使用条件聚合：

SELECT type,
       COUNT(CASE value WHEN 0 THEN 1 END) / COUNT(*) * 100 AS percentage_of_0
FROM   table_name
GROUP BY type

其中，对于示例数据：

CREATE TABLE table_name (type, value) AS
SELECT 'a', 1 FROM DUAL UNION ALL
SELECT 'a', 0 FROM DUAL UNION ALL
SELECT 'a', 1 FROM DUAL UNION ALL
SELECT 'a', 1 FROM DUAL UNION ALL
SELECT 'b', 0 FROM DUAL UNION ALL
SELECT 'b', 1 FROM DUAL UNION ALL
SELECT 'c', 1 FROM DUAL UNION ALL
SELECT 'c', 1 FROM DUAL;

输出：
| 类型|占0的百分比|
| - ------|- ------|
| B|五十|
| （c）秘书长的报告|无|
| 项目a|二十五|
fiddle

如果0和1是仅有的两个可能的值，则可以使用聚合和算术，如下所示：

select type, avg(1 - number) * 100 as percentage_of_0s
from mytable t
group by type

如果有更多可能的值，并且您需要0值的百分比，那么我建议使用avg()和case表达式：

select type, 
    avg(case when number = 0 then 100 else 0 end) as percentage_of_0s
from mytable t
group by type