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CodeIgniter get_where（4个答案）
昨天关门了。
因此，我尝试设置一个条件，如果$lowongan->status == 'active'，它将显示状态为active的数据行，但我得到了错误消息：Attempt to read property "status" on array我如何解决这个问题？？
模型：

public function show_lowongan(){
        return $this->db->get('lowongan');
    }

主计长：

public function lowongan()
    {
        $data['lowongan'] = $this->m_lowongan->show_lowongan()->result();
        $this->load->view('lowongan', $data);
    }

该视图：

<div class="row">
          <?php if($lowongan->status == "draft"){?>
            <?php
              $no = 1;
              foreach($lowongan as $lowong){
            ?>
              <div class="card" style="width: 18rem;">
                <div class="card-body">
                  <h5><?php echo $lowong->title?></h5>
                  <p><?php echo $lowong->lokasi?></p>
                  <p><?php echo $lowong->created_at?></p>
                  <p><?php echo $lowong->level_pekerja?></p>
                  <p><?php echo $lowong->pengalaman_kerja?></p>
                  <p><?php echo $lowong->pendidikan?></p>
                  <p><?php echo $lowong->alamat?></p>
                  <p><?php echo $lowong->no_wa?></p>
                  <p><?php echo $lowong->no_telp?></p>
                  <p><?php echo $lowong->min_gaji?></p>
                  <p><?php echo $lowong->max_gaji?></p>
                  <p><?php echo $lowong->job_desc?></p>
                  <button><a href="<?php echo base_url('lowongan/detail/'.$lowong->id)?>">Detail Lowongan</a></button>
                  <button><a href="<?php echo base_url('lowongan/edit/'.$lowong->id)?>">Edit Lowongan</a></button>
                  <button><a href="<?php echo base_url('lowongan/delete/'.$lowong->id)?>">Hapus Lowongan</a></button>
                </div>
              </div>
            <?php }?>
          <?php }else{?>
            Sorry No Job Posting
          <?php }?>
        </div>