将按钮值从视图传递给控制器(codeigniter)

Question

嗨，我有一个问题，当我传递按钮值…当我点击按钮时，我想要显示依赖于按钮的细节但我猜按钮值不会从视图传递到控制器。帮助…

//view(button)
<div class="marketer">
<?php foreach($marketer_list as $marketer){ ?>
<button  type="button"  class="mid" name= "posting_user_id" value="<?=$marketer->id?>"><?=$marketer->username?></button>
<?php }?> 
</div>

$(document).ready(function () {

$('.mid').click(function(){
var id = $(this).attr('value');
alert(id);
if (id=='<?=$detail->posting_user_id?>')
{
$(".show_posting").show(); 
}else{
alert("no info");
}
});
}); 



//controller
$posting_user_id = $this->input->post('posting_user_id');


$data['daily'] = $result;
$data['posting'] = $this->posting_models->get_manage_hospital_list($posting_user_id);
$this->load->view('posting',$data);

//model
function get_manage_hospital_list($posting_user_id){
$this->db->select('u.hosp_name,p.*');
$this->db->from('users as u');
$this->db->join('posting_plan as p', 'u.id=p.hospital_id');
$this->db->where('posting_user_id',$posting_user_id);
$this->db->order_by('p.report_day','asc');

return $this->db->get()->result();
}

Answer 1

您还没有为按钮添加id，该按钮有一个名为'mid'的类。别忘了在这里放一个唯一的id。您还必须检查$detail->posting_user_id是否在视图中有一个值。