I'm writing a simple Sencha Touch 2 Application with the new MVC Pattern of the Framework - so, there are models, stores, controllers and views and all is tied together with an Application.
Now, there is the possability to use "this.getApplication()" to get the instance of this super uber-class "Application".
And now there is the problem - getApplication() returns only the app-instance, if i'm in a controller - in a view, a model or a store, it is returning "undefined".
Basically, i can understand the idea behind this behavior - your business has to be in controllers, nowhere else.. and so you don't have to know the app-instance in stuff like stores, views or models...
Okey but.. it would be extremely nice to have global-properties living in the main-app.
For example, i want to define my webservice-url globally in the Application and use that variable everywhere i need to - and unfortunately, i need this url in a store too.
Now, the only way i see to access this global variable in a store is the way through my Namespace. Instead of using "this.getApplication().serviceUrl" i found only the solution through the namespace with "NameOfMyApp.app.serviceUrl" - and that could not be the best way to solve this.
Any ideas about that problematic? Is there a better, always working way to get the app instance from everywhere? Or, where should i store global variables if not in the Application?
You can access the application instance on the AppName.app property in the latest Sencha Touch 2 betas.
Ext.application({
name: 'Sencha',
launch: function() {
// Logs the application instance
console.log(Sencha.app);
}
});
looks like Ext.app.Application.appInstance is app name agnostic solution
I work with ExtJS 4.1.1a and Ext.app.Application.instance seems to be the only solution. From a controller you can use this.application.
Related
My application has deep navigational chains of 8+ screens deep (Wizard-style). When i'm editing a View Controller, and i want to make a quick visual change and re-test, As a tester, I have to go through the full flow to end up where I was before recompiling, in the same Data state. Is there a way that Xcode can somehow preserve the application state, and re-run the same view controller?
If there was an automated way to detect the last launched screen, and re-display it after the recompile, that would save a lot of developers a lot of time.
I realize something like this can be built custom. Something like:
IF A Certain Debug Flag is ON:
- Retrieve from NSUserDefaults the Class Name of the last controller used, and redisplay it.
The problem with this is: Data State and Navigation State will not be preserved. Also, all other object state which invokes and depends on your controller will not be included. That's why I need a more universal solution.
IF A Certain Debug Flag is ON: - Retrieve from NSUserDefaults the Class Name of the last controller used, and redisplay it.
The problem with this is: Data State and Navigation State will not be preserved.
But this is exactly the problem that the built-in state saving and restoration mechanism is intended to solve, is it not?
https://developer.apple.com/library/ios/featuredarticles/ViewControllerPGforiPhoneOS/PreservingandRestoringState.html
I recommend you use a User Interface target and then create an XCTestCase that works on your view controller and exercise it. Use the new record button in Xcode 7 to get you started with the test case. Familiarize yourself with some of the new queries it uses for some of the dynamic information you may be creating.
Another option would be to inject your test case's UI state into the application delegate in the setup function of an XCTestCase. Then it will need custom logic to navigate to the correct view controller. Then your test case could focus on just that ViewController once the Application delegate has navigated to it.
Of course you can use the debugger and breakpoints in the Test Case and your View Controller while UI Testing. By "scripting" this you do significantly less manual clicking on a device to test a ViewCOntroller
Also you've now got a Test! Hopefully worth something to you down the road when you change something
I have started to learn about Yii framework about day ago and I came up with the problem.
I have downloaded basic application template (so that you know the structure of my application).
I would like to use one of my classes's function from "models/" in the "views/layouts/main.php", however, I am not sure how I can access them from "main.php". I have searched around the internet and none of the solutions have helped me.
I've read about widget creation, so I can use it in "main.php" in this link - Yii - how to retrieve model data into a layout page?, but this does not provide my version of Yii as I coulnt not find protected folder and etc. So I was kind of confused.
Also read some other solution, but they did not help me as well.
What would you suggest me to do? Because I am currently have no clue about the solution. I am a newbiew in this framework, so, don't judge me hard :)
UPDATED: [SOLUTION]
// use statement on top of main.php like the other use statements
use app\models\Modelname;
// anywhere in the file
....
$myModel = new Modelname;
$myModel->myFunction();
...
// Or if it is a static function:
Modelname::myFunction();
This is how to access model files (classes) from folder models/ from for example views/layouts/main.php.
Here is the deal - view is actually only a "view". View should only represent some data, but not do something with this data. You don't want to have any logic in the view.
Anyway, if you want to show something from your model - you should pass this model through controller, which renders your view.
Controller will look like:
$model = new Device;
$this->render('index', array(
'model' => $model
));
Then you can use it in the view, normally to get some data from the model.
You could also get it directly in the view, but this is not good practice.
Here can you read some basics about models and best practices how to use them.
Update [Yii 1.x]
To use Model-Class specifically in your main.php, import it at the place you want to use it:
Yii::import('application.models.LoginForm');
After that you can normally use functions from your Model Class.
Update 2
Namespaces in Yii2:
use yii\models\LoginForm;
at the start of your main.php layout-File. Look also at the "Layouts" part here.
Good afternoon,
I don't know if what I want to do is possible, so here goes.
I have a module that extends Page_Controller, but I want certain functions to be accessible via the site root.
Eg: getMyDataObjectList();
Currently, they only work if I go through the normal MVC routing structure.
I've found that when I place the function 'getMyDataObjectList' within '/mysite/code/Page.php' it works.The problem is, I don't want to place the code in there. I want it bundled with my Custom Module, but to work the same as though it was in 'mysite/code/Page.php'
[Example Scenario]
site root: http://[somesite].com
By default, the 'Page.ss' template loads.
I would like the theme developer to be able to call my module functions (API) within any template/Layout page, and have the result returned from the site root
Currently, this only works if I move the "API" functions to '/mysite/code/Page.php'
If the code is in my module, then data is only returned when you go to:
http://[somesite].com/[module_controller]
Can this be achieved? If so, how?
Thanks for your assistance.
[Update - Code Solution]
///>MyExtension.php
class MyExtension extends Extension{
public function getMyDataObjectList(){
return 'object list goes here!';
}
}//class
///>[Module] => _config.php
Object::add_extension('Page_Controller', 'MyExtension');
And as always, I do a (/dev/build?flush=1) just in case.
Thanks to: 'simon_w'
Yes, this is relatively straightforward. Simply, create an Extension subclass with your methods in them and then add that to Page_Controller. An Extension subclass is almost exactly the same as a DataExtension, they're just for classes other than DataObjects.
I've finished my first app using sencha touch 2.2.1. Now I uploaded it onto my server and tried to access it with my phone. Everything works well. My Dashboard contains 6 buttons, but only 1 of them is working. Each other throws the following error
TypeError: 'undefined' is not an object (evaluating 'name.substring')
The error occurs in the function parseNamespace. But I don't know what is wrong. I build the app using Sencha Architect and in the preview everything was fine. The testing package was created using the build-button from architect. If anyone could help me, the app is located here: app.ttv-rees-groin.de
Many thanks
This may be issue with class loading. The classes which are referred in the event of button events may not be loaded at the time.
Those classes may be missed when packaging application.
My experience found that Architect's build and package tools created a bloated mess of unnecessary files far exceeding what was required. Technical details: Architect 2 - all builds, Sencha Touch 2.0-2.2.x including all versions in between, Sencha Cmd 3.x
The cleanest and leanest build technique for developing in Architect was to save then fire the build using Sencha Cmd.
sencha app build
This performs the default "production" build.
The difference in output in this case went from a 32MB dump of files in the production folder with all resources, library, extensions etc, to the minimum required files totalling 0.8MB, and no longer requiring the touch library as only the classes needed were compiled into the app.
As for the error at hand, this error has something to do with class namespace, alias and xtype.
(Quick thanks to http://ruidevnotes.wordpress.com/2013/07/25/sencha-ext-js-4-common-typeerror/, saving me quite a lot of typing for these 4 things to check).
Possible solutions:
If class has controller, make sure the controller’s views config match the namespace specified on the class view’s Ext.define. Example: (controller)
views : ['namespace.of.my.View']
When using class on other view as xtype, make sure view’s alias is
widget.[customXtype]
so when adding it as an item to other viems, use
xtype : [customXtype]
Make sure view’s controller is added on app.js controllers.
When class view has no controller and you wanted to use it on other views, make sure to add the namespace of that view on
Ext.require(['class.view.namespace.name']);
and specify the xtype config instead of alias.
On top of these points, I recall an issue with list plugins, that I believe behaves identical to the error you are encountering. Prebuild - would work. Post build, issues and errors. The way I was able to get around this error was via this technique:
requires: [
'Ext.XTemplate',
'Ext.plugin.ListPaging'
],
config: {
..., // other standard configs removed for brevity
plugins: [
{
xclass: 'Ext.plugin.ListPaging',
autoPaging: true,
type: 'listpaging'
}
]
}
The thing to note is the exaggerated plugins declaration. Without this comprehensive declaration, the ListPaging plugin caused all manner of pain and chaos, and solely after a build.
EDIT: spelling.
In .NET I just do something like DataForm.Source = Object and then magic happens. Platform routes data changes from ui fileds to object properties, does validation and so on. Can I do something similar with Cocoa Touch and CoreData objects?
The closest thing in Cocoa is 'Key-Value Observing'. In the desktop Cocoa framework you can use bindings to hook user interface elements up to underlying objects so that changes in the objects or UI elements are reflected in the other.
Whilst Cocoa on iOS doesn't have this sort of UI bindings, you can still use 'Key-Value Observing' to synchronise changes in the data model with UI elements as described here:
http://developer.apple.com/library/iOS/#documentation/General/Conceptual/Devpedia-CocoaApp/KVO.html
I wrote a little open-source library that provides some simple data-binding functionality. It's basically just a wrapper around key-value observing (KVO).
http://github.com/kristopherjohnson/KJSimpleBinding
There are a few other similar libraries on GitHub:
http://github.com/dewind/KeyPathBindings
http://github.com/jonsterling/Observe
http://github.com/mruegenberg/objc-simple-bindings
http://github.com/zeasy/EasyBinding
Probably should also mention Github's Reactive Cocoa, a framework for composing and transforming sequences of values, an objective-C version of .NET's Reactive Extensions (Rx).
Binding mechanics can be done really simple (from the sample):
// RACObserve(self, username) creates a new RACSignal that sends a new value
// whenever the username changes. -subscribeNext: will execute the block
// whenever the signal sends a value.
[RACObserve(self, username) subscribeNext:^(NSString *newName) {
NSLog(#"%#", newName);
}];
Don't forget NSFetchedResultsController.
Not a full blown data bound controller, but makes table views a lot easier to use with Core Data.
If you're using Swift, check out Bond framework: https://github.com/ReactiveKit/Bond
Binding is as simple as:
textField.reactive.text.bind(to: label.reactive.text)
It plays well with functional:
textField.reactive.text
.map { "Hi " + $0 }
.bind(to: label.reactive.text)
And provides simple observations:
textField.reactive.text
.observeNext { text in
print(text)
}
STV (http://sensiblecocoa.com) is a framework that can do that within tableviews
I use CoreDataTableViewController from the Stanford University for my TableViewControllers. It hides a lot of details that you would normally implement in your TableViewController.
Googling for CoreDataTableViewController.h and .m will help you on the road. There are versions from several courses available, the latest does ARC, the old ones don't.
For syncing labels and edit fields with an NSManagedObject, I am still looking for a good solution.
Yes, there is a data binding framework that integrates well into Interface Builder and requires only minimal code overhead (if at all).
Take a look at https://github.com/mutech/aka-ios-beacon
EDIT:
You can for example bind a table view to a fetched results controller simply by setting the data source binding property of the table view in interface builder to:
[ yourResultsController ] { defaultCellMapping: "YourCellId" }
And the only thing you have to do is to define a property yourResultsController in your view controller.
The wiki provides a rather complete documentation and a lot of example use cases.