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c# - 在部分 View 上返回不同的模型

转载 作者:太空宇宙 更新时间:2023-11-04 16:09:21 24 4
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我在将带有模型列表的 ViewBag 返回到 PartialView 时遇到了一些麻烦,PartialView 是与 ParentView 不同类型的模型,但我希望仅在渲染的局部 View 中渲染结果父 View 。或许看看我的代码,会有更好的理解。

这是我的 Controller :

public ActionResult Search()
{
//ViewBag.Usuarios = db.User.ToList();
return View();
}

[HttpGet]
public ActionResult Pesquisar(UserFilter userFilter)
{
List<UserModel> retorno = new List<UserModel>();
ViewBag.Mensagem = "Não foi encontrado registro com os filtros informados";

if (userFilter.Name != null)
{
retorno = db.User.Where(x => x.Name.Contains(userFilter.Name)).ToList();
if (retorno != null)
{
ViewBag.Usuarios = retorno;
return PartialView("Search", ViewBag.Usuarios);
}
return View("Search", ViewBag.Mensagem);
}

if (userFilter.UserID != 0)
{
UserModel retorn = new UserModel();
var id = Convert.ToInt16(userFilter.UserID);
retorn = db.User.FirstOrDefault(x => x.UserID == id);
return View("Details", retorn);
}
return View("Search", ViewBag.Mensagem);
}

这是我的父 View

@model Sistema_ADFP.Filters.UserFilter


<body>
<div class="container">
<h2>Buscar Usuário</h2>
<form role="form" method="GET" action="/User/Pesquisar">
<div class="col-lg-12">
<div class="col-lg-3">
<div class="form-group">
<label>Nome</label>
@Html.EditorFor(model => model.Name, new { htmlAttributes = new { @class = "form-control" } })
</div>
</div>
<div class="col-lg-3">
<div class="form-group">
<label>ID</label>
@Html.EditorFor(model => model.UserID, new { htmlAttributes = new { @class = "form-control" } })
</div>
</div>
<div class="col-lg-3">
<div class="form-group">
<label>CPF</label>
@Html.EditorFor(model => model.CPF, new { htmlAttributes = new { @class = "form-control" } })
</div>
</div>
</div>
<div class="clearfix"></div>
<div class="form-group pull-right">
<label>&ensp;</label>
<input type="submit" class="btn btn-primary"/>
</div>
</form>
</div>
<div class="clearfix"></div>

@{
if (ViewBag.Mensagem == null)
{
<div id="resultadoLista">
@{
Html.RenderPartial("_List");
}
</div>
}
else
{
<label>Nenhum registro encontrado</label>
}
}
</body>

这是我的局部 View

<h4>Usuários</h4>
<div class="table-responsive">
<table class="table">
<tr class="inverse" align="center">
<th>Nome</th>
<th>Sexo</th>
<th>Estado Civil</th>
<th>Educação</th>
<th>Profissão</th>
<th>Voluntário</th>
<th>Data Nascimento</th>
<th>Ações</th>
</tr>

@if (ViewBag.Usuarios != null)
{


foreach (var item in ViewBag.Usuarios)
{
<tr class="active">
<td data-th="Nome"><a class="modal-ajax-link" href="#test-popup">@item.Name</a></td>
<td data-th="Sexo">@item.Sex</td>
<td data-th="EstadoCivil">@item.MaritalStatus</td>
<td data-th="Education" align="center">@item.Education.Description</td>
<td data-th="Education" align="center">@item.Profession.Name</td>
@if (item.Voluntary)
{
<td data-th="Voluntario" align="center">Ativo</td>
}
else
{
<td data-th="Voluntario" align="center">Inativo</td>
}
<td data-th="DataNasc" align="center">@item.BirthDate</td>

@* data-mfp-src="@HttpContext.Current.Request.Url.Host:@HttpContext.Current.Request.Url.Port/User/Details/2" *@

<td data-th="Ações" align="center">
<a class="btn btn-info modal-ajax-link" href='@Html.ActionLink("Editar", "Edit",
new { id = @item.UserID })'><i class="icon_pencil"></i></a>
<a class="btn btn-danger modal-ajax-link" href="#delete-modal"><i class="icon_trash_alt"></i></a>
</td>
</tr>
}
}

</table>
</div>

我得到的错误是:

The model item passed into the dictionary is of type 'System.Collections.Generic.List`1[Sistema_ADFP.Models.UserModel]', but this dictionary requires a model item of type 'Sistema_ADFP.Filters.UserFilter'.

我了解错误原因,但我无法找到重置流程并使其正常工作的方法。有谁知道或知道我能做什么?

最佳答案

解决此问题的方法是不使用 ViewBag 并为您的 View 创建 ViewModel。 ViewModel 是您的 View 的一个类,并将具有您的 View 的所有必要属性。因此,为了举例,您可以有一个名为的 View 模型:

public class WrapperVM
{
public UserFilter Filter {get; set;}
public UserModel Model {get; set;}
}

因此,填充这些属性并将 WrapperVM 传递给父 View (更改为@model WrapperVM),然后您可以将 usermodel 传递给部分 View 。我希望这会有所帮助。

关于c# - 在部分 View 上返回不同的模型,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/35834773/

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