Union all on the same table - sql

Question

There has to be a better way of doing this any one have an idea. there is one table and i have 8 columns i need select all of the columns one on top of each other and for each selected colum i need to count the number of items that are the same
SELECT Col1, count(*) 'Selected'
FROM [Table]
group by temp_id,Col1
UNION ALL
SELECT Col2,count(*)
FROM [Table]
group by temp_id,Col2
having len(ltrim(rtrim(Col2)))<>0
UNION ALL
SELECT Col3,count(*)
FROM [Table]
group by temp_id,Col3
having len(ltrim(rtrim(Col3)))<>0
union all
SELECT Col4,count(*)
FROM [Table]
group by temp_id,Col4
union all
SELECT Col5,count(*)
FROM [Table]
group by temp_id,Col5
having len(ltrim(rtrim(Col5)))<>0
union all
SELECT Col6,count(*)
FROM [Table]
group by temp_id,Col6
having len(ltrim(rtrim(Col6)))<>0
union all
SELECT Col7,count(*)
FROM [Table]
group by temp_id,Col7
having len(ltrim(rtrim(Col7)))<>0
union all
SELECT Col8,count(*)
FROM [Table]
group by temp_id,Col8
having len(ltrim(rtrim(Col8)))<>0

Answer 1

There is. It is called grouping sets (and it is documented here). In your case, you can do something like this (example for first three columns):
select coalesce(col1, col2, col3), count(*) as Selected
from [table]
group by grouping sets ((temp_id, col1), (temp_id, col2), (temp_id, col3));
Your condition using length() and trim() on each column -- you should probably handle that using a having clause.