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HELP: open file dialog with filter
https://www.bbs.xwidget.com/viewtopic.php?f=7&t=6655		 Page 1 of 1
Author:  yereverluvinuncleber [ April 20th, 2017, 6:14 am ]
Post subject:  HELP: open file dialog with filter
Question for you lot here re: dialogs.
When opening a file open dialog box, is there a method to add a file filter to the input of the dialog function?

I want to do something like this:

Code:
var validTypes = [".mpeg",".mp4",".m4a",".mpg",".flv","*.mp3",".aac",".wav",".wma",".aif",".aiff",".au",".snd"];

        fileName = openFileDialog(validTypes);

but it is unclear to me how a traditional filter works with the xwidget dialogs.

The standard way of doing it, if it was a Microsoft dialog box would be:
Code:
openFileDialog.filter= ".mpeg",".mp4",".m4a",".mpg",".flv","*.mp3",".aac",".wav",".wma",".aif",".aiff",".au",".snd"; 
but this does not seem to work.

Instead, I am using a filter function on the returned (output) value as follows:

Code:
var validTypes = ["mpeg","mp4","m4a","mpg","flv","mp3","aac","wav","wma","aif","aiff","au","snd"];

        fileName = openFileDialog("");

        if (validate_fileupload(fileName) === true ) {
           PlayFile(fileName);
        } else {
          alert("incompatible file type");
        }


function validate_fileupload(fileName)
{
    var allowed_extensions = validTypes;
    var file_extension = fileName.split('.').pop();
    for(var i = 0; i < allowed_extensions.length; i++)
    {
        if(allowed_extensions[i]==file_extension)
        {
            return true; // valid file extension
        }
    }
    return false;
}

This does the job but it is not the most elegant approach and so I ask: Is there a better method of adding an input filter to a file open dialog?

Documentation for xwidget is sparse as always and the inline documentation shows little as reference, so please help. It would be nice to know, then I could add it to a tutorial I am writing.
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