【发布时间】:2015-08-24 10:16:00
【问题描述】:
我有两张桌子,一张是table1
id名称价格
1 个名字 1 100
2 名称2 200
table2
id authorityId 机构 start_date end_date
1 1 个机构1 1-2-2015 2-12015
我想在第二个表中搜索对我来说很好的值 但我想根据外键(table1_id)从第一个表中获得名称。我该怎么做。任何帮助将不胜感激?
这是我的代码
控制器:
public function searchResult()
{
$search_term = array(
'authorityId' => $this->input->get('authority'),
'grantVillage' => $this->input->get('village'),
'startDate' => $this->input->get('startDate'),
'endDate' => $this->input->get('endDate'));
//print_r($search_term);
$data['searchResult'] = $this->grant_model->searchResult($search_term);
$this->load->view('searchResult',$data);
}
型号:
public function searchResult($search_term)
{
$this->db->select('*');
$this->db->from('grant_data');
$this->db->like('authorityId', $search_term['authorityId']);
$this->db->like('grantVillage', $search_term['grantVillage']);
$this->db->like('startDate', $search_term['startDate']);
$this->db->like('certificate', $search_term['certificate']);
$this->db->like('endDate', $search_term['endDate']);
$query = $this->db->get();
return $query->result();
}
【问题讨论】:
标签: php ajax codeigniter