这就是如何使用where条件的codeigniter分组。你可以相应地分组。

public function get_onboarded_detail() {

    $supplier_id = 40;
    $query = $this->db->select('*')
                  ->from('job')
                  ->join('job_workforce', 'job_workforce.job_id = job.id', 'Right')
                  ->group_start()
                      ->where('job_workforce.supplier_id',$supplier_id)
                      ->where('job_workforce.status','onboarded')
                  ->group_end()
                  ->group_start()
                      ->or_where('job_workforce.status','job_offer')
                      ->or_where('job_workforce.status','offer_accepted_by_client')
                      ->or_where('job_workforce.status','offer_accepted_by_supplier')
                  ->group_end()
                  ->get();
         $res =  $query->result();          
}

根据查询，条件之一是如果$supplier\u id的值不匹配，则不应显示任何结果。但是，无论供应商id的值是多少，它都会显示所有结果。
如果您正在使用条件或条件之间，即使任何一个条件为真，它也将返回输出。如果要使用提供的所有条件筛选结果，请使用和。

where条件的实际问题是因为它包含“or”条件，所以不管您的供应商id系统返回什么样的结果。见下文example:-
你现在在哪里condition:-

WHERE job_workforce.supplier_id = 40 AND job_workforce.status = 'onboarded' OR job_workforce.status = 'job_offer' OR job_workforce.status = 'offer_accepted_by_client' OR job_workforce.status = 'offer_accepted_by_supplier'

应该是：-

WHERE job_workforce.supplier_id = 40 AND (job_workforce.status = 'onboarded' OR job_workforce.status = 'job_offer' OR job_workforce.status = 'offer_accepted_by_client' OR job_workforce.status = 'offer_accepted_by_supplier')

要使用codeigniter进行更正，请使用：

public function get_onboarded_detail()
{
    $supplier_id = 40;
    $query = $this->db->select('*')

      ->from('job')
      ->join('job_workforce', 'job_workforce.job_id = job.id', 'Right')
      ->where("job_workforce.supplier_id = $supplier_id AND (job_workforce.status = 'onboarded' OR job_workforce.status = 'job_offer' OR job_workforce.status = 'offer_accepted_by_client' OR job_workforce.status = 'offer_accepted_by_supplier')");

      ->get();
      $res =  $query->result();          
}