Postgresql - 在单个查询中过滤掉数据

Question

I have the following data in the booking table.

booking

id      email                 price   type      areaid
1       person1@test1.com     70      type1          1
2       person2@test2.com     60      type2          2
3       person3@test3.com     50      type1          3
4       person4@test4.com     110     type1          3
5       person1@test1.com     90      type2          4
6       person2@test2.com     65      type2          1
7       person3@test3.com     84      type2          2
8       person4@test4.com     84      type1          2

I need to retrieve all email addresses from booking table which have only type2 and no other types. According to the data, only person2 meets this requirement.
How can I achieve this within a single query?

Answer 1

You can use HAVING :

select email
from t
group by email 
having min(type) = 'type2' and max(type) = 'type2'

Answer 2

Select b.email 
from booking b 
group by b.email 
having 
    'type2'= any(array_agg(b.type)) and array_length(array_agg(b.type),1)=1;

You can use above query for your task. First group by clause using email then use having clause to filter the results after get grouped.

Answer 3

One more query for this problem is simple translation English to SQLish:

select distinct email
from test
where type = 'type2'
    and not exists (
        select email from test t2 where test.email = t2.email and t2.type != 'type2' 
    );

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