AutoSelected By Other Slicer

[Bkurtsatar]

Hello,

I have a slicer for a column in table A, and I also have a slicer for another column in table B. I want certain values in the B slicer to be automatically selected based on the value selected in the A slicer. For example, if "3" is selected in A, I want "X," "Z," and "T" to be automatically selected in B, and the filtering should be based on this selection.

How can I do it?

[marcelsmaglhaes]

@Bkurtsatar ,

Is there a relationship between tables A and B? Because if you have a relationship, in theory, slicers should work exactly how you need them to. If you don't have a relationship, well, you're going to need one!

[Bkurtsatar]

Thank you for the response. There is no relationship between the two tables, and there is no common column where a relationship could be established.

[Idrissshatila]

Hello @Bkurtsatar ,

yuou need to create a column in B table to specifiy every value to each number it references in table A, then link this new column with the A column.

in this wayy you can filter from A table.

[Bkurtsatar]

But it's not limited to a single value. So how can I add "X," "Y," "Z" values for the "3" value in a single column? Table B, where X, Y, Z values are, is a static table.

[marcelsmaglhaes]

You need something like that

[Bkurtsatar]

You got quite close, but when the "3" value is selected, I also want the "X," "Y," and "T" values to be selected from the other slicer.

So I believe Table_B will be filtered based on the values X, Y, T in this way.

[marcelsmaglhaes]

No value selected in filter A, all values on screen.

The value 3 selected in A, brings the values of B. Now, if you want the filter B to have a check mark on the values, it is not possible.

[Bkurtsatar]

Actually, my goal is to filter Table B based on "X, Y, T" values when "3" is selected in Table A. So, if A = 3, filter Table B based on X, Y, T values. There's no need to see them selected in the slicer in Table B, but filtering Table B is sufficient. I hope I have explained it correctly and completely.