Question

Something very simple, but I cannot get it working for me 🙂

select x.name, count(x.name) from <table_name> x where ...<complex and long>...

It throws an error, that the x.name isn’t used with group by.

select x.name from <table_name> x where ...<complex and long>...

works fine and returns e.g. 6 names

select count(x.name) from <table_name> x where ...<complex and long>...

works also fine and returns e.g. the number 6

But the combination is not working, when I tried to add the group by:

select x.name, count(x.name) from <table_name> x where ...<complex and long>...
group by x.name

it worked, but all counts were 1 and not 6.

The thing is, I could firstly get the count into variable, then write the long sql statement just to get the names, but I don’t want to write the long complex select statement twice. There has to be some way to do the combination in one select: get all names and by the way tell me how many they were.

Thanks

P.S.

name    bla1    bla2
a       ...     ...     
a       foo     ...     
b       foo     ...     
c       ...     ...     
d       foo     ...     
d       foo     ...     
b       foo     ...     
c       foo     ...     

The result of x.name where bla1 = foo would be:

a
b
d
d
b
c

The result of count(x.name) where bla1 = foo would be:

6

My desired result of:

...variable definitions
select @foundName = x.name, @numOfAllFoundNames = count(x.name)
from <table_name> x where ...<complex and long>...

should be:
@foundName = a (just one of the names, no matter which one)
@numOfAllFoundNames = 6

Answer 1

The simplest way is to use @@rowcount after the query;

select x.name from <table_name> x where ...<complex and long>...

select 'the number of rows is:', @@rowcount

As an aside, if you request a non-aggregate field (name) and an aggregated field (count(name)) you must provide a group by to tell the server what to count; you see 1’s because group by name applies the count to every distinct name in the set – E.g. you would see 1 less row and 2 if a name was repeated.