最快的方法（即，最少的重写量）是简单的UNION您的查询一起这样：

select 'null names' as col1, count(*)
from employees e
where e.name is null
  and e.last_name is null
union
select 'not null', count(*)
from employees e
where e.name is not null
  and e.last_name is not null;

| 列1|计数（*）|
| - -|- -|
| 空名称|一个|
| 不为空|三个|
您还可以使用条件聚合，它使用较少的真实的空间并且非常易于阅读，并使用以下内容在列中生成输出......

select 
  sum(case when e.name is null and e.last_name is null then 1 else 0 end) as null_names, 
  sum(case when e.name is not null and e.last_name is not null then 1 else 0 end) as not_null
from employees e;

| 空名称|非空|
| - -|- -|
| 一个|三个|

SELECT count(*) filter (WHERE e.name IS NOT NULL AND e.last_name IS NOT NULL) as named
     , count(*) filter (WHERE e.name IS NULL AND e.last_name IS NULL) as unnamed
  FROM employees e
 WHERE e.birth_date BETWEEN '2022-10-18 00:00:00' AND '2022-10-18 23:59:59'
;

| 已命名|未命名的|
| - -|- -|
| 三个|五个|
与联合不同，这不需要两个单独的查询。只需要对结果集进行一次传递。
但你要求把结果分开列，所以......

SELECT unnest(ARRAY [
         count(*) filter (WHERE e.name IS NOT NULL AND e.last_name IS NOT NULL)
       , count(*) filter (WHERE e.name IS NULL AND e.last_name IS NULL)
       ])
  FROM employees e
 WHERE e.birth_date BETWEEN '2022-10-18 00:00:00' AND '2022-10-18 23:59:59'
;

| 拆巢|
| - -|
| 三个|
| 五个|
这会将两个结果放入一个数组中，然后将结果取消嵌套到单独的行中，同时仍保持相同的查询效率。