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RLSid17
Helper III
Helper III

Count Values Out Of Total


Hi,

Is there a way to count the values and compare them to another table? Kinda like for example, 24 out of 30 employees responded to the survey.

I have a table that is similar to this:

RLSid17_0-1667303551790.png


So each category there would be the users that have responded alongside there would be the manager column to track. Is there a way to do a distinct count for each category in a single table? So the result would be again like 5 out of 10 or something like that. 

1 ACCEPTED SOLUTION
RLSid17
Helper III
Helper III

Hi @v-yanjiang-msft 

Thank you for your response, I managed to resolve it now by counting the users and then also counting the total number of users.

RLSid17_0-1667491788144.png


There are 2 datasets, where I got the distinct count of responses and then compared it to another table which has the number of employees per manager. 

Just for the benefit of others, all I had to do is to make a relationship between the two datasets to get the required value like the one I'm doing.

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2 REPLIES 2
RLSid17
Helper III
Helper III

Hi @v-yanjiang-msft 

Thank you for your response, I managed to resolve it now by counting the users and then also counting the total number of users.

RLSid17_0-1667491788144.png


There are 2 datasets, where I got the distinct count of responses and then compared it to another table which has the number of employees per manager. 

Just for the benefit of others, all I had to do is to make a relationship between the two datasets to get the required value like the one I'm doing.

v-yanjiang-msft
Community Support
Community Support

Hi @RLSid17 ,

According to your description, if you just want to count the numbers of each category, simply put the Category column in a visual twices then select Count in drop-down box. Get the result:

vkalyjmsft_2-1667374413587.png

Best Regards,
Community Support Team _ kalyj

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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