i have one problem. In my project i have one NEXT button. I when i click the next button i need to save all the records of page and also go to the next page at the same instance. Similar in next page also same process repeated. I have used single view and used the pagination.
I am little bit confused how to proceed. I have different Action result for Save and Next page.
SO any good way to move forward
Since you have 2 action method which is save and next, why dont you call the Save action when the "NEXT" button is clicked, and then in your save action, just return the call to the Next action?
Hope I understand you right :)
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I have used Fresh MVMM. I need to pass a data from one page model to another page model without using init and reverse init.
Eg: I have one toggle button in first content page.
I need to toggled the button based on the selection in second popup page without popup the second PopPopupPageModel.
ReverseInit is worked when popup the second PopPopupPageModel and can easily pass data.
But I need to toogled the button in first page without popup the second page.
Please help...
I have a value of a field on page 1 that I need to pass to page 2 (modal dialog).
The field is changed on page 1. and I can see it's got updated on the page and when I print its value in debug messages.
When I'm clicking the button, with the action "Redirect to Page in application", the value on page 2 is the initial value of the field, and not the updated one, is this a bug?
what can I do?
I don't want to use dynamic action or things like "prepare_url" because of many reasons.
Thanks.
That is not how a link works. What you are seeing is not a bug, it is expected behaviour. The link is rendered when the page is rendered and will have the item values at render time and a calculated checksum (depends on settings) based on those values. It will not pick up any changes.
The "traditional" way to do this is the following. Assumptions: Need to navigate from page 1 to page 2 and set P2_ITEM on page 2 to value of P1_ITEM on page 1.
Create button (let's say button name is MYBUTTON) of type "Submit Page"
Under "Processing", create a branch to page 2.
For "Page or URL" pick page 2 and set P2_ITEM to value P1_ITEM.
For "Process point" pick "After Submit"
Under "Server Side Condition", select "MYBUTTON" under "When Button Pressed".
Since this is a branch your page has been submitted and all the changes you made to P1_ITEM will be picked up. The fact that page 2 is modal does not affect this, APEX will take care of that for you.
As an alternative to Koen Lostries excellent explanation, in some cases it is also possible to have the target page load the item from the source page.
Ensure that the item is being pushed to the session, for example by using a dynamic action with the items to submit option, but do nothing else in this dynamic action.
Then, when page 2 opens, simply add a pre-header-process, that makes something like :P2_ITEM := :P1_ITEM;
In my experience this approach is easier than doing a redirect in a dynamic action, as there is no declarative option for that and you often need hand-written javascript to do that.
So, I am using ui-router for my app's routing. There is a scenario, I clicked like button on the item detail page. Then I hit back button, it goes back to the main page, it is a list of items. The item has a likes counter. But I found the counter did not increase because back button did not fire the controller for main page.
I want to understand how can I fire the controller while clicking back button.
You should isolate your data in a provider to manage the data sync between your views. If all of the data is contained in a service, you can update it from one view, and it will always be synced with the other.
In both of the controllers, require 'ItemRepository' which will provide the data to the views.
I have one issue in my program, i have three pages in my program, Main page, first page and second page. I'm able to navigate from the main page to the first page and from the first page to the second page. I have a stack panel in the first page. It consists of children which are dynamically created from the db. This page contains a OnNavigateTo method, for bringing the elements from the main page and displaying it, And i wrote my codings in this method to add dynamically children to the stackpanel. when navigating back from the second page to the first page, the compiler will goes to the OnNavigateTo method and regenerate the children of the stackpanel once again.
My problem is:
i want to show the previously displayed first page with all the data, when the back button is pressed. I doesn't want the page to regenerate again. I want the static page which i have previously when i go back from the second page to the first page.
How to do it? please help me with some piece of code... Thank You..
Do the logic in the constructor, because it doesn't get called when navigating back (unless the app has been tombstoned). OnNavigatedTo gets called every time no matter how you navigate to the page.
Read this for more info:
WP7 Development Tip of the Day: Page Startup: The Constructor
2 days of reading every posting i could get my hands on with no luck. Here is what I am trying to do:
I have a standard split view controller that loads a new view controller on the Right for each item you choose on the Left. Standard easy stuff. Now the "impossible part". I have added a big button in the middle of each detail view controller. I want to click the button and have it automatically advance to the next view (simulating someone actually clicking the next item in the root view in the left).
A bonus would be to actually have the left root view side also highlight the next row in the list when the button is hit.
I'd pay fifty bucks for working sample code "not much I know".
Thanks.
I think you should probably read the ViewController Programming tutorial...
http://developer.apple.com/library/ios/#featuredarticles/ViewControllerPGforiPhoneOS/NavigationControllers/NavigationControllers.html%23//apple_ref/doc/uid/TP40007457-CH103-SW1