Getting the file name without extension in a Windows Batch Script

I'm trying to create a right-click context menu command for compressing JavaScript files with YUI compressor. My ultimate goal is to try to get this to run on a context menu:

java.exe -jar yuicompressor-2.4.2.jar -o <filename>.min.js <filename>.js

I know I can use the variable %1 to reference the file name being opened. I can't figure out how to get this command into a batch file syntax and haven't been able to find any answers online.

Update:
Jeremy's answer (+comments) worked. For anyone who stumbles upon this, here is what I had to do:

In the action I created for the JavaScript file, I used this as the command:

minify.bat "%1"

Which calls my batch script, which looks like this:

java.exe -jar yuicompressor-2.4.2.jar -o "%~dpn1.min.js" %1

For the batch script, keep in mind that the code above assumes the directories for java.exe & yuicompressor are both added to your PATH variables. If you don't add these to your path, you'll have to use the full path for the files.

The sequence %~dpn is used to get:

1. %~d - The drive
2. %~p - The path
3. %~n - The file name

Solution 1:

Change the action to call a batch file:

RunCompressor.bat "%1"

Use %~n1 to get the filename without the extension in RunCompressor.bat:

start javaw.exe -jar yuicompressor-2.4.2.jar -o "%~n1.min.js" "%1"

Helpful article

start javaw.exe closes the command window when running the batch file.

Solution 2:

echo path of this file name is: %~dp0
echo file name of this file without extension is:%~n0
echo file extention of this file is:%~x0
echo The file name of this file is: %~nx0