Question

需要在隐藏字段中获取所选值的id，但它始终返回undefined。控制器动作和脚本如下。

我在这里做错了什么。我完全是这个UI自动完成的新手。

所以需要帮助解决这个问题。

('#PanelSearchKeyword').autocomplete({
    source: function(request, response) {
        var params = {};
        params.supplieTyperName = document.getElementById('PanelSearchKeyword').value;
        $.ajax({
            type: 'POST',
            dataType: "json",
            url: '/Home/GetSupplierTypeNameList',
            data: AddAntiForgeryToken(params),
            success: function(data) {
                var array = data.map(function(element) {
                    return {
                        value: element['SupplierTypeName'],
                        id: element['SupplierTypeId']
                    };
                });
                response(array);
            },
            error: function(xhr, ajaxOptions, thrownError) {
                logError(ajaxOptions, thrownError);
            }
        });
    },
    select: function(event, ui) {
        $("#SuppTypeId").val(ui.item.SupplierTypeId); // save selected id to hidden input

        var a = $("#SuppTypeId").val();
        alert(a);
        //return false;
    },
    minLength: 2
});

我的控制器动作看起来像

[HttpPost]
    [ValidateAntiForgeryToken]
    public JsonResult GetSupplierTypeNameList(string supplieTyperName)
    {
        try
        {
             List<SupplierTypeViewModel> supplierTypeNameList = new List<SupplierTypeViewModel>();
             supplierTypeNameList = (from supplierType in db.PPSupplierTypes
                                    where   supplierType.PPSupplierTypeName.ToUpper().StartsWith(supplierTypeName.ToUpper())
                                    orderby supplierType.PPSupplierTypeName
                                    select new SupplierTypeViewModel
                                    {
                                        SupplierTypeId = supplierType.PPSupplierTypeId,
                                        SupplierTypeName = supplierType.PPSupplierTypeName
                                    }).ToList();
            return supplierTypeNameList;
        }
        catch (Exception)
        {
            throw;
        }
        return Json(suppplierTypeResult, JsonRequestBehavior.AllowGet);
    }

Answer 1

就我迄今为止所做的一样........ 在视图文件

中

         public function search() {
    if (isset($_POST['userA'])) {
        $userPart = $_POST['userA'];
    } else {
        $userPart = "";
    }

    $result = $this->dbmodel->search($userPart);
    $list = array();

    foreach ($result as $finaldata) {
        $dataN = $finaldata->name;
        $dataK = $finaldata->id;
        array_push($list, $dataN);
    }

    if ($userPart != "" && $userPart != NULL) {
        echo json_encode($result);
    }
}

在控制器中

        function search($value) {
    $this->db->select('id, name');
    $this->db->where("status", "1");
    $this->db->where("verification_status", "1");
    $this->db->where("suspension_status", "0");
    $where = "(name LIKE '%$value%' 
      OR address LIKE '%$value%' OR contact LIKE '%$value%')";
    $this->db->where($where);
    $result = $this->db->get('hotel_info');

    return $result->result();
}

和dbmodel

RootAction

希望它能帮到你

Jquery UI - 自动完成自定义错误

1 个答案: