Testing a column in collection has data

[Anonymous]

Hello PowerApps,

This Code: IsEmpty(Filter(DataViewer, 'Column' = Blank() || 'Column' = "0")) is not working. It does not correclty tell me if there is something in the collection. It shows ture for all occasions.

Does anyone have a better solution to test if a column is blank or not or has any fixs.

Thanks,

AskingQuestions

[RandyHayes]

@Anonymous 

Consider changing your formula to the following:

CountRows(Filter(DataViewer, 'Column' = Blank() || 'Column' = "0"))=0

The above will be true if the columns are all blank or 0 and false if there is data in them.

I hope that is helpful for you.

[Anonymous]

@RandyHayes ,

The My powerapps never returns false for any columns that do have values.

[RandyHayes · Accepted solution]

@Anonymous 

Sorry, had that a little backwards...

In re-reading, you want to determine if the column is blank or is a 0 and thus you consider the colun to be empty.  If that assumption is correct then what you want is:

CountRows(
   Filter(DataViewer, !(IsBlank('Column') || 'Column' = "0"))
   )

Using that formula, if the count is 0, then the column is empty or "0", if it is 1 or more, then there is something besides a blank or a "0".

I hope that is clearer.

[Anonymous]

Thanks @RandyHayes ,

The Formula seems to be working.

Just to be in the same page. This formula filters out the blanks and zeros in a column then counts the rows present in the column. if the result of the formula is zero then that means the column is empty. If the result is greater than zero then that mean there is something in that column.

Thanks,

AskingQuestions

[RandyHayes · Accepted solution]

@Anonymous 

Yes, as the formula reads - if NOT 'Column' is blank or is "0" then it will match the filter and thus count as a row.  If all are blank or "0" then the filter results will have no records and thus the countrows will be 0.

If any count other than 0 comes back, then some row had something other than a blank or a "0" - and thus had data.

[Anonymous]

Thank You @RandyHayes for helping me out.

Once again Thanks,

AskingQuestions