How to get the highest value in a row together with the column name

Question

I have a table with 10 columns, eight of them are columns with dates. My desired outcome is to get max date for each row (what I have already accomplished) but I would also

Answer 1

XML has great abilities to deal with generic queries:

DECLARE @tbl TABLE(ID INT IDENTITY,d1 DATE, d2 DATE, d3 DATE);
INSERT INTO @tbl VALUES
 ('20180101','20180102','20180103')  --one max value
,('20170101','20190102','20190102'); --two max values

SELECT TOP 1 WITH TIES
       t.ID 
      ,y.value('text()[1]','date') d
      ,y.value('local-name(.)','varchar(100)') c
FROM @tbl t
CROSS APPLY(SELECT d1,d2,d3 FOR XML PATH('d'),TYPE) A(x)
CROSS APPLY x.nodes('/d/*') B(y) 
ORDER BY DENSE_RANK() OVER(PARTITION BY ID ORDER BY y.value('text()[1]','date') DESC);

UPDATE: Some explanation

The first CROSS APPLY will create an XML which looks like this:

  2018-01-01
  2018-01-02
  2018-01-03

The second CROSS APPLY uses .nodes() to return all nodes within . With .value() we can get the element's name (local-name()) and its content.

The trick with DENSE_RANK and TOP 1 WITH TIES will return all rows which get a 1 (which are the highest per ID).