我正在创建一个可以根据MVC中的条件访问的表单。我有第一个带下拉列表和提交按钮的视图,我想要在单击提交按钮时,传递下拉列表中的值并与为该值设置的条件进行比较,如果条件不正常,则显示警报而不是处理到形式。
这是我的代码:
public ActionResult ChooseType()
{
var x = DataAccess.GetEmployee(@User.Identity.Name);
var lists = new SelectList(RuleAccess.GetAllRule(), "ID", "TypeDetail");
ViewBag.CategoryId = lists;
/*rule*/
ViewBag.comp1 = Logic.AnnualToogle(@User.Identity.Name);
if (x.EmpSex == "F" && x.EmpMaritalSt == "NIKAH")
{ ViewBag.comp2 = 1; }
else ViewBag.comp2 = 0;
return View();
}
[HttpGet]
public ActionResult Create(int lv_type)
{
var type = RuleAccess.GetTypeByID(lv_type);
ViewBag.type = type;
var model = new LeaveApplicationViewModels();
model.X = DataAccess.GetEmployee(@User.Identity.Name);
model.C = DataAccess.GetLeaveApp(@User.Identity.Name);
/*disable*/
ViewBag.dis = DataAccess.GetDisabledDate(@User.Identity.Name);
/*max*/
var max= RuleAccess.GetMaxByID(lv_type);
ViewBag.c = max;
if (lv_type == 1)
{
var used = RuleAccess.CountUsedAnnual(@User.Identity.Name);
var rem = max - used;
ViewBag.a = used;
ViewBag.b = rem;
}
else
{
ViewBag.b = max;
}
return View(model);
}
我使用了Viewbag.comp 1& 2在我看来:
<script type="text/javascript">
var x = @ViewBag.comp1;
var y = @ViewBag.comp2;
function validatecreate()
{
var value= document.getElementById("lv_type").value;
if (value==1)
{
if(x==1)
document.getElementById('validatecreate').submit();
else { alert('Action cant be done. You either have another annual leave application in pending status or you have reach the limit of annual leave'); }
}
else if(value==2)
{
if(y==1)
document.getElementById('validatecreate').submit();
else { alert('Action cant be done. You either are Male or Not Married Yet'); }
}
else if(value==3)
{
document.getElementById('validatecreate').submit();
}
else {
document.getElementById('validatecreate').submit();
//alert('Http Not Found');
}
}
@Html.DropDownList(
"lv_type", (SelectList) ViewBag.CategoryId,
"--Select One--",
new{ //anonymous type
@class = "form-control input-sm"
}
)
我觉得我做错了,特别是因为如果有人手动将网址设置为?lv_type = 2,他们就不会验证并且可以直接转到表单。但我需要lv_type bcs的值,我在我的视图中使用它。请赫尔普:(
答案 0 :(得分:1)
验证必须始终在服务器上完成,客户端验证应该只被认为是一个很好的奖励,可以最大限度地减少对服务器的调用需求。并在下拉列表中向用户显示选项,然后告诉他们不能选择该选项是一种糟糕的用户体验。相反,您应该只显示适用于用户的那些选项(并删除您显示的所有脚本)。
在RuleAccess
课程中创建一个额外的方法,比如说GetEmployeeRules(Employee employee)
,它只返回适用于该员工的规则,例如
public static List<Rule> GetEmployeeRules(Employee employee)
{
// Get the list of all rules
if (employee.EmpSex == "F" && employee.EmpMaritalSt == "NIKAH")
{
// Remove the appropriate Rule from the list
}
.....
// Return the filtered list
}
此外,您应该在视图中使用视图模型
public class LeaveTypeVM
{
[Required(ErrorMessage = "Please select a leave type")]
public int SelectedLeaveType { get; set; }
public IEnumerable<SelectListItem> LeaveTypeList { get; set; }
}
然后在ChooseType()
方法
public ActionResult ChooseType()
{
var employee = DataAccess.GetEmployee(@User.Identity.Name);
var rules = RuleAccess.GetEmployeeRules(employee);
var model = new LeaveTypeVM()
{
LeaveTypeList = new SelectList(rules, "ID", "TypeDetail")
};
return View(model);
}
并在视图中
@model LeaveTypeVM
@using (Html.BeginForm())
{
@Html.DropDownListFor(m => m.SelectedLeaveType, Model.LeaveTypeList, "--Select One--", new { @class = "form-control input-sm" }
@Html.ValidationMessageFor(m => m.SelectedLeaveType)
<input type="submit" value="Submit" />
}
并提交一个POST方法,该方法允许您在视图无效时轻松返回视图,或重定向到Create
方法。
[HttpPost]
public ActionResult ChooseType(LeaveTypeVM model)
{
if (!ModelState.IsValid)
{
model.LeaveTypeList = .... // as per GET method
}
return RedirectToAction("Create", new { leaveType = model.SelectedLeaveType });
和Create()
方法
public ActionResult Create(int leaveType)
{
var employee = DataAccess.GetEmployee(@User.Identity.Name);
var rule = RuleAccess.GetEmployeeRules(employee).Where(x => x.ID == leaveType).FirstOrDefault();
if (rule == null)
{
// Throw exception or redirect to an error page
}
var model = new LeaveApplicationViewModels();
....
return View(model);
}
请注意,您的LeaveApplicationViewModels
应包含其他属性,以便您可以避免所有这些ViewBag
属性并生成强类型视图。