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Struckuter 所以我想当我在 tproject 上输入 blablabla 时，从 id project 到 tproject_pic 的自动 id 但是当我使用 blabla 时，tproject_pic 上没有记录，这是错的吗？

控制器 

$data = array(
  'id' => '',
  'division' => $division,
  'client'=> $client,
  'siteid' => $siteid,
  'sitename' => $sitename,
  'lat' => $lat,
  'lng' => $lng,
  'location' => $location,
  'address' => $address,
  'description' => $description,
  'account' => $account,
  'datestart' => $datestart,
  'dateclose' => $dateclose,
  'pono' => $pono,
  'podate' => $podate,
  'povalue' => $povalue,
  'status' => $status,
  'status_remark' => $status_remark,
  'datestamp' => date('YmdHis')
);

$data1 = array (
  'pid' => $id,
  'uid' => $uid,
  'title'=> "user"
);

helper_log("add", "menambahkan data");
$this->form_validation->set_rules('siteid', 'Site ID', 'trim|required|alpha_dash|xss_clean|callback_checksite');
//$data['client_dropdown'] = $this->main_model->client_dropdown;
$this->session->set_flashdata('message','<div class="teal">Data telah di update</div>');
$this->main_model->check_site($siteid);
$this->main_model->input_data($data,'tproject');
$this->main_model->input_data($data1,'tproject_pic');
print_r($data1);
exit;
$data['dd_client'] = $this->main_model->dd_client();
$data['dd_lokasi'] = $this->main_model->dd_lokasi();
redirect('page/master');

模型

function input_data($data,$table){
  $this->db->insert($table, $data);
  $id = $this->db->insert_id();
  return (isset($id)) ? $id : FALSE;
}
4

1 回答 1

1

当您在项目表中成功插入数据时，将返回的值存储在变量中

$project_id=$this->main_model->input_data($data,'tproject');

并将该 ID 存储在 $data1 数组中：

$data1 = array (
'pid' => $project_id,
'uid' => $uid,
'title'=> "user"
);

并将其插入为：

$this->main_model->input_data($data1,'tproject_pic');
于 2018-01-29T09:05:41.960 回答