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I have a draggable editor where the user selects an audio file drag it and drop it n that editor where a waveform is open. here is the code for that:

this.defineDropHandler = function defineDropHandler(dropContainerElement)
{
    // init event handlers
    dropContainerElement.addEventListener("dragenter", this.skipEventHandler, false);
    dropContainerElement.addEventListener("dragexit", this.skipEventHandler, false);
    dropContainerElement.addEventListener("dragover", this.skipEventHandler, false);
    dropContainerElement.addEventListener("drop", this.dropHandler, false);
    dropContainerElement.masterObj = this; // need to define this controller for further events
};

what I actually want is to have an open file button which opens a window where the user double click on an audio file and that file is directly dropped in the editor(not drag and drop required). the path of the folder to open is: C:\wamp\www\JSSoundRecorder\upload

souna
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  • When you double click in the open file dialog, that should provide a result (ie. the file selected). You should then be able to use that path, to open the file and display the result in the control you talk about. – askrich Aug 12 '14 at 13:36
  • how is that answer supposed to help??? – souna Aug 12 '14 at 13:36
  • Souna, without much more information, that is all we can provide. I don't see how my comment is 'unhelpful'.. – askrich Aug 12 '14 at 13:45
  • my question is simply how to get the path??i sorry but im new to javascript so I will be thankful if you help me with the coding – souna Aug 12 '14 at 13:53
  • See a similar question: http://stackoverflow.com/questions/18874729/get-the-file-name-after-click-open-button-in-file-browse-dialog-box-using-javasc – askrich Aug 12 '14 at 14:13

0 Answers0