Question

请帮助我，我有问题使用ajax显示数据..我尝试更改变量，但那不起作用 enter image description here

查看此

<form>
        <select name="status" onchange="showStatus(this.value)" class="form-control" style="width: 125px">
        <option value="">Pilih Status:</option>
        <option value="1">Lulus</option>
        <option value="0">Tidak Lulus</option>
        </select>
        </form>

AJAX

  var status = $("#showStatus").change(function(){
  $.ajax({
      type:"POST",
      url:'<?php echo base_url("sales/status") ?>'+status,
      data:"salesstatus="+status,
      dataType:'json',
      success:function(data){
        $("#example1").html(data);
      },
      error:function(XMLHttpRequest){
        alert(XMLHttpRequest.responseText);
      }
  });
});

控制器

public function status()
{
    $user=$this->session->userdata('user');
    $status = $this->view->post('showStatus');
    $this->load->view('v_header');
    $data = $this->modelsales->status($status); 
    $this->load->view($this->folder.'/view', $data);
    $this->load->view('v_footer');
}

模型

function status($s)
{
    $this->db->where('salesstatus',$s);
    return $query->result();
}

查看全部

Answer 1

错误说明

首先：SqlConnection conn; string conectionStrign; public Inregistrare() { InitializeComponent(); } private void Inregistrare_Load(object sender, EventArgs e) { //geting the conectionstring conectionStrign = ConfigurationManager.ConnectionStrings["Atestat.Properties.Settings.bazaDeDate"].ConnectionString; } private void label1_Click(object sender, EventArgs e) { } private void button1_Click(object sender, EventArgs e) { //creating a conecton to the database conn = new SqlConnection(conectionStrign); conn.Open(); //query for inserting the data from the EditBox string query = "INSERT INTO Utilizatori (userName,password) VALUES (@nume,@parola)"; SqlCommand comanda = new SqlCommand(query, conn); //adding value for parameters in command comanda.Parameters.Add("nume", SqlDbType.VarChar).Value = numeDeUtilizator.Text; comanda.Parameters.Add("parola", SqlDbType.VarChar).Value = parola.Text; //execute the comand comanda.ExecuteReader(); conn.Close(); }您在这里没有提到身份证明。

第二次：<select name="status" onchange="showStatus(this.value)" class="form-control" style="width: 125px">获取下拉值的错误方法

我修复了这些错误

var status = $("#showStatus").change(function(){

Answer 2

id未定义首先定义这样的id。并且不要在选择字段中调用函数。试试这个

<form>
    <select name="status" id="showStatus" class="form-control" style="width: 125px">
    <option value="">Pilih Status:</option>
    <option value="1">Lulus</option>
    <option value="0">Tidak Lulus</option>
    </select>
    </form>

然后你可以尝试这样的脚本

$('#showStatus').change(function(){
var status= $('#showStatus').val();
if(status){
$.ajax({
        url: '<?php echo base_url();?>sales/status/' + status,
        success: function(response)
        {
            jQuery('#example1').html(response);
        }
    });
}
});

Answer 3

看起来你正面临着ajax响应的问题。您的电话可能无法获得响应html。您应该更改ajax响应的控制器代码。

$html = $this->load->view('v_header',array(),true);
$data = $this->modelsales->status($status); 
$html .= $this->load->view($this->folder.'/view', $data,true);
$html .= $this->load->view('v_footer',array(),true);

return $html; //echo $html;

如何使用ajax select显示数据

3 个答案: