我想你想 in :

SELECT 
    id_dept, 
    COUNT(DISTINCT CASE WHEN e.statut IN ('pub', 'priv') THEN id_patients END) AS nb_patients_pub_and_venues
FROM venues
GROUP BY id_dept

注意，我添加了一个 GROUP BY 子句，这是最初丢失的（这是几乎所有数据库中的语法错误）。
根据你的数据，这可能不完全是你想要的；如果一个给定的 id_patient 如果两种状态都有，那么它只会被计数一次，而您的代码在每种状态中只计数一次 count(distinct ...) . 如果是这样的话，那么你只需要保留两个分开的计数，然后求和：

SELECT 
    id_dept, 
    COUNT(DISTINCT CASE WHEN e.statut IN = 'pub' THEN id_patients END) 
        + COUNT(DISTINCT CASE WHEN e.statut IN = 'priv' THEN id_patients END) 
        AS nb_patients_pub_and_venues
FROM venues
GROUP BY id_dept

如果您对当前代码满意，那么sum（使用 + )那些 count s、 或者将该查询用作cte（或内联视图）并

with test as
  (SELECT id_dept, 
       count(DISTINCT case when e.statut='pub'  then id_patients end) 
         AS nb_patients_pub,
       count(DISTINCT case when e.statut='priv' then id_patients end) 
         AS nb_patients_priv
   FROM venues
   GROUP BY id_dept
  )   
select id, nb_patients_pub + nb_patients_priv as result
from test;