Question

我正在做一份报告，以显示我们每周从学校获得的申请人数量。我们在15所不同的学校招聘。我目前的疑问是这样的。问题是，如果学校没有获得申请人，那么他们就不会出现在报告中。即使没有申请人，我怎么能让学校出现？

SELECT 
weekofyear(tbl_job_applications.ja_date) AS 'Week #',
tbl_universities.univ_name,
Count(tbl_job_applications.ja_date) AS 'Applicants'
FROM
tbl_job_applications
    Inner JOIN
tbl_universities ON tbl_job_applications.univ_id = tbl_universities.univ_id
    Inner JOIN
tbl_positions ON tbl_job_applications.pos_id = tbl_positions.pos_id
where year(tbl_job_applications.ja_date) = 2014
Group by tbl_universities.univ_name , weekofyear(tbl_job_applications.ja_date)
Order by weekofyear(tbl_job_applications.ja_date) DESC , tbl_universities.univ_name ASC;

Answer 1

使用LEFT JOIN或RIGHT JOIN。 INNER JOIN排除了非加入值，您的WHERE条件也需要更改为不排除没有ja_date值的记录：

SELECT  weekofyear(a.ja_date) AS 'Week #'
      , u.univ_name
      , COUNT(a.ja_date) AS 'Applicants'
FROM    tbl_universities u
LEFT JOIN tbl_job_applications a
    ON a.univ_id = u.univ_id
LEFT JOIN tbl_positions p
    ON a.pos_id = p.pos_id
WHERE   YEAR(a.ja_date) = 2014  OR a.ja_date IS NULL
GROUP BY u.univ_name
      , weekofyear(a.ja_date)
ORDER BY weekofyear(a.ja_date) DESC
      , u.univ_name ASC;

注意，这会返回所有univ_name值，但它不会每周返回univ_name个值，如果这是您需要的后的值要创建具有所有周和univ_name组合的驱动程序，最简单的方法是使用子查询中的CROSS JOIN（在此示例中，它仅限于至少有一个学校的申请人数周，如果你希望所有几周都不得不以不同的方式进行设置：

SELECT sub.*,COUNT(a.ja_date)
FROM (SELECT DISTINCT
             weekofyear(a.ja_date) AS week_
            ,u.univ_name
            ,u.univ_id      
      FROM tbl_job_applications a
      CROSS JOIN tbl_universities u
      ) as sub
LEFT JOIN tbl_job_applications a
    ON a.univ_id = sub.univ_id
    AND weekofyear(a.ja_date) = sub.week_
LEFT JOIN tbl_positions p
    ON a.pos_id = p.pos_id
GROUP BY sub.univ_name
        ,sub.week_
ORDER BY week_
         ,sub.univ_name

两者的演示：SQL Fiddle

Answer 2

更改查询，以便首先列出大学并使用外部联接：

FROM tbl_universities 
LEFT JOIN tbl_job_applications
  ON tbl_job_applications.univ_id = tbl_universities.univ_id
INNER JOIN tbl_positions 
 ON tbl_job_applications.pos_id = tbl_positions.pos_id

GROUP BY显示0

2 个答案: