What does the percent sign (% and %%) in a batch file argument mean?

I think I'm missing something because I can't seem to find what this means.

Example:

for /D %%A in (*) do "\7za.exe" u -t7z -m9=LZMA2 "%%A.7z" "%%A"

That line was supposed to use a command line version of 7zip to compress individual folders, but I'm stumped as to what %%A means in this context.

Solution 1:

The for command needs a placeholder so you can pass along variables for use later in the query, we are telling it use the placeholder %A, the reason the code you saw uses %%A is because inside a batch file (which I assume is where you found this) the % has a special meaning, so you must do it twice %% so it gets turned in to a single % to be passed to the for command

To actually break apart what the command is doing, there is two parts to the command:

 for /D %%A in (*) do .....

What this part says is for every folder in the current folder execute the following command replacing %%A with the name of the currently processing folder.

..... "\7za.exe" u -t7z -m9=LZMA2 "%%A.7z" "%%A"

What this part says is execute the command "\7za.exe" u -t7z -m9=LZMA2 "%%A.7z" "%%A" and replace the two %%A's with the current record we are processing.