Question

我的表单需要插入以前的值....我使用ajax，但成功：函数（数据）不会让我转到下一页... 这是我的代码

HTML

<form>
  <input type="text" name="id_1" id="id_1>
  <input type="text" name="id_2" id="id_2>
  <input type="text" name="id_3" id="id_3>
  <button type="button" onclick="next();">
</form>
<div id="tabelna"></div>

JQuery的

var id_1 = $('#id_1').val();
var id_2= $('#id_2').val();
var id_3= $('#id_3').val();

var datana = 'id_1='+id_1+'&id_2='+id_2+'&id_3='+id_3; 
var urlna="<?=base_url()?>something/something/something";
$.ajax({
      type: 'POST',
      url: urlna,
      data: datana,
      beforeSend:function(data){
         },
          message:"<center>><h3>Loading Data. . .</h3></center>" 
          }); 
      },
      error: function(data) {
        jAlert('Failed');
      },
      success: function(data) {
        load();
      }    
  })
  return false;
  }
function load()
{
  $('#tabelna').load('<?=base_url()?>something/something/something') (This is my mistake)
}

CONTROLLER

function set_value()
{
 extract($_POST);
 $d['id1']  = $this-db->query('SELECT * FROM TBL1 where id='.$id_1);
 $d['id2']  = $this-db->query('SELECT * FROM TBL2 where id='.$id_2);
 $d['id3']  = $this-db->query('SELECT * FROM TBL3 where id='.$id_3);

 $this->load->view('something/v_add',$d);  (this is my mistake)   
}

如何将提交的值传递给控制器并显示新表单？

Answer 1

我们可以使用window.location调用控制器函数

function load()
{
window.location.href = "<?php echo site_url('controller_d/login/admin_link_delete_user');?>";
 }

使用ajax提交表单并将提交的值传递给新页面

1 个答案: