Is there any visual that could accomplish this?

[emma313823]

Hi

I have a table that includes lots of info like the customer, part number, sales, commissions, etc. My salespeople often need to search on multiple part number lead-offs. By this I mean a part number could be EGG.TT.123456 or FGG.TT.56789. I need to see if there is a visual that would allow the ability to filter off on just EGG.TT, select all that appears, then a second search for FGG.TT, select all. The end result would allow the user to have all the EGG.TT and FGG.TT part numbers showing, then they could export that data.

The part number column is set to TEXT.

I thought the slicer would work, but it does not seem to allow for this type of scenario. I also found a Filter by List visual, but this seems to require the entire part number not just the Starts With or lead off of the part number.

Emma

[v-stephen-msft · Accepted solution]

Hi @emma313823 ,

Here is a sample data I made. I believe that this measure could accomplish what you described in the forum.

First, I created a table with ETG, number and index. And the ETG column is the only column that matters in this problem.

I used the following measure:

Measure = 
VAR FilterGG = 
    
   IF( CONTAINSSTRING(MAX('Table'[ETG]),"EGG.TT") || CONTAINSSTRING(MAX('Table'[ETG]),"FGG.TT")=TRUE(),1)
   
RETURN FilterGG

This will return 1 for every row with its name begins with EGG.TT and FGG.TT

Now, we can go to the filter, and drag Measure into the filter. I circled those parts in the below picture.

Next, fill the blank with the constraints mentioned in the measure function. In this case, show items when the value is 1. Then click on “APPLY FILTER”.

And now we get a table all starts with EGG or FGG followed by the data.

The essential part of this method is to create a Boolean function to identify the beginning three letters and make sure that only those valid rows are showed.

Best Regards,

Stephen Tao

If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.