1. #1
    Sencha User
    Join Date
    Dec 2013
    Posts
    84
    Vote Rating
    48
    Malte123 is infamous around these parts Malte123 is infamous around these parts

      1  

    Default Cmd generated Base App - Ext.application() cause error on "config.name"

    Cmd generated Base App - Ext.application() cause error on "config.name"


    Ext version tested:
    • Ext 4.2.1
    Browser versions tested against:
    • Safari 4
    Description:
    • Using Ext.application() causes in
      TypeError: 'undefined' is not an object (evaluating 'config.name')
    Steps to reproduce the problem:
    • Generate an App with Sencha CMD
    • Execute Ext.application(); within a Class-Constructor
    • Instantiate and take a look on Console
    The result that was expected:
    • Get Application Name to go forward with Ext.application().getController('controllername') for example
    The result that occurs instead:
    • TypeError: 'undefined' is not an object (evaluating 'config.name')

  2. #2
    Sencha Premium Member Fredric Berling's Avatar
    Join Date
    Sep 2007
    Location
    Sweden
    Posts
    186
    Vote Rating
    15
    Fredric Berling has a spectacular aura about Fredric Berling has a spectacular aura about

      2  

    Default


    Not sure if you are supposed to use it that way.

    To access a controller from a controller you would simply use this.getController('ControllerName')

    To get a controller instance from the window scope you can do: MyApp.app.getController('ControllerName').
    To get your application instance from the controller scope you can do: MyApp.app.getApplication()
    To get your application instance from the window scope you can do: MyApp.app

    "MyApp" would be the top namespace/application name in your app.

    The Ext.application() function is only used in app.js in the application folder.

  3. #3
    Sencha User
    Join Date
    Dec 2013
    Posts
    84
    Vote Rating
    48
    Malte123 is infamous around these parts Malte123 is infamous around these parts

      0  

    Default


    first: thanks for thinking about !

    The Ext.application() function is only used in app.js in the application folder.
    I am developing an UX and I am needed to get access to Application-Events independent from what the Programmer wants to do with my Fassade-Class. So it should be really easy to use.
    like this.
    PHP Code:
    var service Ext.widget('AliasNameOfClass',{
    param1'value'
    }).on('readyLoadingAndDecryptingTimeIntenciveRemoteData',function(){
     
    console.log(service);
    }) 
    I cannot find where I have read about to use Ext.application() to go forward with getController() instead of the known Name of Application like "MyApp".

    I have seen the Problem maybe causes of the Structuring of Code Sencha CMD builds.
    I have not checked whether I can get what I want when I use a self made initial Codestructur.
    (Because it should be unimportant)

    So the Question is:

    How can I get Application-Name out of UX ?

Thread Participants: 1

Tags for this Thread

Turkiyenin en sevilen filmlerinin yer aldigi xnxx internet sitemiz olan ve porn sex tarzi bir site olan mobil porno izle sitemiz gercekten dillere destan bir durumda herkesin sevdigi bir site olarak tarihe gececege benziyor. Sitenin en belirgin ozelliklerinden birisi de Turkiyede gercekten kaliteli ve muntazam, duzenli porno izle siteleri olmamasidir. Bu yuzden iste. Ayrica en net goruntu kalitesine sahip adresinde yayinlanmaktadir. Mesela diğer sitelerimizden bahsedecek olursak, en iyi hd porno video arşivine sahip bir siteyiz. "The Best anal porn videos and slut anus, big asses movies set..." hd porno faketaxi