我正在尝试创建一个表单,为创建新用户创建一个Ajax帖子。我的目标是显示一个弹出窗口,指出操作的结果。当前版本的动作方法会在模型状态出错时向模型状态添加错误,如果成功则重定向。 AdminController.cs内如何使用Ajax调用此后期操作方法?
操作方法:
[HttpPost]
public async Task<ActionResult> Create(CreateModel model)
{
if (ModelState.IsValid)
{
AppUser user = new AppUser { UserName = model.Name, Email = model.Email };
IdentityResult result = await UserManager.CreateAsync(user,
model.Password);
if (result.Succeeded)
{
return RedirectToAction("Index");
}
else
{
AddErrorsFromResult(result);
}
}
return View(model);
}
的观点:
@model IdentityDevelopment.Models.CreateModel
@{ ViewBag.Title = "Create User";}
<h2>Create User</h2>
@Html.ValidationSummary(false)
@using (Html.BeginForm("Create", "Admin", new { returnUrl = Request.Url.AbsoluteUri }))
{
<div class="form-group">
<label>Name</label>
@Html.TextBoxFor(x => x.Name, new { @class = "form-control" })
</div>
<div class="form-group">
<label>Email</label>
@Html.TextBoxFor(x => x.Email, new { @class = "form-control" })
</div>
<div class="form-group">
<label>Password</label>
@Html.PasswordFor(x => x.Password, new { @class = "form-control" })
</div>
<button type="submit" onclick="return showDiv();" class="btn btn-primary">Create</button>
@Html.ActionLink("Cancel", "Index", null, new { @class = "btn btn-default" })
}
<button id="usercreatebutton">Create</button>
从上面可以看出,与 “usercreatebutton” 按钮的id是我发展的按钮,我希望把AJAX功能在:
$("#usercreatebutton")
.button()
.click(function (event) {
alert("ajax post call");
});
其他创建按钮用于常规表单提交。
的CreateModel:
public class CreateModel
{
[Required]
public string Name { get; set; }
[Required]
public string Email { get; set; }
[Required]
public string Password { get; set; }
}
基于Shyju的回应,我得到了这个工作。下面我将发布更新我的代码:
在视图中,我修改了BeginForm声明给予形式的ID和移动里面的提交按钮:
@model IdentityDevelopment.Models.CreateModel
@{ ViewBag.Title = "Create User";}
<h2>Create User</h2>
@Html.ValidationSummary(false)
@using (Html.BeginForm("Create", "Admin", new { returnUrl = Request.Url.AbsoluteUri }, FormMethod.Post, new { @id = "signupform" }))
{
<div class="form-group">
<label>Name</label>
@Html.TextBoxFor(x => x.Name, new { @class = "form-control" })
</div>
<div class="form-group">
<label>Email</label>
@Html.TextBoxFor(x => x.Email, new { @class = "form-control" })
</div>
<div class="form-group">
<label>Password</label>
@Html.PasswordFor(x => x.Password, new { @class = "form-control" })
</div>
<!-- <button type="submit" onclick="return showDiv();" class="btn btn-primary">Create</button> -->
<button type="submit" id="usercreatebutton">Create</button>
@Html.ActionLink("Cancel", "Index", null, new { @class = "btn btn-default" })
}
控制器代码进行了修改,是Shyju的回应之一。
最后,JavaScript代码是:
$("form#signupform").submit(function (event) {
event.preventDefault();
var form = $(this);
$.post(form.attr("action"), form.serialize(), function (res) {
if (res.status === "success") {
alert(res.message);
}
else {
alert(res.message);
}
});
});
你看过jquery'$ .ajax()'的文档吗? HTTP://api.jquery。com/category/ajax/ –
@JonathanM是的,我有和多次尝试没有奏效。 – ITWorker
请向我们展示您使用'$ .ajax()'进行的最新尝试,我相信我们可以帮助您实现它。 –