COUNT 与 IF FROM DISTINCT 值 MySQL答案

【问题标题】：COUNT with IF FROM DISTINCT values MySQLCOUNT 与 IF FROM DISTINCT 值 MySQL
【发布时间】：2019-09-16 23:24:10
【问题描述】：

在对 UserID 执行 DISTINCT 之后，我试图计算有多少用户是女性和多少男性

SELECT
COUNT(IF("Gender" = 'female', 1, NULL)) as Ufemale,
COUNT(IF("Gender" = 'male', 1, NULL)) as Umale
FROM (SELECT DISTINCT UserID FROM user_stats where Year='2019' and Account='P') as UID

如果我执行

SELECT DISTINCT UserID FROM user_stats where Year='2019' and Account='P'

它返回唯一的用户 ID。但是，如果我将它与计数性别部分结合起来，它会返回零。

这是值的样子

UserID            |  Gender
-----------------------------
2018359084885123  |  male
1925823664195671  |  female
2033134076795519  |
2122445674469149  |  female
2315129265210413  |  female
2018359084885123  |  male
2122445674469149  |  female

目的是展示

Ufemale  |  Umale
-------------------
   3     |    1

【问题讨论】：

你可以把SELECT DISTINCT UserID FROM user_stats where Year='2019' and Account='P'改成SELECT DISTINCT UserID, gender FROM user_stats where Year='2019' and Account='P'
“我正在尝试计算女性用户数和男性用户数” 那么基于当前数据的预期结果将是 4 和 2？此外，您的示例数据缺少年份和帐户列。

标签： mysql count distinct-values

【解决方案1】：

试试这个：

表格

drop table if exists test;

create table test (userid bigint, gender char(6));

insert into test values
(2018359084885123,'male')
,(1925823664195671,'female')
,(2033134076795519, null)
,(2122445674469149,'female')
,(2315129265210413,'female')
,(2018359084885123,'male')
,(2122445674469149,'female');

查询

select
    sum(case when gender = 'female' then 1 else 0 end) as ufemale,
    sum(case when gender = 'male' then 1 else 0 end) as umale
from
(select distinct userid, gender from test) x

结果

ufemale umale
3       1

示例：https://rextester.com/JLQ19855

【讨论】：

【解决方案2】：

子查询结果中没有Gender列：

SELECT DISTINCT UserID FROM user_stats where Year='2019' and Account='P'

如你所想。
你正在做的是比较 2 个字符串："Gender" 和 'female' 这显然不相等，所以这个：

IF("Gender" = 'female', 1, NULL)

返回NULL 和COUNT(NULL) 返回0。
Gender 列在哪里？
它是在user_stats 表中还是在另一个表中，例如Users？
如果它在user_stats 中，那么您需要像这样编写查询：

select
  sum(t.gender = 'female') ufemale,
  sum(t.gender = 'male') umale
from (
  select distinct userid, gender 
  from user_stats 
  where Year='2019' and Account='P'
) t

如果它在users 表中，那么您首先需要一个连接：

select
  sum(u.gender = 'female') ufemale,
  sum(u.gender = 'male') umale
from (
  select distinct userid, gender 
  from user_stats 
  where Year='2019' and Account='P'
) t inner join users u
on u.userid = t.userid

【讨论】：