如何在Razor中编写"Html.BeginForm"

Question

如果我这样写:

form action ="Images"method ="post"enctype ="multipart/form-data"

有用.

但是在使用'@'的Razor中,它不起作用.我犯了什么错误吗？

@using (Html.BeginForm("Upload", "Upload", FormMethod.Post, 
                             new { enctype = "multipart/form-data" }))
{
    @Html.ValidationSummary(true)

    <fieldset>

        Select a file <input type="file" name="file" />
        <input type="submit" value="Upload" />

    </fieldset>
}

我的控制器看起来像这样:

[AcceptVerbs(HttpVerbs.Post)]
public ActionResult Upload() 
{
    foreach (string file in Request.Files)
    {
        var uploadedFile = Request.Files[file];
        uploadedFile.SaveAs(Server.MapPath("~/content/pics") + 
                                      Path.GetFileName(uploadedFile.FileName));
    }

    return RedirectToAction ("Upload");
}

Answer 1

以下代码工作正常:

@using (Html.BeginForm("Upload", "Upload", FormMethod.Post, 
                                      new { enctype = "multipart/form-data" }))
{
    @Html.ValidationSummary(true)
    <fieldset>
        Select a file <input type="file" name="file" />
        <input type="submit" value="Upload" />
    </fieldset>
}

并按预期生成:

<form action="/Upload/Upload" enctype="multipart/form-data" method="post">    
    <fieldset>
        Select a file <input type="file" name="file" />
        <input type="submit" value="Upload" />
    </fieldset>
</form>

另一方面,如果您在其他服务器端构造的上下文中编写此代码,if或者foreach您应该删除@之前的using.例如:

@if (SomeCondition)
{
    using (Html.BeginForm("Upload", "Upload", FormMethod.Post, 
                                      new { enctype = "multipart/form-data" }))
    {
        @Html.ValidationSummary(true)
        <fieldset>
            Select a file <input type="file" name="file" />
            <input type="submit" value="Upload" />
        </fieldset>
    }
}

就服务器端代码而言,以下是如何继续:

[HttpPost]
public ActionResult Upload(HttpPostedFileBase file) 
{
    if (file != null && file.ContentLength > 0) 
    {
        var fileName = Path.GetFileName(file.FileName);
        var path = Path.Combine(Server.MapPath("~/content/pics"), fileName);
        file.SaveAs(path);
    }
    return RedirectToAction("Upload");
}