如何在控制器中获取用户文件的名称以添加到数据库

Question

这是到目前为止的代码。我有 2 张图片正在上传并添加到所需的地图中。我的问题是，如何从图像中获取两个名称，以便我可以将名称与其他表单信息一起添加到数据库中。

public 函数 CreateTypesForGamma() {

        $type              = new stdClass();

        $type->TypeID     = $this->input->post('typeID') == 0 ? null : $this->input->post('typeID');
        $type->Titel        = $this->input->post('titel');
        $type->TechnischeFiche      = $this->input->post('technischeFiche');
        $type->GammaID      = $this->input->post('gammaID');
        $type->ExtLink      = $this->input->post('extLink');
        $type->InfoNL       = $this->input->post('infoNL');
        $type->InfoFR       = $this->input->post('infoFR');
        $type->InfoEN       = $this->input->post('infoEN');

        $files = $_FILES;
            $cpt = count($_FILES['userfile']['name']);
             for($i=0; $i<$cpt; $i++)
            {
            $_FILES['userfile']['name']= $files['userfile']['name'][$i];
            $_FILES['userfile']['type']= $files['userfile']['type'][$i];
            $_FILES['userfile']['tmp_name']= $files['userfile']['tmp_name'][$i];
            $_FILES['userfile']['error']= $files['userfile']['error'][$i];
            $_FILES['userfile']['size']= $files['userfile']['size'][$i];
            $this->upload->initialize($this->set_upload_options());
            $this->upload->do_upload();
            $fileName = $_FILES['userfile']['name'];
            if ($i == 0)
             $images[] = $fileName;
            }
              $fileName = implode(',',$images);
              $this->gamma_model->upload_image($fileName);
            system.out.println($images);
        if ($type->TypeID == 0) {
            $type->TypeID = $this->type_model->insert($type);
        } else {
            $this->type_model->update($type);
        }
        redirect('gamma/viewAdminGamma');

}
private function set_upload_options()
    { 
      // upload an image options
        $config = array();
        $config['upload_path'] = APPPATH . 'images/types'; //give the path to upload the image in folder
        $config['allowed_types'] = 'gif|jpg|png';
        $config['max_size'] = '5120';
        $config['overwrite'] = FALSE;
        return $config;
    }

Answer 1

将代码更改为以下

 $cpt = count($_FILES['userfile']['name']);
             for($i=0; $i<$cpt; $i++)
            {
            $_FILES['userfile']['name']= $files['userfile']['name'][$i];
            $_FILES['userfile']['type']= $files['userfile']['type'][$i];
            $_FILES['userfile']['tmp_name']= $files['userfile']['tmp_name'][$i];
            $_FILES['userfile']['error']= $files['userfile']['error'][$i];
            $_FILES['userfile']['size']= $files['userfile']['size'][$i];
            $this->upload->initialize($this->set_upload_options());
            $this->upload->do_upload();
            $fileName = $_FILES['userfile']['name'];

             $images[] = $fileName;
             $type->TypeFoto=$fileName[0];// this gives name of first            
             image 
             $type->Lastentabel =$fileName[1];
            }
              $fileName = implode(',',$images);
              $this->gamma_model->upload_image($fileName);
            system.out.println($images);
        if ($type->TypeID == 0) {
            $type->TypeID = $this->type_model->insert($type);
        } else {
            $this->type_model->update($type);
        }
        redirect('gamma/viewAdminGamma');

}
private function set_upload_options()
    { 
      // upload an image options
        $config = array();
        $config['upload_path'] = APPPATH . 'images/types'; //give the path to upload the image in folder
        $config['allowed_types'] = 'gif|jpg|png';
        $config['max_size'] = '5120';
        $config['overwrite'] = FALSE;
        return $config;
    }

如何在控制器中获取用户文件的名称以添加到数据库

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