SQL：如何显示通过和失败学生的百分比？

我有一个表格，里面有学生的数据和他们的成绩

ID  Name  Result
1   St1   FAIL
2   St2   PASS
3   St3   PASS
4   St4   PASS
5   St5   FAIL
6   St6   PASS

我希望通过和不通过考试的学生的总百分比四舍五入到小数点后2位。我如何才能做到这一点？

SQL:

SELECT 
    ROUND(
        SUM(
            CONVERT(
                FLOAT,
                CASE WHEN Result = 'PASS' THEN 1 
                     ELSE 0 
                     END
            )
        )
        /COUNT(*) * 100,
        2
    ) AS 'Percent Pass',
    ROUND(
        SUM(
            CONVERT(
                FLOAT,
                CASE WHEN Result = 'FAIL' THEN 1 
                     ELSE 0 
                     END
            )
        )
        /COUNT(*) * 100,
        2
    ) AS 'Percent Fail' 
FROM StudentsResults

结果:

Pass    Fail
66.67   33.33

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只是另一个没有计数的选项

范例

返回

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用GROUPBY子句怎么样

select round(cast(count(Name) as decimal)/cast((select count(*) from StudentPassFail) as decimal)*100,2), result from StudentPassFail 
group by result;

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我没有得到想要的输出。我得到的两个值都是0。请显示您执行的sql。选择roundcountName/从StudentPassFail*100中选择count*，2，结果从StudentPassFail按结果分组；Sry，忘记了十进制的类型转换。答案中已更正

Pass    Fail
66.67   33.33

select round(cast(count(Name) as decimal)/cast((select count(*) from StudentPassFail) as decimal)*100,2), result from StudentPassFail 
group by result;