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  1. #1
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    amuniz is on a distinguished road

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    Default Application "name" required checking

    Hello.

    I'm getting this error when creating an Ext.app.Application object using Ext.create:
    Code:
    Ext.app.Application.constructor(): [Ext.app.Application] Name property is required
    This error is shown only when using ext-dev.js or ext-dev-all.js (it's not thrown using ext-all.js or ext-all-debug.js).

    This is my code:

    Code:
    Ext.create('Ext.app.Application', {
        name: 'MyApp',
        appFolder: 'ui/app',
        launch: function() {
            ...
        }
    });
    The name attribute is actually set, so I think this is a bug in Ext.app.Application constructor (only in dev builds):

    Code:
    constructor: function(config) {
        var me = this;
    
        //<debug>
        if (Ext.isEmpty(me.name)) {
            Ext.Error.raise("[Ext.app.Application] Name property is required");
        }
        //</debug>
    
        me.callParent(arguments);
    
        me.doInit(me);
    
        me.initNamespace();
        me.initControllers();
        me.onBeforeLaunch();
          
        me.finishInitControllers();
    }
    The "debug" section is checking the name BEFORE the callParent, so "me" does not have the "config" object applied yet.

    Regards,
    Antonio.

  2. #2
    Sencha - Support Team
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    Default

    Thanks for the report. Can you please post a test case which reproduces the issue? You can use our Fiddle if you like.


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  3. #3
    Sencha User
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    Default

    Hello Gary,

    Sure.
    How can I include ext-dev.js in your Fiddle?

    Regards,
    Antonio.

  4. #4
    Sencha - Ext JS Dev Team nohuhu's Avatar
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      1  

    Default

    @amuniz You're not supposed to create an instance of Ext.app.Application; subclass it and instantiate the derived class instead:
    PHP Code:
    Ext.define('MyApp.Application', {
         
    extend'Ext.app.Application',
        
         
    name'MyApp'
        ...
    });

    Ext.create('MyApp.Application'); // Or just `new MyApp.Application()` 
    Regards,
    Alex.
    Regards,
    Alex.

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