Unanswered

How can I jump to an assigned/customized Form from a View list?

(0) Share

Report

Posted on by Tiamo_Liu

Hi PowerApps professionals:

I have a model-driven app in PowerApps, an Entity called "ProjectInfoList". Within the Entity, there's a customized Views called "NewView". And there's a default Main Form under Forms when the Entity is created.

On the front-end page, I could see the data list of records through "NewView". By clicking the link text of one data record, it opens the default Main Form to show record details.

Later, I created a customized Main Form called "NewForm". On top of this change, when I click the link text of one data record on the front-end page, it automatically opens the Main Form "NewForm" that I created to show the record details. It has different set of fields and different format to the Entity's default Main Form.

What I would like to get is: when I click the link text of one record, it could open the Form that I assigned, either the default Main Form or the created "New Form". In another words, I expect View A can navigate to Form A, View B can navigate to Form B, all of them are from the same Entity.

Can anyone share the way how to achieve this goal?

I searched many resources online but failed to see a proper reference.

Thanks in advance!

Categories:

Building Power Apps

I have the same question (0)

All responses (2)

Answers (0)

GarethPrisk 2,828 on at

Like (0)

Report
What you are describing is not quite possible. However, there are a few ways to accomplish this.
There are functions you can execute on a view's column, and you could execute the openForm function. This would need to be setup per view, for a specific form, and for multiple columns. This likely would not lead to a consistent user experience.

If it is absolutely necessary to switch Forms based on the record, then you can accomplish this with a form script on either of the forms.
Create a web resource to put on each of the Forms
Determine the logic (similar to view logic) to determine what kind of record opens what kind of form
Execute an OnLoad function that calls
https://docs.microsoft.com/en-us/powerapps/developer/model-driven-apps/clientapi/reference/xrm-navigation/openform
https://docs.microsoft.com/en-us/powerapps/developer/model-driven-apps/clientapi/reference/xrm-utility/getpagecontext
You can use their samples to build up the function, and dynamically determine if the form is the correct one, otherwise open the other form
This is not a great approach, I realize, but it's going to be the simplest way to accomplish the desired behavior.

A few items for consideration:
Are the forms really that much different? Can you simply show/hide based on the same criteria that determines the views?
How often will users change types? If it's minimal, then the approach above should be fine, since the double-navigation won't happen every time

Was this reply helpful? Yes No
Tiamo_Liu 23 on at

Like (0)

Report

Hi @GarethPrisk ,

Thanks a million for your reply. It gives couple of suggestions. Among those, I believe "There are functions you can execute on a view's column, and you could execute the openForm function" is what I'm eager to pursuit. Due to my light knowledge base, could you please kindly share a detailed example of how to execute functions on a view's column, and where to execute the openForm? I read through the openForm function doc, it seems like the "formId" has to be assigned to the openForm in order to display the desired form instance, could you please also comment where to find the "formId"?

Thanks again!

Was this reply helpful? Yes No