![]() ![]() const path require ('path') const filename 'hello.html' path.parse (filename).name //> 'hello' path.parse (filename).ext //> '.html' path.parse (filename).base //> 'hello.html'. Can you exclude it from displayName property or add a new property for that filename. If you don't add these to your path, you'll have to use the full path for the files. In node.js, the name of the file without the extension can be obtained as follows. Overview The find command is a convenient utility when searching files in the Linux command line. These THUMB files are also known as JAlbum thumbnail files, which are generated by the JAlbum application.db is a hidden, protected operating system file. I need to get just file name without extension part. Which calls my batch script, which looks like this: java.exe -jar yuicompressor-2.4.2.jar -o "%~" %1įor the batch script, keep in mind that the code above assumes the directories for java.exe & yuicompressor are both added to your PATH variables. In the action I created for the JavaScript file, I used this as the command: minify.bat "%1" For anyone who stumbles upon this, here is what I had to do: 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. I know I can use the variable %1 to reference the file name being opened. ![]() Now I display the filename using f.Name (picture.tif) I want to display only the filename without the extension (picture). 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 without extension thread707-1530586 Forum Search FAQs Links MVPs wmbb (TechnicalUser) (OP) 13 Feb 09 03:15 I have a variable called f and defined as file. idw files in IV10 I want to have the filename property get on my drawings in a few places so I have made symbols that have text that use the property type of FILENAME. ![]() ![]()
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