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daw7560
Regular Visitor

Filter Count By Date

I am trying to figure out the formula to get the total price of the count of filter ID's by select date. The current formula will just multiple the total count of filter ID's by the price, but won't change based on dates selected in the slicer. I am hoping to show the total price of count of filter ID's by Quarter or by year when those dates are select. So, the total price will change depending on how many filter ID's we used per quarter or per year. Let me know if I need to clarify this some more. 

 

Is there a way to write the formula to do this? 

 

daw7560_0-1708524657539.png

 

1 ACCEPTED SOLUTION
v-zhouwen-msft
Community Support
Community Support

Hi @daw7560 ,

The table ‘LabDataRTI’ data is shown below:

vzhouwenmsft_0-1708585392466.png

Please follow these steps:
1. Use the following DAX expression to create a measure named ‘Total price’

Total Price = 
 VAR _a = COUNTROWS('LabDataRTI')
 VAR _b = _a * 68.96
 RETURN _b

2. Final output

vzhouwenmsft_1-1708585454491.png

 

vzhouwenmsft_2-1708585463433.png

 

vzhouwenmsft_3-1708585472839.png

vzhouwenmsft_4-1708585479777.png

vzhouwenmsft_5-1708585492574.png

 

vzhouwenmsft_6-1708585502177.png

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

View solution in original post

1 REPLY 1
v-zhouwen-msft
Community Support
Community Support

Hi @daw7560 ,

The table ‘LabDataRTI’ data is shown below:

vzhouwenmsft_0-1708585392466.png

Please follow these steps:
1. Use the following DAX expression to create a measure named ‘Total price’

Total Price = 
 VAR _a = COUNTROWS('LabDataRTI')
 VAR _b = _a * 68.96
 RETURN _b

2. Final output

vzhouwenmsft_1-1708585454491.png

 

vzhouwenmsft_2-1708585463433.png

 

vzhouwenmsft_3-1708585472839.png

vzhouwenmsft_4-1708585479777.png

vzhouwenmsft_5-1708585492574.png

 

vzhouwenmsft_6-1708585502177.png

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

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