Hello,
Please, I am currently unable to apply two filters to my Gal Animal table, I have tried several Power FX expressions but I still encounter errors such as "Unsupported comparison types". I tried to also filter my animal table view to Ready for Foster View but it is still not working.
I will appreciate a solution to fixing this issue please.
Attached is what my formula currently looks like, I can only filter to the shelter location successfully at the moment, but can't filter the Shelter Status to "Ready for Foster".
This is what i used to filter successfully
Thanks @kashjoyin
I tried using this formula as well, but it keeps giving me an error message 😥
After a few hours of research, I finally found the formula that worked for me. So I used a combo box and the formula works in a way that when nothing is selected it returns my full Ready to Foster view. It should work for dropdown as well. Combo box data is my Shelters table.
You should try this Dianah.
For your gallery
Items = If(
IsBlank(cmbShelterName.Selected),
Filter(
'Animal Details',
'Animal Details (Views)'.'Ready for Foster'
),
Filter(
'Animal Details',
'Animal Details (Views)'.'Ready for Foster',
'Shelter Name' = cmbShelterName.Selected.'Shelter Name'
)
)
Then for the combo box
Items = 'Shelter Info'.'Shelter Name'
Nothing selected
Selected
Thanks @Oyin It helped me as well. but when I use the view from dataverse table, I am not able to see the image of the pets. Any suggestion on how to display images here ?
PS: i have uploaded the photos of the pets.
Hi @Poovarasan27 ,
Go to insert, search for image and drag into your gallery. Highlight it, then use the formula:
Image = ThisItem.'Photo URL'
Depending on your column name, Photo URL or Photo.
sorry to hijack this but looks like you are working on the same thing I am. I have been trying to get my form to update shelter status once someone picks an animal to foster but haven't been able to get it to work, do you have any suggestions?
Hi @jsl6v8 ,
What I did was add the choice option 'Claimed for Foster' in the Shelter Status column of the animals table and use the patch formula below.
OnSuccess = Patch(
'Animal Details',
{
'Animal Detail': Form1.LastSubmit.'Animal Detail',
'Shelter Status': 'Shelter Status (Animal Details)'.'Claimed for Foster'
}
);
Refresh('Animal Details');
Navigate('Home Screen');
Notify(
"You have succesfully claimed an animal to foster",
NotificationType.Success
)
Hi @Dianah_Ebere ,
Try setting the Items property of the Drop Down control used to select the shelter status to:
Choices(AnimalFamilies.'Shelter Status')
Then in the Items property of the gallery, include a filter condition like this:
Filter(
AnimalFamilies,
'Shelter Status' = DropDown.Selected.Value
)
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