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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.

1 Answer

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.

It's a variable.

That particular example uses the directory option of a FOR loop, iterating through the directories and assigning them to %%A.

That's also not a command-line example, but a batch file example. In batch files, you need to use %%A, while on the command-line, you'd just use %A.

In your scenario, the %%A is a placeholder for what the "for" loop is iterating over (which the /D indicates directories). So each iteration of the loop, %%A is one of the directories.

You'll see %% instead of % in batch code. You'll see % instead of %% used in your command prompt.

So know that if you copy over a batch file code into a command prompt and run it with %% being used, it will error, and vice versa.