SELECT TO_CHAR(a_date_time,'FMDay') day_of_week
      ,COUNT(*) number_of_appointments
  FROM appointment
 GROUP BY TO_CHAR(a_date_time,'FMDay')
 ORDER BY number_of_appointments ASC
 FETCH FIRST ROW WITH TIES

| 星期几|任命人数|
| - ------|- ------|
| 星期日|1个|
| 星期三|1个|
| 星期六|1个|

您可以生成一周中所有日期的列表，然后对约会表执行LEFT OUTER JOIN，然后按约会数对行执行ORDER，并在Oracle 12中执行FETCH FIRST ROW WITH TIES以查找最小值：

WITH days_of_week (day) AS (
  SELECT TO_CHAR(TRUNC(SYSDATE, 'IW') + LEVEL - 1, 'DY')
  FROM   DUAL
  CONNECT BY LEVEL <= 7
)
SELECT d.day,
       COUNT(a.app_id) AS num_appointments
FROM   days_of_week d
       LEFT OUTER JOIN appointment a
       ON d.day = TO_CHAR(a.A_DATE_TI, 'DY')
GROUP BY d.day
ORDER BY
       num_appointments ASC
FETCH FIRST ROW WITH TIES;

其中，对于示例数据：

CREATE TABLE appointment (APP_ID, A_DATE_TI, PET_ID, VN_ID) AS
SELECT  1, DATE '2023-01-20', 10001, 801 FROM DUAL UNION ALL
SELECT  2, DATE '2023-01-20', 10002, 803 FROM DUAL UNION ALL
SELECT  3, DATE '2023-01-20', 10003, 804 FROM DUAL UNION ALL
SELECT  4, DATE '2023-01-20', 10004, 803 FROM DUAL UNION ALL
SELECT  5, DATE '2023-01-15', 10005, 801 FROM DUAL UNION ALL
SELECT  6, DATE '2023-01-14', 10006, 803 FROM DUAL UNION ALL
SELECT  7, DATE '2023-01-13', 10007, 804 FROM DUAL UNION ALL
SELECT  8, DATE '2023-01-12', 10008, 803 FROM DUAL UNION ALL
SELECT  9, DATE '2023-02-01', 10009, 801 FROM DUAL UNION ALL
SELECT 10, DATE '2023-02-02', 10010, 803 FROM DUAL UNION ALL
SELECT 11, DATE '2023-02-03', 10011, 804 FROM DUAL;

输出：
| 日|任命人数|
| - ------| - ------|
| 星期二|无|
| 周一|无|
fiddle

您可以使用以下命令从日期列中确定星期几

TO_CHAR(date A_DATE_TI, 'DAY')

因此，您可以使用此查询来查找每个工作日发生了多少约会。

SELECT COUNT(*) appointments, 
       TO_CHAR(A_DATE_TI, 'DAY') weekday
  FROM appointment
 GROUP BY TO_CHAR(A_DATE_TI, 'DAY')

这将查找约会最多的一天。

SELECT COUNT(*) appointments, 
       TO_CHAR(A_DATE_TI, 'DAY') weekday
  FROM appointment
 GROUP BY TO_CHAR(A_DATE_TI, 'DAY')
 ORDER BY COUNT(*) DESC
 LIMIT 1