finding no of non empty columns from a table in %

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SQL-Server-2005

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Hello friends,

I am stuck at a problem, not sure on how to go about writing a query that will return as a percentage the number of fields in a row that are null on the base of a user's id.

For instance, a row from my table:
Row1 : field1 field2 field3

If field3 is empty or null, my query should return 67%.

So far I have gotten the number of fields:
select count(1) from information_schema.columns where table_name='myTable'
I am using MS SQL 2008.
Thanks in avance friends

Posted 5-Jul-11 2:08am

pawanvats

Updated 5-Jul-11 3:09am

v2

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OriginalGriff · Answer 1 · 2011-07-05T03:02:00

Try:

SQL

SELECT 
   (SELECT (COUNT(1)*100) from myTable where ISNULL(myColumn,'')='') 
   / 
   (SELECT COUNT(1) FROM myTable)

"Thanks sir for your reply, but i need to count all the columns(columns having value) of a single row and show it in %"

Your original question asked for:
"return as a percentage the number of fields in a row that are null"

If you have 6 rows, with null in two of them, the query above will return "33" - ie, 33% or 1/3
To invert it, and return the percentage that are not null, put a "NOT" in front of the ISNULL:

SQL

SELECT 
   (SELECT (COUNT(1)*100) from myTable where NOT ISNULL(myColumn,'')='') 
   / 
   (SELECT COUNT(1) FROM myTable)