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phersan_sme
New Contributor III

nullcounts of all fields

Hi all,

I was wondering if there was an easy way to leverage the system field, $Field, in order to set up a table that lists the field name and its nullcount

I want the output to be something like:

FieldName          NullCount

ID                        0

Age                    10

Region                 3

Name                   2

Address              47

Email                  40

SSN               28437

I can write script and loop through all the field names and get the nullcounts that way, but I would like to keep the association on the calculation as well. For example, if I pick USA as the region, obviously the NullCount for Region would go down to 0, but I would be able to see the nullcounts of all other fields where Region is USA. Is there a way to do this, without brute force?

Thanks in advance,

Paulo

1 Reply
MVP
MVP

Re: nullcounts of all fields

This seems to work with $Field as the dimension.  You have to list each table you care about separately since you have a different ID to count for each table.  I suspect in your case all the fields you care about are on a single table so you can simplify that part.  There's probably a simpler way than this.

pick(match($Table,'MainTable','OtherTable')
,count(ID)
-pick(match($Field,'$(=concat($Field,chr(39)&','&chr(39)))')
$(=concat(',count({<['&$Field&']-={}>} ID)')))
,count(ID2)
-pick(match($Field,'$(=concat($Field,chr(39)&','&chr(39)))')
$(=concat(',count({<['&$Field&']-={}>} ID2)'))))

Edit: OK, here's a little simpler.  I still think it should be easier than this.

pick(match($Table,'MainTable','OtherTable'),count(ID),count(ID2))

-pick(match($Field,'$(=concat($Field,chr(39)&','&chr(39),$FieldNo))')

,count([$(=concat($Field,']),count([',$FieldNo))]))

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