用笨從形式

Question

我的文章數據更新到mysql數據庫多個值就要爲

Array 
(
    [radiogroup_1] => 1 
    [radiogroup_2] => 4 
    [radiogroup_4] => 11 
    [submit] => submit 
) 

和我的模型是

function add_votes(){ 

    foreach($POST[] as $k=> $v){ 
    $id = $v; 
    } 

    $this->db->where('answerid','radiogroup_'. $id); 

    $this->db->set('votes', 'votes+1',FALSE); 
    $this->db->update('vote_table'); 


} 

顯然，它不工作。

Answer 1

也許

function add_votes(){ 
    foreach($_POST[] as $k=> $v){ 
     $this->db->where('answerid','radiogroup_'. $v); 
     $this->db->set('votes', 'votes+1',FALSE); 
     $this->db->update('vote_table'); 
    } 
} 

或

function add_votes(){ 
    foreach($_POST[] as $k=> $v){ 
     $this->db->where('answerid', $v); 
     $this->db->set('votes', 'votes+1',FALSE); 
     $this->db->update('vote_table'); 
    } 
} 

Answer 2

不要需要的foreach更新。只需使用更新表格一次。

function add_votes(){ 
$radiogroup_1=$this->input->post('radiogroup_1'); 
$radiogroup_2=$this->input->post('radiogroup_2'); 
$radiogroup_4=$this->input->post('radiogroup_4'); 
$data=array('radiogroup_1'=>$radiogroup_1,'radiogroup_2'=>$radiogroup_2,'radiogroup_4'=>$radiogroup_4); 
$this->db->where('answerid','radiogroup_'. $id); 
$result=$this->db->update('vote_table',$data); 
}