use ANY with OVER (PARTITION BY …)?

Question

For example I have the following data:

select *
into t1 
from (
          select 'apple' as company, 'Tom' as people
    union select 'apple' as company, 'Jessi' as people
    union select 'fb' as company, 'Alex' as people
    union select 'fb' as company, 'Nick' as people
) as a

select * from t1

the data looks like:

company     people
apple       Jessi
apple       Tom
fb          Alex
fb          Nick

I want to see if each company has Tom - at each row level.

Is there something similar to:

   select *,
        any(people = 'Tom') over (partition by company order by company) as find_tom
    from t1

---

A work-around:

-- step 1: binary indicator has_tom
select *
    ,case when people = 'Tom' then 1 else 0 end
    as has_tom
into t2
from t1

-- use max(has_tom) to see if tom exists for each partition
select *,
    case when max(has_tom) over (partition by company order by company) = 1 then 'has tom' else 'no tom' end
    as find_tom
from t2

this yields the result I want:

company people  has_tom find_tom
apple   Jessi   0       has tom
apple   Tom     1       has tom
fb      Alex    0       no tom
fb      Nick    0       no tom

Answer 1

is this helpful?

Select t1.*,
 CASE WHEN 
    MAX(CASE WHEN people='Tom' THEN 1 ELSE 0 END)
    OVER(Partition by company Order by (select null))  =1 THEN 'has Tom'
 ELSE 'no Tom' END
FROM t1

SQl Fiddle: http://sqlfiddle.com/#!18/8ad91/19

Answer 2

I would just use case like this:

select t1.*,
       (case when sum(case when people = 'tom' then 1 else 0 end) over (partition by company) > 0
             then 'has tom' else 'no tom'
        end) as find_tom
from t1;