[英]Submit form using ajax and pass the value submitted to new page
i have form that need the previous value inserted....i'm using ajax,but the success: function(data) wont let me moved to the next page... here is my code 我有需要插入先前值的表格....我正在使用ajax,但成功了:函数(数据)不会让我移至下一页...这是我的代码
HTML 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 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()
{
$('#tabelna').load('<?=base_url()?>something/something/something') (This is my mistake)
}
CONTROLLER CONTROLLER
function set_value()
{
extract($_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)
}
How can i pass the submitted value to the controller and shows new form ? 如何将提交的值传递给控制器并显示新表格?
we can call controller function using window.location 我们可以使用window.location调用控制器功能
function load()
{
window.location.href = "<?php echo site_url('controller_d/login/admin_link_delete_user');?>";
}
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