This Item. Combo Box values doesn't work, what's is the right way?

[LuisGuzman98]

Hi, 

I Have this gallery where I select items from a  sharepoint list and the values of each column are then showed of on another gallery. 

Everything is working ok with all the columns, but I have one value which comes from a combo box, which means multiple values are stored on that column. 

Actually using the formula: 

ThisItem.'Additional Resources'.Value

Which works with the other values... How could I display all the values present in this field?

Thanks 

[Jeff_Thorpe · Accepted solution]

If you are displaying the combo box values in a text field then you need to convert it to a string like the example below or you could change you display field to list box or gallery which can display a record table.

Example: Concat(ThisItem.'Additional Resources', Value, "; ")

[LuisGuzman98]

Hey @Jeff_Thorpe , 

Just tried what you suggested but not quite sure if I did it right. 

Obviously it's not this way but that's sort of what I understand:

Concat(ThisItem.'Additional Resources',ThisItem.'Additional Resources'.Value, ";")

I did this in flow, I have to create a loop for gathering all the values that are contained in the field into one string. I need to get this done into power apps. 

Thanks

[Jeff_Thorpe]

What happen when you tried my example without any changes to it (assuming I had the field names correct)?

[rebeccas]

What is making the values show in the second gallery...are you creating a collection?

If the second gallery has the same data source (not a collection) then you can just select the item in the first gallery and then filter for it in the second gallery without having to carry over any values manually. 

[LuisGuzman98]

Hi @rebeccas, 

No I'm not using any collection. 

I'm using two gallerys and calling each item manually. But actually the method of @Jeff_Thorpe is correct and it worked fine for my purpose, I was just doing it incorrectly when I tried. 

The concat funtion expects a table, a value and a separator. I was confused because I didn't realize that the value I was calling was actually a table data type. 

I call it on the gallery just like you said @Jeff_Thorpe  and it works perfectly. 

 Thanks you both!