FIND_IN_SET有两个字符串

问题描述

我有这个雇员列表的EMPLOYEE表

I have this EMPLOYEE table of employees list

+-----+---------------+-------------+
| ID  |EMPLOYEE_ID    | SKILLS      |
+-----+---------------+-------------+
|  1  |       1       |   3,4       |
+-----+---------------+-------------+
|  2  |       2       |   3,5,2     |
+-----+---------------+-------------+
|  3  |       3       |  1,5        |
+-----+---------------+-------------+

和表POSTED_JOB列出作业

and table POSTED_JOB listing jobs

+-----+---------------+-------------+
| ID  |POSTED_JOB_ID  |  JOB_SKILLS |
+-----+---------------+-------------+
|  1  |       1       |   1,2,3     |
+-----+---------------+-------------+
|  2  |       2       |   3,4       |
+-----+---------------+-------------+
|  3  |       3       |   5,4       |
+-----+---------------+-------------+
|  4  |       4       |   5,6       |
+-----+---------------+-------------+

我如何通过laravel查询发布具有与员工技能相对应的技能的所有职位.

How can I get all jobs posted with skills corresponding to the skills of employees with laravel query.

例如，对于具有employee_id 1的员工，工作将是1,2和3.

For example for employee with employee_id 1, the jobs would be 1,2, and 3.

我尝试过使用find_in_set，但这两个都是列表. DB :: raw("find_in_set(EMPLOYEE.SKILLS，POSTED_JOB.JOB_SKILLS)")，DB :: raw('')，DB :: raw(''))

I tried with find_in_set but here both are lists. DB::raw("find_in_set(EMPLOYEE.SKILLS , POSTED_JOB.JOB_SKILLS)"), DB::raw(''), DB::raw(''))

推荐答案

$skills = 'select the employee skills';
$skl_arr = explode(',',$skills);
$skl_length = count($skl_arr); 

/*查询*/

$rows->orwhere(DB::raw("find_in_set('$skl_arr[0]','post_job.skills')"));

for ($i=1; $i < $skl_length ; $i++) { 
                $rows->$join->on(DB::raw("find_in_set('$skl_arr[$i]','post_job.skills')",DB::raw(''),DB::raw('')));

}

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