MySql的选择where列值是唯一

Question

我的表people_jobs：

+--------+--------+ 
| NAME | JOB | 
+--------+--------+ 
| John | Actor | 
+--------+--------+ 
| Jane | Driver | 
+--------+--------+ 
| Bill | Actor | 
+--------+--------+ 
| John | Cook | 
+--------+--------+ 

我希望选择具有职业演员，其中名称列将是独特的全名。这里所需的查询输出将只是Bill。

喜欢的东西：

SELECT name FROM people_jobs WHERE job LIKE "actor" AND COUNT(SELECT * FROM people_jobs WHERE name LIKE name) = 1; 

这显然是错误语法，我不能让GROUP BY工作...思考？

Answer 1

为什么选择？

SELECT name, COUNT(*) AS cnt 
FROM people_jobs 
WHERE job='Actor' 
GROUP BY name 
HAVING cnt = 1 

---

好了，现在我看到的问题。试试这个：

SELECT name, SUM(JOB='Actor') AS actor_cnt, COUNT(*) as job_cnt 
FROM people_jobs 
GROUP BY name 
HAVING (actor_cnt = 1) AND (job_cnt = 1) 

这会找出有多少人是演员，并计算有多少作业有，并且只返回其唯一的工作人行事。

Answer 2

SELECT name, COUNT(*) c 
FROM people_jobs 
WHERE job = 'Actor' 
GROUP BY name HAVING c = 1 

Answer 3

select name 
from people_jobs 
where name in (
    select name 
    from people_jobs 
    group by name 
    having count(name) = 1 
) and job like 'actor' 

这个例子是简单的理解，但我喜欢另一个问题：

SELECT name 
FROM people_jobs 
GROUP BY name 
HAVING COUNT(name) = 1 AND sum(job = 'Actor') = 1 

Answer 4

SELECT DISTINCT x.* 
      FROM people_jobs x 
      LEFT 
      JOIN people_jobs y 
      ON y.name = x.name 
      AND y.job <> x.job 
      WHERE x.job = 'actor' 
      AND y.name IS NULL; 

Answer 5

我会用NOT EXISTS在这种情况下：

SELECT name 
FROM people_jobs 
WHERE job = 'actor' 
     AND NOT EXISTS (SELECT * FROM people_jobs pj 
         WHERE pj.name = people_jobs.name AND pj.job != 'actor') 

Answer 6

更容易并解决你的问题：

SELECT distinct name FROM people_jobs WHERE job = "Actor"