【发布时间】:2020-09-19 08:55:36
【问题描述】:
我需要将一个函数的 id 传递给其他函数,我尝试过这种方式,但我无法得到结果。任何人都可以发现错误并给我一个解决方案。这是我正在使用的确切代码
public function splstaff($id)
{
$staff = DB::select("SELECT staff_id FROM special_labs WHERE sno={$id}");
$this->splstafflist($staff);
}
public function splstafflist($staff)
{
$data=DB::connection('mysql3')->select("SELECT MD5(CONCAT(a.staff_id,a.dob,a.cat_id)) staffid,a.staff_id,a.dep_id,
LOWER(CONCAT(a.legend,' ',a.staff_name))staff_name,a.photo,b.desi_name,IFNULL(c.degree,'')degree,b.desi_orderno,
IFNULL(MIN(sp.date_promotion),a.doj) doj FROM staff_personal_copy a LEFT JOIN staff_designationmaster b ON a.desi_id=b.desi_id
LEFT JOIN staff_qualification_view c ON c.staff_id=a.staff_id LEFT JOIN (camps.staff_promotion sp INNER JOIN staff_designationmaster sd
ON sp.to_designation=sd.desi_id AND sp.status>0 INNER JOIN staff_desigmaster sd1 ON sp.from_design=sd1.desi_id AND
sd1.desi_orderno!=sd.desi_orderno ) ON sp.staff_id=a.staff_id AND b.desi_orderno=sd.desi_orderno WHERE a.staff_id IN ('{$staff}')
AND a.status=1 AND a.cat_id=1 GROUP BY a.staff_id,sd.desi_orderno ORDER BY b.desi_orderno,doj,a.staff_id");
return view('research.labs.staff',['splfac'=>$data]);
}
【问题讨论】:
标签: php mysql laravel controller