Filter list by variable

Community Discussions
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1

Hi!
I have some issue with filter data using variable. The value is transferred from the previous screen and stored in a variable in the current screen. To filter the list, I use the function:

=FILTER(table,"table[Column] = $[Variable]")

When I use a variable in the function as above, I get the answer "No results to show". However, when I specify the value it contains instead of a variable as below, filtering works.

=FILTER(table,"table[Column] = ""Value""")

What could be the problem?

2

Hi @Andr-576d ,

Thank you for your post, it's nice to hear from you again!

Here are some possible solutions:

  1. Can you add a text field and display =$[Variable] and see whether it returns a value or not?
  2. Try this formula =FILTER(table,"table[Column] = %", $[Variable])

I hope this helps!

Thanks,
Evelyn

3

Hi @Evelyn ,
Thank you for your response.

  1. I have a field containing the value of a variable on the second screen. My expected value is displayed here.
  2. I tried this formula and I still get the message "No results to show".

When I copied the displayed value and used it in a filter function instead of a variable, this function works fine. I am sending a view containing a piece of the configuration of the second screen. I used your suggestion from point two here.

obraz
obraz1186×519 28.5 KB

4

Hi,
Anyone else have a proposal to solve his problem?

5

Hi @Andr-576d,

Here are some additional suggestions for our engineering team:

  1. Create a table with a column named something like 'variable'
  2. Create a button attach and then create automation on that button - add the action “Add a row to” table: “temp” from: =$[Variable], to: Temp[Variable]
  3. Update the filter to use this as the source: =FindRow(Temp)[Variable]
  4. Alternatively to 3, write this into another column: =Filter(Table1,"Table1[Column1]=%",FindRow(Temp)[Variable])

I hope this helps!

Thank you!
Evelyn

6

Hi @Evelyn,
Thanks for the help. I was able to fix the problem yesterday. A hint from your engineering team was also a good lead. The problem was the variable passed to the filter function. The value of the variable was displayed correctly in the content box, however, it was a row and required the name of the column to be indicated.
Thanks again for your help!

1 Like
7

Hi @Andr-576d,

That's great to hear!

Please let us know if you have any other questions of concerns.

Thank you,
Evelyn