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Find a SQL statement ~~~ Master help ~~~

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 China

Post time: 2020-2-21 01:00:01
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Records in the table allip

id ip sid
1 127.0.0.1 a
2 127.0.0.2 a
3 127.0.0.3 a
4 127.0.0.1 a
5 127.0.0.4 b
6 127.0.0.1 a
7 127.0.0.5 c
8 127.0.0.6 d

Now I find the sum of the number of all unique IPs with sid = a (that is, an IP can only appear once)
The resulting value should be 3

I use
select count (ip) as c from allip where sid = 'a' group by ip
The answer is the number of each IP instead of their sum. How to write SQL? Ask for advice ~~~~~~~~~~~~~~~~
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 China

Post time: 2020-4-29 12:00:01
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select count (distinct ip) as c from allip where sid = 'a'
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Post time: 2020-4-29 14:30:01
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select count (distinct ip) as c from allip where sid = 'a' group by ip
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 Brazil

Post time: 2020-4-29 17:15:01
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select count (ip) as c from allip where sid = 'a' and ip in (select distinct ip from allip)
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 China

Post time: 2020-5-1 00:30:01
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--Test Data
create table t (id int identity (1,1), ip nvarchar (20), sid char (1))
insert t (ip, sid)
select '127.0.0.1', 'a' union all
select '127.0.0.2', 'a' union all
select '127.0.0.3', 'a' union all
select '127.0.0.1', 'a' union all
select '127.0.0.4', 'b' union all
select '127.0.0.3', 'a' union all
select '127.0.0.5', 'c' union all
select '127.0.0.6', 'd'
select count (distinct ip) from t where sid = 'a'
drop table t
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 China

Post time: 2020-5-1 13:00:02
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select count (distinct ip) from #temp where sid = 'a'
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 China

Post time: 2020-5-2 02:15:01
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select count (ip) from (select ip from allip where sid = 'a' group by ip having count (ip) = 1) m
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 China

Post time: 2020-5-2 08:30:01
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Learn it!
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 United States

Post time: 2020-5-2 14:45:01
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select count (distinct ip) from allip where sid = 'a'
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