.Thanks to Clint Priest for the base code.