Question

我正在尝试创建一个查询，该查询将返回特定公司发布的所有作业，以及应用于此作业的总人数。第一部分工作正常 - 我得到了所有的工作和我需要的一切：

$query = "SELECT *,j.job_id as jid, c.name as city_name ".
         "FROM jobs j JOIN areas a ON a.area_id = j.job_area ".
         "JOIN positions p ON p.position_id = j.job_position ".
         "JOIN fields f ON f.id = j.job_field ".
         "JOIN cities c ON j.job_city = c.id ".
         "JOIN jobTypes jt ON j.job_type = jt.job_id " .
         "JOIN companies comp ON j.job_company = comp.company_id ".
         "LEFT JOIN jobApplications ja ON ".
              "ja.user_id = '".$_SESSION['user_id']."' AND ".
              "j.job_id = ja.job_id WHERE j.job_company='$company_id'";

问题是，我想从jobApplications表中添加每个结果行的作业申请人数...我试图在查询中添加COUNT列，这本身很好用：

SELECT COUNT(*) FROM jobApplications ja WHERE ja.job_id=j.job_id

当添加到第一个大查询时，我甚至无法在语法级别上使其工作，所以我不确定它是否可以正常工作...

我尝试将最后一个查询添加到主查询的选择区域，但我总是在计数查询结束时'ja.job_id = j.job_id'后面出现语法错误...

这甚至可能吗？

我希望问题很明确，我知道这里有很多表...... 感谢您的时间和帮助！

Answer 1

试试这样：

$query = "SELECT (SELECT COUNT(ja2.job_id) FROM jobApplications ja2 WHERE ja2.job_id=j.job_id group by j.job_id), *,j.job_id as jid, c.name as city_name FROM jobs j JOIN areas a ON a.area_id = j.job_area" .
" JOIN positions p ON p.position_id = j.job_position JOIN fields f ON f.id = j.job_field "
 ." JOIN cities c ON j.job_city = c.id JOIN jobTypes jt ON j.job_type = jt.job_id " .
 "JOIN companies comp ON j.job_company = comp.company_id LEFT JOIN jobApplications ja ON ja.user_id = '".$_SESSION['user_id']."' AND j.job_id = ja.job_id WHERE j.job_company='$company_id'";

Answer 2

我不知道你的PK应用程序，但这可能有效。

$query = "SELECT *,j.job_id as jid, c.name as city_name, COUNT(ja.<primary key>) ".
     "FROM jobs j JOIN areas a ON a.area_id = j.job_area ".
     "JOIN positions p ON p.position_id = j.job_position ".
     "JOIN fields f ON f.id = j.job_field ".
     "JOIN cities c ON j.job_city = c.id ".
     "JOIN jobTypes jt ON j.job_type = jt.job_id " .
     "JOIN companies comp ON j.job_company = comp.company_id ".
     "LEFT JOIN jobApplications ja ON ".
          "ja.user_id = '".$_SESSION['user_id']."' AND ".
          "j.job_id = ja.job_id WHERE j.job_company='$company_id' ".
     "GROUP By jid";

Answer 3

您必须使用GROUP

SELECT *,j.job_id as jid, c.name as city_name, COUNT(jobApplications.*)
FROM jobs j JOIN areas a ON a.area_id = j.job_area 
JOIN positions p ON p.position_id = j.job_position 
JOIN fields f ON f.id = j.job_field 
JOIN cities c ON j.job_city = c.id 
JOIN jobTypes jt ON j.job_type = jt.job_id 
JOIN companies comp ON j.job_company = comp.company_id 
LEFT JOIN jobApplications ja ON 
ja.user_id = '".$_SESSION['user_id']."' AND 
j.job_id = ja.job_id WHERE j.job_company='$company_id'
GROUP BY jid

希望有所帮助：）

更大的查询中的MySQL COUNT语法

3 个答案: