尝试以下查询

SELECT admission_date, count(admission_date) as adm_day,
 count(admission_date) - LAG(count(admission_date)) OVER(ORDER BY admission_date) AS admission_count_change 
FROM admissions
 group by admission_date

LAG（）是SQL中的一个窗口函数，它提供对当前行之前的指定偏移值处的行的访问。这意味着在LAG（）函数的帮助下，我们可以访问当前行上任何先前行的数据。
更多了解请参考这里：mysql window function和SQLshack

您可以使用

• GROUP BY获取每天的计数
• LAG窗口函数获取前一行的值

代码（Oracle语法）：

with tmp_dat(pk, admission_date) as (
  select 1, date '2018-06-06' from dual union all
  select 2, date '2018-06-06' from dual union all
  select 3, date '2018-06-07' from dual union all
  select 4, date '2018-06-08' from dual
),
tmp_aggr as (
  SELECT
    admission_date,
    count(admission_date) as adm_day
  from tmp_dat
  group by admission_date
)
select a.*, 
       lag(adm_day) over (partition by 1 order by admission_date) as prev_adm_day,
       adm_day - lag(adm_day) over (partition by 1 order by admission_date) as adm_day_change
 from tmp_aggr a
 order by admission_date;

就我个人而言，我更喜欢在应用程序端（或者使用查询结果的任何地方）处理计算，因为我不喜欢引用另一行本身的查询。
但是一个可能的解决方案是使用子选择，其中日期用于连接。
由于没有标记DBMS，因此我使用PostgreSQL 15。除了连接的日期计算外，它应该可以用于大多数DBMS。

SELECT 
   admissions.admission_date, 
   count(admissions.admission_date), 
   count(admissions.admission_date)-calc.prevday as "admission_count_change" FROM admissions
LEFT JOIN (
    SELECT 
       admission_date, 
       count(admission_date) as "prevday" 
    FROM admissions
    GROUP BY admission_date
) as calc ON admissions.admission_date = ( calc.admission_date - interval '1 day')
GROUP BY admissions.admission_date, calc.prevday;

左连接是必要的，这样最新的日期仍然会出现（admission_count_change null）。
我希望这个查询在大表中的伸缩性不是很好。
db<>fiddle to try it out yourself