I use filepicker.io to handle upload. But I want the user to provide additional data related to the uploaded file, so basically I would need to add to the modal some comboboxes and text inputs. Is it possible by using the widget approach with the filepicker.constructWidget call?
I'm afraid it's not possible, constructWidget only accepts addition options listed here https://www.filepicker.com/documentation/file_ingestion/widgets/programming?v=v1 and a type which can be 'filepicker' or 'filepicker-dragdrop'. You would need to try a different approach i.e. add the text inputs after the modal is called and files uploaded.
Related
Is there any way to add custom javascript?
I want to add custom javascript to some of the pages but it seems i am not able to.
Each default Backpack operation has its own CSS and JS file, in:
public/vendor/backpack/crud/css
public/vendor/backpack/crud/js
If you don't find one there, you can create one, and Backpack will pick it up in that operation's view (e.g. create.css or list.js).
Read more: https://backpackforlaravel.com/docs/3.5/crud-how-to#customize-css-and-js-for-default-crud-operations
experts,
These are two very basic connected questions.
I'm studying SAPUI5 and I cannot find means to position my UI elements on the screen.
In my view.js file I create, let's say, a button, a datepicker and a text field.
If I do something like:
aControls=[];
<Define the button - oButton>
aControls.push(oButton);
<Define the datepicker - oDatePicker>
aControls.push(oDatePicker);
<Define the text field - oText>
aControls.push(oText);
return aControls;
then I get all three elements positioned in a row one right after another.
I cannot use css, because I pass all those objects in one array and all of them are placed into a common div on index.html.
How do I position these? A link to any good tutorial/examples is very welcome.
Also, how do I refresh UI elements?
For example, I have situation, when on button press I make a call to the server, get response and put it into a using something like:
response.placeAt('some-id');
The button is created in the view.js and the call is processed in controller.js.
The response is added every time I press the button and I have no idea how to replace the old response with the new one.
A good link is very welcome.
Thanks.
There are a lot of layout controls of SAPUI5 you can use: Grid, HorizontalLayout, VerticalLayout, MatrixLayout,etc. You can check the examples and see how you need to layout your views.
You are currently doing UI5 JS view which implements createContent method to define views, this is one approach. Another common approach is to use XML views, it is declarative and more straightforward, also needs less code. See this simple example in JSBin of defining XML view and controller to refresh UI.
SAP UI5 is all about Model(JSONModel/ODataModel)-View(JSView/XMLView)-Controller. You are highly recommend to read this MVC example, though it is based on SAP UI5 mobile, the content is relevant to SAP UI5 desktop as well.
Hope you will get some hints.
I would like to know what is the best method to have an action return different views. Let's say you have a form for submitting data, but you want to choose the view depending on what data is submitted. I would prefer not using a redirection, since there is stuff I want to be displayed in the data that is posted.
An example of this would be to have an Edit form that displays a Details view when clicking on Save, but without using a redirection.
I know this could be done with a single view containing a conditional if statement to display this or that, but there are cases where I would prefer my views to stay simple without too much code in them. If the controller could just choose the view to display once the data is posted, this would be great.
There is an overload to the View() method that allows you to specify the name of the View you want to return.
return View("DetailsView", model);
You should be using the Post/Redirect/Get pattern. You can still "display stuff." You can pass an ID on the URI and look them up in the new GET or use TempData.
Attempting to circumvent Post/Redirect/Get is not a good solution. Among other things, it breaks the back button.
I`m using asp.net mvc 2.0 and trying to create reusable web site parts, that can be added at any page dynamically.
The problem I have is how to load a partial view with all related js and data? Ive tried the following ways to do that:
Use partial view and put all the js into it. In main view use render partial. But to initialize partial view I need to add model to current action method model to be able to make RenderPartial("MyPartialView", Model.PartialViewModel).
Also I do not have a place to put additional data I need to fill my form(like drop down lists values, some predefined values etc).
Use RenderAction, but it seems it have same problems as RenderPartial, except for I do not need to add anything to any other model.
Any other oprions are greatly appreciated.
As I understand it, RenderAction performs the full pipeline on the action, then renders the result - so what is rendered is the same as what you'd see if you'd browsed to the action.
I've used RenderAction to render 'widgets' throughout a site, but in my view they should be independent of the page rendering them, otherwise they're not really widgets and should be part of the rendering page's code instead. For instance, if there's a log in form, you will always take the user to a page that can process the information, no matter what page they are currently on, so this makes for a good widget. Other ways I've used it is to show a shopping basket or advertising. Neither of which are dependent on the page being shown.
Hope this helps a little!
I am new to grails and i got stuck with another issue.
I have two form's in my single GSP search.gsp and have two actions in my controller serach and results.
Now when i click on search button in one of my GSP file it takes me to search action which renders me search.gsp.At this time it should display me only first form in it. when i click results button in that form it will take me to results action.which has code line.
redirect(action:"search",params:[merchants:merchant,address:address])
this will take me back to search action but now i want to display 2nd form in search.gsp..
My problem is
how can i make search action once to run with out parameter's and once with parameter's?
how to determine in GSP from which action its been called?
with Advance thanks.
Depending on how different your forms are, you may want to consider having two separate GSP files (e.g., search.gsp and results.gsp). Use render(view:'action', model:[...]) to render a different view in the controller. This is often clearer that a single file with lots of conditionals.
Otherwise, you can find out the action using ${params.action}, so for example:
<g:if test="${params.action == 'search'}">
Text to show if the action is search
</g:if><g:else>
Text to show if the action is results
</g:else>
I would suggest you to separate your result page as template (_search.gsp), and render it from your result action. So that's how you will have different forms in different files.
By the way template is nothing but an ajax response, google it for detail about template in grails.