我进行了一次民意调查，结果来自用户的投票。例如：
您想要菲亚特吗？
id | answer
1 yes
2 no
3 yes
...
25 no
count = 20 yes / 5 no
(20 * 100) /25 = 80% Yes
(5 * 100) /25 = 20% No
因此，80％的人想要菲亚特，20％的人不想要。显然我可以做类似的事情：
select answer,count(*) as total from fast_survey group by answer;
但是，这将显示计数，我正在寻找相对百分比。知道我该怎么做吗？
SELECT round(count(*) FILTER (WHERE answer) * 100.0 / count(*),2) AS pct_yes,round(count(*) FILTER (WHERE NOT answer) * 100.0 / count(*),2) AS pct_no
FROM fast_survey;
pct_yes | pct_no --------+------- 80.00 | 20.00
db 提琴here
我乘以100.0（不是 ）以避免整数除法。结果是类型100 numeric，可以将其输入round()进行美化。参见：
假设answer为boolean。否则，请适应。
Postgres 9.4引入了聚合FILTER子句。参见：
应该尽可能快。
,您可以COUNT在分区上获取每种答案类型的值：
SELECT DISTINCT answer,COUNT(*) OVER (partition BY answer) AS total,COUNT(*) OVER (partition BY answer) * 100 /
COUNT(*) OVER () AS percentage
FROM fast_survey
如果您要提高百分比精度（对于上述查询，它是整数除法），请将第一个COUNT强制转换为FLOAT：
SELECT DISTINCT answer,CAST(COUNT(*) OVER (partition BY answer) AS FLOAT) * 100 /
COUNT(*) OVER () AS percentage
FROM fast_survey
怎么样？
SELECT
t1.answer,t1.votes / t2.total_votes
FROM
(SELECT answer,count(*) AS votes FROM fast_survey GROUP BY answer) AS t1,(SELECT count(*) AS total_votes FROM fast_survey) AS t2
;
如果您的调查有多个答案，这也将起作用。
,只需将其与您在其中查找所有记录的表连接即可，无需对答案进行除法并计算百分比...类似这样的事情
select answer,count(*) / max(totalCount) * 100 as total from fast_survey group by answer
left join (select count(*) FROM fast_survey as totalCount) as all on true