How To Add The Details in MessageBox? eg: i link to forms after fillin the second form i want a messagebox showing the details entered by da user and to b shown in messagebox
You should create a new form and show it as a modal dialog using Form.ShowDialog(). Not messagebox.
Here is a list of all posible way to form a MessageBox...
But I think that for your need you should just create a form, that get as parameter all info that you want to display and then call it
wanabemb wanabemb = new wanabemb(string info_here);
wanabemb.Show(); //or wanabemb.ShowDialog();
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I have a Insert form for Account Entity.
Once user create Record, I want to move to another page and show same record as Edit Form.
Does anyone have idea way to achieve this?
You should be able to redirect the user to desired page using OnSuccess of form submission.
On success: Select one of the following options:
Show success message: Requires a message to be displayed to the
user on successful submission of the form. You can also select Hide
form on success to hide the form upon successful submission.
Redirect to webpage: Redirects the user to the selected webpage
in the portal. You must select a webpage from the Redirect to webpage
list.
Redirect to URL: Redirects the user to the specified URL. You
must enter a URL in the Redirect to URL field.
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Out of Box configuration helps!!!
There is a field on "Entity Form" records named "Append Record ID to Query String". This is to configure passing the id of the record to the page after form submission. This screenshot shows it in the Redirect Section in the classic interface
You can do this several ways. What I do, is use Navigate() to go to the new screen, and I pass in a context variable to the new edit screen.
Assuming you're using OnSelect() on a button to call SubmitForm(), you can do the Navigate() call via the OnSuccess() property of the form.
And within Navigate(), you just pass whatever data you need to in order for the next screen to properly bind it to the Edit Form. This is the UpdateContextRecord optional param. You'd have the Edit form take its values from a Context Variable.
Passing variables between screens has been discussed here, as well.
I am creating an application on Oracle Apex version 18 that takes survey from users.
I have used form on table for that purpose. However, users are able to submit the form multiple times. Is there a way to freeze the submit button once the form is submitted?
Yes; you have to be able to find out whether it was already submitted (for example, record in a table gets its ID; or, survey_date gets is set to sysdate).
Then create a condition for that button. A simple option is to use a function body that returns Boolean, e.g.
return :P1_ID is null;
which will render the button only if :P1_ID item is empty. If it is not, you won't see the button any more.
In the first instance I would try setting the page attribute Enable duplicate page submissions to No.
It seems like you don't want users to be able to answer the same survey multiple times.
In order to achieve this you can use a server side condition as Littlefoot suggested, but you seem to have problems with setting up the server side condition properly.
If you use a hidden item on your page, which holds the value of the current survey the user has to answer like "P1_SURVEY_ID", you can add a server side condition that doesn't allow the user to answer it again, if he already has:
Type: No rows returned
SQL Query: select * from YOUR_SURVEYS where SURVEY_ID = :P1_SURVEY_ID;
This way the user can only click the button if he has not answered this specific survey yet.
This is a easy solution:
https://community.oracle.com/thread/4167853?start=15&tstart=0
Simply add the "Page Submit" TRUE Action to your Dynamic Action and
set the Show Processing "o Yes. Then you don't even need any Enable or
Disable Actions on your button.
I have a CodeIgniter webpage containing (i) a list of items and (ii) a form to add a new item or edit the details of a selected item.
To select an item for editing, I have an "edit" anchor tag with a hyperlink ending with ?id=nnn. This results in a GET request that populates the detail fields of my selected item and redisplays the list. My URL now contains site_address/index.php/country?id=151.
When I submit my changes the list is updated correctly but my URL remains unchanged. What do I need to do in my controller to remove "?id=151 after my update was successful?
Thanks for your help.
I'm not sure of everything you are doing, so I'll just answer directly:
in your controller, after you update, you can:
redirect('site_address/index.php/country');
Does that fit what you mean?
Through php there's no way to change URL after request expect redirecting
redirect('controller/method');
and you can success message using session flash message after redirecting, if that what makes you want to reload the view
I am creating dynamic textbox and button in codeigniter view page through foreach loop. Then I want to retrieve the value of those textboxes on buttton click. please help me. Thank you.
Add a counter and attach the counter at end of textbox names so that dynamically created textboxes have names like textbox1, textbox2 etc. In controller's submit function run the loop again and fetch the values.
Regards
iijb
<delete_gamecode translate="label">
<frontend_type>button</frontend_type>
<value>button</value>
So in my system.xml, I created a button, but it only shows a button with no text. How do I add a value to the button?
Also, I want the button to delete the game code as well. How can I do this?
Is there a way for me to attach JavaScript to these buttons so I can do some JavaScript confirmations? For example, after I click the button, it will say "Are you sure you want to delete this gamecode?"
You need a <frontend_model>. Please refer to following files for the example:
app/code/core/Mage/Core/etc/system.xml
app/code/core/Mage/Adminhtml/Block/System/Config/System/Storage/Media/Synchronize.php
app/design/adminhtml/default/default/template/system/config/system/storage/media/synchronize.phtml