2 Replies Latest reply on Jul 3, 2008 5:54 AM by Jed Schneider

    specifying where the WindowedApplication opens

    Jed Schneider Level 1
      Hi,
      I am having trouble figuring out what properties will move the main application window to dock on the top and flush on left (0.0) to OS menu bar. Right now when I run the application from the Flex Builder in Eclipse the application opens about in the middle of my display with the upper left corner centered in the display. There seems to be no x,y property for the mxml tag windowedApplication. Here is my properties definition (obviously I have a close tag at the end omitted from this exert):


      quote:

      <mx:WindowedApplication
      xmlns:mx = " http://www.adobe.com/2006/mxml"
      xmlns:login = "com.ameritest.flex.login.*"
      xmlns:loginAS = "com.ameritest.actionscript.login.*"
      xmlns:handlers = "com.ameritest.actionscript.handlers.*"
      xmlns:video = "com.ameritest.actionscript.videocontrols.*"
      xmlns:skins = "com.ameritest.actionscript.skins.*"
      xmlns:dbTools = "com.ameritest.actionscript.databasetools.*"
      xmlns:window = "com.ameritest.actionscript.windows.*"
      xmlns:project = "com.ameritest.flex.project.*"
      xmlns:projectAS = "com.ameritest.actionscript.project.*"
      width = "1280"
      height = "768"
      layout = "absolute"
      creationComplete = "initApp()" >


      the initApp() function hides a menubar and project window until the user is authenticated. That would be a good place to resize or move the menu with actionscript if needed. What am I missing. Here is a screenshot of the runtime:
      http://img.skitch.com/20080702-chum1psbibxkhfdbtm3nftgiyy.jpg