I want to create a breadcrumb when one of the entries is shown but not clickable.
for example:
page1 \ page2 \ page3 \ page4
page1 and page3 are clickable but page2 is not (page4 is the current page).
Defining the breadcrumb entry's target as "No target" keeps it clickable (redirect to the same page)
so this solution not good enough for me.
Apex version is 19.1.
Thanks.
On the page where you have the breadcrumbs rendered, you could remove the href using jquery.
In the page attributes > execute when page loads, put:
$('a.t-Breadcrumb-label').filter(function(){
return this.innerHTML == 'page2';
}).removeAttr("href");
Related
I've three pages first showing List of my items on tap I'm redirecting it to a History Page with Object defined with static property with
await Shell.Current.GoToAsync($"/{nameof(DeviceHistoryPage)}");
then from this page i'm again redirecting user to show more details of each history item to next page with data as static property
await Shell.Current.GoToAsync($"/{nameof(DeviceHistoryDetailsPage)}");
I'm able to achieve this navigation and able to move back and forth from first page to second page but from second page once redirected to third page and i'm not able return back to previous page with back button present in navbar
Also I've tried registering routes as independent as well as a child in AppShell as follows
Routing.RegisterRoute($"{nameof(ListPage)}/{nameof(HistoryPage)}", typeof(HistoryPage));
Routing.RegisterRoute($"{nameof(HistoryPage)}/{nameof(HistoryDetailsPage)}", typeof(HistoryDetailsPage));
This is my issue: I have a text field item in a normal page (P7_PRODUCT), in this page there is a button that opens a modal page and, in the modal page there is another text field item (P7_PRODUCT). I would like that when I write a product in P7_PRODUCT, the same product appears in P8_PRODUCT. I tried with several dynamic actions, but none of them worked.
Here's one option.
on page 7, there are: P7_PRODUCT text item and P7_GOTO_8 button
let button Submit the page
create branch
name it e.g. "Goto 8"
it should redirect to page 8
in "Link Builder", set P8_PRODUCT item to &P7_PRODUCT. (note leading ampersand and trailing dot)
run page 7
enter something into the P7_PRODUCT item
press the button
modal dialog page opens and its P8_PRODUCT contains value you entered on page 7
A good blog to read is this one, by John Snyders from the apex development team. As he explains, it's not advised to branch to a modal dialog - instead you should link to it.
Here is an example based on techniques described in [this] (https://jeffkemponoracle.com/2022/02/reusable-region-as-a-modal-page/) blog.
I have 2 pages: Page 108 (normal page) and page 109 (modal page). Page 108 has a region (Region 1) with page item P108_NEW and a button called TO_MODAL. Page 109 has page item P109_NEW. When a user enters a value in P108_NEW and clicks the button, page 109 should open with the value from P108_NEW shown in P109_NEW.
The technique generates the url of the modal page and then redirects to it.
Give the region on page 108 a static id: "region1". This is used if you want to create a dynamic action on close of the dialog.
Create a new hidden item on page 108 (P108_URL_TO_MODAL1) to hold the value of the modal page url. Source type: Expression, source:
apex_page.get_url(
p_page => 109,
p_triggering_element => '$(''#region1'')'
)
Make sure to set "Used" to "Always, replacing any existing value in session state".
Create a dynamic action on click of button on page 108.
Add an action to the dynamic action of type pl/sql to set the value on page 109.
Action 1: Execute Server-side Code, Pl/SQL Code:
:P109_NEW := :P108_NEW;
Values to submit: P108_NEW
Action 2: Execute Javascript code:
apex.navigation.redirect("&P108_URL_TO_MODAL1.");
This should be it.
When a user double clicks on a row on my page, I navigate them to an edit screen which shows them some important information. When they submit that patient I gerimander jqGrid to take them page to the page that they were just viewing...
serializeGridData: function (postData) {
//debugger;
var pagInfo = $(this).se
debugger;
if ((rowFromTemp != "") && (pageFromTemp != "")) {
//debugger;
postData.page = pageFromTemp;
pageFromTemp = "";
rowFromTemp = "";
}
return postData;
},
I basically edited the postData in the serializeGridData function.
My question is, when I navigate to a different page lets say page 3 of 10 after I have edited information and returned to the correct page, when I click refresh of the whole page...
I want the user to return to the page that they just navigated to. Right now, it will return them back to the page that they initially returned from after editing information. So If I were to construct a timeline of events...
1) user see's paged information...
2) user selects row to edit
3) user navigates to edit page and does his thing then submits
4) user now is sent back to page where he came from...
5) user goes to a different page (either next or last)
6) user refreshes page
7) !! user is sent to page X from step 3 and 4 and Not 5 !!
I was hoping to to some kind of refresh of page event and set the correct page in there.
Is my methodology or thinking in correct here? Is there a way to ensure that the correct page is selected no matter what?
I just checked again. It would seem that if a user
1) navigates to page X
3) refreshes a page
he will be sent to page 1.
Maybe I am setting an unrealistic expectation. Maybe if they reload the whole page, they should expect to go back to page 1.
Well even so, how could I accomplish that task with the first workflow
In my opinion the real problem which you have is the usage of step 3: "user navigates to edit page and does his thing then submits". The step makes all the problems which you try to solve. Instead of going to another page one can create dynamically some controls which allows to edit the selected row. jqGrid provide inline editing or form editing for the purpose. You need just include editable: true property in the columns which you need to edit implement starting of inline/form editing on select of row or on double click. Alternatively you can use buttons from the navigator bar (see navGrid and inlineNav) or in addition column of the grid (see formatter: "actions").
If you still don't want to use any editing mode which provide jqGrid you can just place the whole content of the body of the main page inside of a div. You can hide the div using jQuery.hide instead of redirection to "edit page". You can load the content of "edit page" on the same main page (per jQuery.load for example). Then you can remove the loaded "edit page" and show the original main page. As the result the main page will be shown in exactly the same form as it was before starting of editing.
im new here.
i would like to ask about partial view.
I created 2 partial view( 1 for Link to another pages, and another 1 is pages)
i call those 2 partial view in to 1 view and put it into my master page.
now i want to click the link in partial view and show it in master page. how to do this?
sorry for my bad english.
hope u guys understand..
Best Regards,
Sen
If I understood your question correctly ,you organized the partial views and your master page in a wrong way and you expect something wrong.
You should create a MasterPager (or _Layout.cshtml file in Razor) with a just one PartialView for your menus/lins(lets say it MenuBar). Your _Layout.cshtml also should contains a #RenderBody which is responsible for different page contents.
After that you just create your pages and point them to user your masterpage (it is done by default by Razor). After that whenever you click on each link on your MenuBar , you redirected to the page which is pointed to.
At server and on rendering phase, The MVC starts to render the page from your master page (so it contains the Menu Bar automatically) and when it reachs the #RenderBody line , it renders your main page content instead.
EDIT
you can check this link for a full tutorial about layouting : http://weblogs.asp.net/scottgu/archive/2010/10/22/asp-net-mvc-3-layouts.aspx or
http://www.mikesdotnetting.com/Article/164/Nested-Layout-Pages-with-Razor
But
this is supposed to be simple
i have two pages A and B what i want to do is add a button (a hyperlink or image will do) in page A that when clicked go to page B
how can i do that ?
thanks
If you don't want a button, simply use a html href to f?p=&APP_ID.:<your page number to link to>:&SESSION.
Go to page 2! takes you to your page number 2...
There are many ways. Perhaps the simplest is to create an Apex button using the Create Button wizard. Follow the instructions and when you get to the "Action When Clicked" page, choose "Redirect to Page in this Application" as the Action and enter the Page number you want to go to when the button is clicked. You can also enter other details such as Request, Clear Cache, Set These Items.