![](/img/trans.png)
[英]How to get values from another table in codeigniter using the Active Record?
[英]How to search for multiple values in a single field of mysql table using codeigniter active record?
我必須在codeigniter中使用mysql在一個字段中搜索多個值。 這是我的代碼。
在控制器中
public function vpsearch()
{
$data['info'] = $this->psearch_m->emp_search_form();
$this->load->view("employer/result",$data);
}
IN模型
public function emp_search_form()
{
$skill = $this->security->xss_clean($this->input->post('ps_skills'));
$jrole = $this->input->post('ps_jobrole'));
if ( $jrole !== NULL)
{
return $this->db->get('js_edu_details');
$this->db->like('js_skills','$skill');
}
}
考慮到 (../雇主/結果)
foreach($info->result() as $row)
{
echo $row->js_id."<br/><br/>" ;
}
但是,我正在獲取“ js_edu_details”表中的所有記錄,而不是搜索“技能”的字段。
我要去哪里錯了? 任何幫助,謝謝,謝謝。
嘗試:
public function emp_search_form()
{
$skill = $this->security->xss_clean($this->input->post('ps_skills'));
//$skill = $this->input->post('ps_skills', true); other short way of getting the above result with `xss clean`
if ( $jrole !== NULL)
{
$this->db->like('js_skills',$skill); #remove the single quote around the `$skill`
$res = $this->db->get('js_edu_details');
echo $this->db->last_query(); #try to print the query generated
return $res;
}
}
Return
語句應在like
語句之后
您應該像這樣正確安排代碼
public function emp_search_form()
{
$ps_skills = $this->input->post('ps_skills')
$skill = $this->security->xss_clean($ps_skills);
if ( $jrole !== NULL)
{
$this->db->like('js_skills','$skill');
return $this->db->get('js_edu_details');
}
}
您還應該注意,條件永遠不會滿足。 它將始終給出錯誤的undefined variable $jrole
聲明:本站的技術帖子網頁,遵循CC BY-SA 4.0協議,如果您需要轉載,請注明本站網址或者原文地址。任何問題請咨詢:yoyou2525@163.com.